8. Divergence, Curl and Potentials

Exercises

    a. The Del Operator

    Compute the gradient of each scalar field.

  1. f=xyzf=xyz

    Hint

    ∇⃗f=⟨∂xf,∂yf,∂zf⟩ \vec\nabla f =\left\langle \partial_x f,\partial_y f,\partial_z f\right\rangle

    [×]

    Answer

    ∇⃗f=⟨yz,xz,xy⟩\vec\nabla f =\left\langle yz,xz,xy\right\rangle

    [×]

    Solution

    ∇⃗f=⟨∂x(xyz),∂y(xyz),∂z(xyz)⟩=⟨yz,xz,xy⟩\begin {aligned} \vec\nabla f &=\left\langle \partial_x(xyz),\partial_y(xyz),\partial_z(xyz)\right\rangle \\ &=\left\langle yz,xz,xy\right\rangle \\ \end{aligned}

    [×]

  2. f=x2y3z4f=x^2y^3z^4

    Answer

    ∇⃗f=⟨2xy3z4,3x2y2z4,4x2y3z3⟩\vec\nabla f =\left\langle 2xy^3z^4,3x^2y^2z^4,4x^2y^3z^3\right\rangle

    [×]

    Solution

    ∇⃗f=⟨∂x(x2y3z4),∂y(x2y3z4),∂z(x2y3z4)⟩=⟨2xy3z4,3x2y2z4,4x2y3z3⟩\begin {aligned} \vec\nabla f &=\left\langle \partial_x(x^2y^3z^4),\partial_y(x^2y^3z^4),\partial_z(x^2y^3z^4)\right\rangle \\ &=\left\langle 2xy^3z^4,3x^2y^2z^4,4x^2y^3z^3\right\rangle \\ \end{aligned}

    [×]

  3. f=x2+y2+z2f=x^2+y^2+z^2

    Answer

    ∇⃗f=⟨2x,2y,2z⟩\vec\nabla f =\left\langle 2x,2y,2z\right\rangle

    [×]

    Solution

    ∇⃗f=⟨∂x(x2+y2+z2),∂y(x2+y2+z2),∂z(x2+y2+z2)⟩=⟨2x,2y,2z⟩\begin {aligned} \vec\nabla f &=\left\langle \partial_x(x^2+y^2+z^2), \partial_y(x^2+y^2+z^2), \partial_z(x^2+y^2+z^2)\right\rangle \\ &=\left\langle 2x,2y,2z\right\rangle \\ \end{aligned}

    [×]

  4. f=zsin⁡xcos⁡yf=z\sin x\cos y

    Answer

    ∇⃗f=⟨zcos⁡xcos⁡y,−zsin⁡xsin⁡y,sin⁡xcos⁡y⟩\vec\nabla f =\left\langle z\cos x\cos y,-z\sin x\sin y,\sin x\cos y\right\rangle

    [×]

    Solution

    ∇⃗f=⟨∂x(zsin⁡xcos⁡y),∂y(zsin⁡xcos⁡y),∂z(zsin⁡xcos⁡y)⟩=⟨zcos⁡xcos⁡y,−zsin⁡xsin⁡y,sin⁡xcos⁡y⟩\begin {aligned} \vec\nabla f &=\left\langle \partial_x(z\sin x\cos y), \partial_y(z\sin x\cos y), \partial_z(z\sin x\cos y)\right\rangle \\ &=\left\langle z\cos x\cos y,-z\sin x\sin y,\sin x\cos y\right\rangle \\ \end{aligned}

    [×]

    5. b. The Divergence Operator

    Compute the divergence of each vector field.

  5. F⃗=⟨xy,yz,zx⟩\vec F=\left\langle xy,yz,zx\right\rangle

    Hint

    ∇⃗⋅F⃗=∂xF1+∂yF2+∂zF3 \vec\nabla\cdot\vec F =\partial_x F_1+\partial_y F_2+\partial_z F_3

    [×]

    Answer

    ∇⃗⋅F⃗=y+z+x\vec\nabla\cdot\vec F=y+z+x

    [×]

    Solution

    ∇⃗⋅F⃗=∂x(xy)+∂y(yz)+∂z(zx)=y+z+x\begin{aligned} \vec\nabla\cdot\vec F &=\partial_x(xy)+\partial_y(yz)+\partial_z(zx) \\ &=y+z+x \end{aligned}

    [×]

  6. F⃗=⟨yz,xz,xy⟩\vec F=\left\langle yz,xz,xy\right\rangle

    Answer

    ∇⃗⋅F⃗=0\vec\nabla\cdot\vec F=0

    [×]

    Solution

    ∇⃗⋅F⃗=∂x(yz)+∂y(xz)+∂z(xy)=0+0+0=0\begin{aligned} \vec\nabla\cdot\vec F &=\partial_x(yz)+\partial_y(xz)+\partial_z(xy) \\ &=0+0+0=0 \end{aligned}

    [×]

  7. F⃗=⟨12z2x,12x2y,12y2z⟩\vec F =\left\langle \dfrac{1}{2}z^2x,\dfrac{1}{2}x^2y,\dfrac{1}{2}y^2z\right\rangle

    Answer

    ∇⃗⋅F⃗=12(x2+y2+z2)\vec\nabla\cdot\vec F =\dfrac{1}{2}(x^2+y^2+z^2)

    [×]

    Solution

    ∇⃗⋅F⃗=∂∂x(12z2x)+∂∂y(12x2y)+∂∂z(12y2z)=12(z2+x2+y2)\begin{aligned} \vec\nabla\cdot\vec F &=\dfrac{\partial}{\partial x}\left(\dfrac{1}{2}z^2x\right) +\dfrac{\partial}{\partial y}\left(\dfrac{1}{2}x^2y\right) +\dfrac{\partial}{\partial z}\left(\dfrac{1}{2}y^2z\right) \\ &=\dfrac{1}{2}(z^2+x^2+y^2) \end{aligned}

    [×]

  8. F⃗=⟨xcos⁡z,ycos⁡z,sin⁡z⟩\vec F=\left\langle x\cos z,y\cos z,\sin z\right\rangle

    Answer

    ∇⃗⋅F⃗=3cos⁡z\vec\nabla\cdot\vec F=3\cos z

    [×]

    Solution

    ∇⃗⋅F⃗=∂x(xcos⁡z)+∂y(ycos⁡z)+∂z(sin⁡z)=cos⁡z+cos⁡z+cos⁡z=3cos⁡z\begin{aligned} \vec\nabla\cdot\vec F &=\partial_x(x\cos z)+\partial_y(y\cos z)+\partial_z(\sin z) \\ &=\cos z+\cos z+\cos z=3\cos z \end{aligned}

    [×]

  9. F⃗=⟨zcos⁡xcos⁡y,−zsin⁡xsin⁡y,sin⁡xcos⁡y⟩\vec F =\left\langle z\cos x\cos y,-z\sin x\sin y, \sin x\cos y\right\rangle

    Answer

    ∇⃗⋅F⃗=−2zsin⁡xcos⁡y\vec\nabla\cdot\vec F=-2z\sin x\cos y

    [×]

    Solution

    ∇⃗⋅F⃗=∂x(zcos⁡xcos⁡y)+∂y(−zsin⁡xsin⁡y)+∂z(sin⁡xcos⁡y)=−zsin⁡xcos⁡y−zsin⁡xcos⁡y+0=−2zsin⁡xcos⁡y\begin{aligned} \vec\nabla\cdot\vec F &=\partial_x(z\cos x\cos y) +\partial_y(-z\sin x\sin y) \\ &\quad+\partial_z(\sin x\cos y) \\ &=-z\sin x\cos y-z\sin x\cos y+0 \\ &=-2z\sin x\cos y \end{aligned}

    [×]

  10. F⃗=⟨x3,y3,z3⟩\vec F=\left\langle x^3,y^3,z^3\right\rangle

    Answer

    ∇⃗⋅F⃗=3x2+3y2+3z2\vec\nabla\cdot\vec F=3x^2+3y^2+3z^2

    [×]

    Solution

    ∇⃗⋅F⃗=∂x(x3)+∂y(y3)+∂z(z3)=3x2+3y2+3z2\begin{aligned} \vec\nabla\cdot\vec F &=\partial_x(x^3)+\partial_y(y^3)+\partial_z(z^3) \\ &=3x^2+3y^2+3z^2 \end{aligned}

    [×]

  11. F⃗=⟨ex,ey,ez⟩\vec F=\left\langle e^x,e^y,e^z\right\rangle

    Answer

    ∇⃗⋅F⃗=ex+ey+ez\vec\nabla\cdot\vec F=e^x+e^y+e^z

    [×]

    Solution

    ∇⃗⋅F⃗=∂x(ex)+∂y(ey)+∂z(ez)=ex+ey+ez\begin{aligned} \vec\nabla\cdot\vec F &=\partial_x(e^x)+\partial_y(e^y)+\partial_z(e^z) \\ &=e^x+e^y+e^z \end{aligned}

    [×]

  12. F⃗=⟨ln⁡(xyz),ln⁡(yxz),ln⁡(zxy)⟩\vec F=\left\langle \ln\left(\dfrac{x}{yz}\right), \ln\left(\dfrac{y}{xz}\right), \ln\left(\dfrac{z}{xy}\right)\right\rangle

    Answer

    ∇⃗⋅F⃗=1x+1y+1z\vec\nabla\cdot\vec F=\dfrac{1}{x}+\dfrac{1}{y}+\dfrac{1}{z}

    [×]

    Solution

    ∇⃗⋅F⃗=∂xln⁡(xyz)+∂yln⁡(yxz)+∂zln⁡(zxy)=yzx1yz+xzy1xz+xyz1xy=1x+1y+1z\begin{aligned} \vec\nabla\cdot\vec F &=\partial_x\ln\left(\dfrac{x}{yz}\right) +\partial_y\ln\left(\dfrac{y}{xz}\right) +\partial_z\ln\left(\dfrac{z}{xy}\right) \\ &=\dfrac{yz}{x}\dfrac{1}{yz} +\dfrac{xz}{y}\dfrac{1}{xz} +\dfrac{xy}{z}\dfrac{1}{xy} \\ &=\dfrac{1}{x}+\dfrac{1}{y}+\dfrac{1}{z} \end{aligned}

    ∇⃗⋅F⃗=∂x(ln⁡x−ln⁡y−ln⁡z)+∂y(ln⁡y−ln⁡x−ln⁡z)+∂z(ln⁡z−ln⁡x−ln⁡y)=1x+1y+1z\begin{aligned} \vec\nabla\cdot\vec F &=\partial_x(\ln x-\ln y-\ln z) +\partial_y(\ln y-\ln x-\ln z) \\ &\quad+\partial_z(\ln z-\ln x-\ln y) \\ &=\dfrac{1}{x}+\dfrac{1}{y}+\dfrac{1}{z} \end{aligned}

    [×]

  13. Compute the divergence of the gradient of f=x2y3z4f=x^2y^3z^4, i.e. ∇⃗⋅∇⃗f\vec\nabla\cdot\vec\nabla f.

    Answer

    ∇⃗⋅∇⃗f=2y3z4+6x2yz4+12x2y3z2\vec\nabla\cdot\vec\nabla f=2y^3z^4+6x^2yz^4+12x^2y^3z^2

    [×]

    Solution

    The gradient of ff is: ∇⃗f=⟨2xy3z4,3x2y2z4,4x2y3z3⟩ \vec\nabla f=\left\langle 2xy^3z^4, 3x^2y^2z^4, 4x^2y^3z^3\right\rangle Then the divergence of the gradient is: ∇⃗⋅∇⃗f=∂x(2xy3z4)+∂y(3x2y2z4)+∂z(4x2y3z3)=2y3z4+6x2yz4+12x2y3z2\begin{aligned} \vec\nabla\cdot\vec\nabla f &=\partial_x(2xy^3z^4)+\partial_y(3x^2y^2z^4)+\partial_z(4x^2y^3z^3) \\ &=2y^3z^4+6x^2yz^4+12x^2y^3z^2 \end{aligned}

    [×]

    Remark

    The divergence of the gradient is an operation called the Laplacian, ∇2=∇⃗⋅∇⃗\nabla^2=\vec\nabla\cdot\vec\nabla, which we will study on a later page. It can be computed directly using the formula: ∇2f=∇⃗⋅∇⃗f=d2fdx2+d2fdy2+d2fdz2 \nabla^2f=\vec\nabla\cdot\vec\nabla f =\dfrac{d^2f}{dx^2}+\dfrac{d^2f}{dy^2}+\dfrac{d^2f}{dz^2}

    [×]

  14. Compute the divergence of the gradient of f=xyzf=xyz, i.e. ∇⃗⋅∇⃗f\vec\nabla\cdot\vec\nabla f.

    Hint

    The divergence of the gradient is an operation called the Laplacian, ∇2=∇⃗⋅∇⃗\nabla^2=\vec\nabla\cdot\vec\nabla, which we will study on a later page. It can be computed directly using the formula: ∇2f=∇⃗⋅∇⃗f=d2fdx2+d2fdy2+d2fdz2=∂x2f+∂y2f+∂z2f\begin{aligned} \nabla^2f&=\vec\nabla\cdot\vec\nabla f \\ &=\dfrac{d^2f}{dx^2}+\dfrac{d^2f}{dy^2}+\dfrac{d^2f}{dz^2} \\ &=\partial_x^2f+\partial_y^2f+\partial_z^2f \end{aligned}

    [×]

    Answer

    ∇⃗⋅∇⃗(xyz)=0\vec\nabla\cdot\vec\nabla(xyz)=0

    [×]

    Solution

    The Laplacian is: ∇2f=∇⃗⋅∇⃗f=∂x2f+∂y2f+∂z2f\begin{aligned} \nabla^2f &=\vec\nabla\cdot\vec\nabla f \\ &=\partial_x^2f+\partial_y^2f+\partial_z^2f \end{aligned} We compute the Laplacian directly: ∇2(xyz)=∇⃗⋅∇⃗(xyz)=∂x2(xyz)+∂y2(xyz)+∂z2(xyz)=0+0+0=0\begin{aligned} \nabla^2(xyz) &=\vec\nabla\cdot\vec\nabla(xyz) \\ &=\partial_x^2(xyz)+\partial_y^2(xyz)+\partial_z^2(xyz) \\ &=0+0+0=0 \end{aligned}

    [×]

    15. c. The Curl Operator

    Compute the curl of each vector field.

  15. F⃗=⟨xy,yz,zx⟩\vec F=\left\langle xy,yz,zx\right\rangle

    Hint

    ∇⃗×F⃗=∣ı^ȷ^k^∂x∂y∂zF1F2F3∣ \vec\nabla\times\vec F =\begin{vmatrix} \hat\imath & \hat\jmath & \hat k \\ \partial_x & \partial_y & \partial_z \\ F_1 & F_2 & F_3 \end{vmatrix}

    [×]

    Answer

    ∇⃗×F⃗=⟨−y,−z,−x⟩\vec\nabla\times\vec F =\left\langle -y,-z,-x\right\rangle

    [×]

    Solution

    ∇⃗×F⃗=∣ı^ȷ^k^∂x∂y∂zxyyzzx∣=ı^(0−y)−ȷ^(z−0)+k^(0−x)=⟨−y,−z,−x⟩\begin{aligned} \vec\nabla\times\vec F &=\begin{vmatrix} \hat\imath & \hat\jmath & \hat k \\ \partial_x & \partial_y & \partial_z \\ xy & yz & zx \end{vmatrix} \\ &=\hat\imath(0-y)-\hat\jmath(z-0)+\hat k(0-x) \\ &=\left\langle -y,-z,-x\right\rangle \end{aligned}

    [×]

  16. F⃗=⟨yz,xz,xy⟩\vec F=\left\langle yz,xz,xy\right\rangle

    Answer

    ∇⃗×F⃗=0⃗≡⟨0,0,0⟩\vec\nabla\times\vec F =\vec0\equiv\left\langle 0,0,0\right\rangle

    [×]

    Solution

    ∇⃗×F⃗=∣ı^ȷ^k^∂x∂y∂zyzxzxy∣=ı^(x−x)−ȷ^(y−y)+k^(z−z)=⟨0,0,0⟩=0⃗\begin{aligned} \vec\nabla\times\vec F &=\begin{vmatrix} \hat\imath & \hat\jmath & \hat k \\ \partial_x & \partial_y & \partial_z \\ yz & xz & xy \end{vmatrix} \\ &=\hat\imath(x-x)-\hat\jmath(y-y)+\hat k(z-z) \\ &=\left\langle 0,0,0\right\rangle=\vec0 \end{aligned}

    [×]

  17. F⃗=⟨12z2x,12x2y,12y2z⟩\vec F =\left\langle \dfrac{1}{2}z^2x,\dfrac{1}{2}x^2y, \dfrac{1}{2}y^2z\right\rangle

    Answer

    ∇⃗×F⃗=⟨yz,xz,xy⟩\vec\nabla\times\vec F =\left\langle yz,xz,xy\right\rangle

    [×]

    Solution

    ∇⃗×F⃗=∣ı^ȷ^k^∂x∂y∂z12z2x12x2y12y2z∣=ı^(yz−0)−ȷ^(0−zx)+k^(xy−0)=⟨yz,xz,xy⟩\begin{aligned} \vec\nabla\times\vec F &=\begin{vmatrix} \hat\imath & \hat\jmath & \hat k \\ \partial_x & \partial_y & \partial_z \\ \dfrac{1}{2}z^2x & \dfrac{1}{2}x^2y & \dfrac{1}{2}y^2z \end{vmatrix} \\ &=\hat\imath(yz-0)-\hat\jmath(0-zx)+\hat k(xy-0) \\ &=\left\langle yz,xz,xy\right\rangle \end{aligned}

    [×]

  18. F⃗=⟨xcos⁡z,ycos⁡z,sin⁡z⟩\vec F=\left\langle x\cos z,y\cos z,\sin z\right\rangle

    Answer

    ∇⃗×F⃗=⟨ysin⁡z,−xsin⁡z,0⟩\vec\nabla\times\vec F =\left\langle y\sin z,-x\sin z,0\right\rangle

    [×]

    Solution

    ∇⃗×F⃗=∣ı^ȷ^k^∂x∂y∂zxcos⁡zycos⁡zsin⁡z∣=ı^(ysin⁡z)−ȷ^(0−−xsin⁡z)+k^(0−0)=⟨ysin⁡z,−xsin⁡z,0⟩\begin{aligned} \vec\nabla\times\vec F &=\begin{vmatrix} \hat\imath & \hat\jmath & \hat k \\ \partial_x & \partial_y & \partial_z \\ x\cos z & y\cos z & \sin z \end{vmatrix} \\ &=\hat\imath(y\sin z)-\hat\jmath(0--x\sin z)+\hat k(0-0) \\ &=\left\langle y\sin z,-x\sin z,0\right\rangle \end{aligned}

    [×]

  19. F⃗=⟨zcos⁡xcos⁡y,−zsin⁡xsin⁡y,sin⁡xcos⁡y⟩\vec F =\left\langle z\cos x\cos y,-z\sin x\sin y, \sin x\cos y\right\rangle

    Answer

    ∇⃗×F⃗=0⃗≡⟨0,0,0⟩\vec\nabla\times\vec F =\vec0\equiv\left\langle 0,0,0\right\rangle

    [×]

    Solution

    ∇⃗×F⃗=∣ı^ȷ^k^∂x∂y∂zzcos⁡xcos⁡y−zsin⁡xsin⁡ysin⁡xcos⁡y∣=ı^(−sin⁡xsin⁡y−−sin⁡xsin⁡y)−ȷ^(cos⁡xcos⁡y−cos⁡xcos⁡y)+k^(−zcos⁡xsin⁡y−−zcos⁡xsin⁡y)=⟨0,0,0⟩=0⃗\begin{aligned} \vec\nabla\times\vec F &=\begin{vmatrix} \hat\imath & \hat\jmath & \hat k \\ \partial_x & \partial_y & \partial_z \\ z\cos x\cos y & -z\sin x\sin y & \sin x\cos y \end{vmatrix} \\ &=\hat\imath(-\sin x\sin y--\sin x\sin y) \\ &\quad-\hat\jmath(\cos x\cos y-\cos x\cos y) \\ &\quad+\hat k(-z\cos x\sin y--z\cos x\sin y) \\ &=\left\langle 0,0,0\right\rangle=\vec0 \end{aligned}

    [×]

  20. F⃗=⟨−y,x,0⟩\vec F=\left\langle -y,x,0\right\rangle

    Answer

    ∇⃗×F⃗=⟨0,0,2⟩\vec\nabla\times\vec F =\left\langle 0,0,2\right\rangle

    [×]

    Solution

    ∇⃗×F⃗=∣ı^ȷ^k^∂x∂y∂z−yx0∣=ı^(0−0)−ȷ^(0−0)+k^(1−−1)=⟨0,0,2⟩\begin{aligned} \vec\nabla\times\vec F &=\begin{vmatrix} \hat\imath & \hat\jmath & \hat k \\ \partial_x & \partial_y & \partial_z \\ -y & x & 0 \end{vmatrix} \\ &=\hat\imath(0-0)-\hat\jmath(0-0)+\hat k(1--1) \\ &=\left\langle 0,0,2\right\rangle \end{aligned}

    [×]

  21. F⃗=⟨eyz,exz,exy⟩\vec F=\left\langle e^{yz},e^{xz},e^{xy}\right\rangle

    Answer

    ∇⃗×F⃗=⟨x(exy−exz),y(eyz−exy),z(exz−eyz)⟩\vec\nabla\times\vec F =\left\langle x(e^{xy}-e^{xz}), y(e^{yz}-e^{xy}), z(e^{xz}-e^{yz})\right\rangle

    [×]

    Solution

    ∇⃗×F⃗=∣ı^ȷ^k^∂x∂y∂zeyzexzexy∣=ı^(xexy−xexz)−ȷ^(yexy−yeyz)+k^(zexz−zeyz)=⟨x(exy−exz),y(eyz−exy),z(exz−eyz)⟩\begin{aligned} \vec\nabla\times\vec F &=\begin{vmatrix} \hat\imath & \hat\jmath & \hat k \\ \partial_x & \partial_y & \partial_z \\ e^{yz} & e^{xz} & e^{xy} \end{vmatrix} \\ &=\hat\imath(xe^{xy}-xe^{xz}) -\hat\jmath(ye^{xy}-ye^{yz}) +\hat k(ze^{xz}-ze^{yz}) \\ &=\left\langle x(e^{xy}-e^{xz}), y(e^{yz}-e^{xy}), z(e^{xz}-e^{yz})\right\rangle \end{aligned}

    [×]

  22. F⃗=⟨ln⁡(xyz),ln⁡(yxz),ln⁡(zxy)⟩\vec F=\left\langle \ln\left(\dfrac{x}{yz}\right), \ln\left(\dfrac{y}{xz}\right), \ln\left(\dfrac{z}{xy}\right)\right\rangle

    Answer

    ∇⃗×F⃗==⟨1z−1y,1x−1z,1y−1x⟩\vec\nabla\times\vec F= =\left\langle \dfrac{1}{z}-\,\dfrac{1}{y}, \dfrac{1}{x}-\,\dfrac{1}{z}, \dfrac{1}{y}-\,\dfrac{1}{x}\right\rangle

    [×]

    Solution

    ∇⃗×F⃗=∣ı^ȷ^k^∂x∂y∂zln⁡x−ln⁡y−ln⁡zln⁡y−ln⁡x−ln⁡zln⁡z−ln⁡x−ln⁡y∣=ı^(−1y−−1z)−ȷ^(−1x−−1z)+k^(−1x−−1y)=⟨1z−1y,1x−1z,1y−1x⟩\begin{aligned} \vec\nabla\times\vec F &=\begin{vmatrix} \hat\imath & \hat\jmath & \hat k \\ \partial_x & \partial_y & \partial_z \\ \ln x-\ln y-\ln z & \ln y-\ln x-\ln z & \ln z-\ln x-\ln y \end{vmatrix} \\ &=\hat\imath\left(-\,\dfrac{1}{y}--\,\dfrac{1}{z}\right) -\hat\jmath\left(-\,\dfrac{1}{x}--\,\dfrac{1}{z}\right) +\hat k \left(-\,\dfrac{1}{x}--\,\dfrac{1}{y}\right) \\ &=\left\langle \dfrac{1}{z}-\,\dfrac{1}{y}, \dfrac{1}{x}-\,\dfrac{1}{z}, \dfrac{1}{y}-\,\dfrac{1}{x}\right\rangle \end{aligned}

    [×]

  23. Compute the curl of the gradient of f=x2y3z4f=x^2y^3z^4, i.e. ∇⃗×∇⃗f\vec\nabla\times\vec\nabla f.

    Answer

    ∇⃗×∇⃗(x2y3z4)=0⃗≡⟨0,0,0⟩\vec\nabla\times\vec\nabla(x^2y^3z^4)=\vec0\equiv\left\langle 0,0,0\right\rangle

    [×]

    Solution

    The gradient of ff is: ∇⃗f=⟨2xy3z4,3x2y2z4,4x2y3z3⟩ \vec\nabla f=\left\langle 2xy^3z^4, 3x^2y^2z^4, 4x^2y^3z^3\right\rangle Then the curl of the gradient is: ∇⃗×∇⃗f=∣ı^ȷ^k^∂x∂y∂z2xy3z43x2y2z44x2y3z3∣=ı^(12x2y2z3−12x2y2z3)−ȷ^(8xy3z3−8xy3z3)+k^(6xy2z4−6xy2z4)=⟨0,0,0⟩=0⃗\begin{aligned} \vec\nabla\times\vec\nabla f &=\begin{vmatrix} \hat\imath & \hat\jmath & \hat k \\ \partial_x & \partial_y & \partial_z \\ 2xy^3z^4 & 3x^2y^2z^4 & 4x^2y^3z^3 \end{vmatrix} \\ &=\hat\imath(12x^2y^2z^3-12x^2y^2z^3) \\ &\quad-\hat\jmath(8xy^3z^3-8xy^3z^3) \\ &\quad+\hat k (6xy^2z^4-6xy^2z^4) \\ &=\left\langle 0,0,0\right\rangle=\vec0 \end{aligned}

    [×]

  24. Compute the curl of the gradient of f=xyzf=xyz, i.e. ∇⃗×∇⃗f\vec\nabla\times\vec\nabla f.

    Answer

    ∇⃗×∇⃗(xyz)=0⃗≡⟨0,0,0⟩\vec\nabla\times\vec\nabla(xyz)=\vec0\equiv\left\langle 0,0,0\right\rangle

    [×]

    Solution

    The gradient of ff is: ∇⃗f=⟨yz,xz,xy⟩ \vec\nabla f=\left\langle yz, xz, xy\right\rangle Then the curl of the gradient is: ∇⃗×∇⃗f=∣ı^ȷ^k^∂x∂y∂zyzxzxy∣=ı^(x−x)−ȷ^(y−y)+k^(z−z)=⟨0,0,0⟩=0⃗\begin{aligned} \vec\nabla\times\vec\nabla f &=\begin{vmatrix} \hat\imath & \hat\jmath & \hat k \\ \partial_x & \partial_y & \partial_z \\ yz & xz & xy \end{vmatrix} \\ &=\hat\imath(x-x)-\hat\jmath(y-y)+\hat k(z-z) \\ &=\left\langle 0,0,0\right\rangle=\vec0 \end{aligned}

    [×]

    Remark

    It is no coincidence that the curl of the gradient is 0⃗\vec0 in both this and the previous problem. We will see on a later page that ∇⃗×∇⃗f=0⃗\vec\nabla\times\vec\nabla f=\vec0 for any function ff.

    [×]

  25. Compute the divergence of the curl of F⃗=⟨xy,yz,zx⟩\vec F=\left\langle xy,yz,zx\right\rangle, i.e. ∇⃗⋅∇⃗×F⃗\vec\nabla\cdot\vec\nabla\times\vec F.

    Answer

    ∇⃗⋅∇⃗×⟨xy,yz,zx⟩=0\vec\nabla\cdot\vec\nabla\times\left\langle xy,yz,zx\right\rangle =0

    [×]

    Solution

    The curl of F⃗\vec F is: ∇⃗×F⃗=∣ı^ȷ^k^∂x∂y∂zxyyzzx∣=⟨−y,−z,−x⟩ \vec\nabla\times\vec F =\begin{vmatrix} \hat\imath & \hat\jmath & \hat k \\ \partial_x & \partial_y & \partial_z \\ xy & yz & zx \end{vmatrix} =\left\langle -y,-z,-x\right\rangle Then the divergence of the curl is: ∇⃗⋅∇⃗×F⃗=∂x(−y)+∂y(−z)+∂z(−x)=0+0+0=0\begin{aligned} \vec\nabla\cdot\vec\nabla\times\vec F &=\partial_x(-y)+\partial_y(-z)+\partial_z(-x) \\ &=0+0+0=0 \end{aligned}

    [×]

  26. Compute the divergence of the curl of F⃗=⟨eyz,exz,exy⟩\vec F=\left\langle e^{yz},e^{xz},e^{xy}\right\rangle, i.e. ∇⃗⋅∇⃗×F⃗\vec\nabla\cdot\vec\nabla\times\vec F.

    Answer

    ∇⃗⋅∇⃗×F⃗=0\vec\nabla\cdot\vec\nabla\times\vec F=0

    [×]

    Solution

    The curl of F⃗\vec F is: ∇⃗×F⃗=∣ı^ȷ^k^∂x∂y∂zeyzexzexy∣=⟨x(exy−exz),y(eyz−exy),z(exz−eyz)⟩\begin{aligned} \vec\nabla\times\vec F &=\begin{vmatrix} \hat\imath & \hat\jmath & \hat k \\ \partial_x & \partial_y & \partial_z \\ e^{yz} & e^{xz} & e^{xy} \end{vmatrix} \\ &=\left\langle x(e^{xy}-e^{xz}), y(e^{yz}-e^{xy}), z(e^{xz}-e^{yz})\right\rangle \end{aligned} Then the divergence of the curl is: ∇⃗⋅∇⃗×F⃗=∂x[x(exy−exz)]+∂y[y(eyz−exy)]+∂z[z(exz−eyz)]=[(exy−exz)+x(yexy−zexz)]+[(eyz−exy)+y(zeyz−xexy)]+[(exz−eyz)+z(xexz−yeyz)]=0\begin{aligned} \vec\nabla\cdot\vec\nabla\times\vec F &=\partial_x[x(e^{xy}-e^{xz})] \\ &\quad+\partial_y[y(e^{yz}-e^{xy})] \\ &\quad+\partial_z[z(e^{xz}-e^{yz})] \\ &=[(e^{xy}-e^{xz})+x(ye^{xy}-ze^{xz})] \\ &\quad+[(e^{yz}-e^{xy})+y(ze^{yz}-xe^{xy})] \\ &\quad+[(e^{xz}-e^{yz})+z(xe^{xz}-ye^{yz})] \\ &=0 \end{aligned}

    [×]

    Remark

    It is no coincidence that the divergence of the curl is 00 in both this and the previous problem. We will see on a later page that ∇⃗⋅∇⃗×F⃗=0\vec\nabla\cdot\vec\nabla\times\vec F=0 for any vector field F⃗\vec F.

    [×]

  27. Compute the curl of the curl of F⃗=⟨xy,yz,zx⟩\vec F=\left\langle xy,yz,zx\right\rangle, i.e. ∇⃗×∇⃗×F⃗\vec\nabla\times\vec\nabla\times\vec F.

    Answer

    ∇⃗×∇⃗×⟨xy,yz,zx⟩=⟨1,1,1⟩\vec\nabla\times\vec\nabla\times\left\langle xy,yz,zx\right\rangle =\left\langle 1,1,1\right\rangle

    [×]

    Solution

    The curl of F⃗\vec F is: ∇⃗×F⃗=∣ı^ȷ^k^∂x∂y∂zxyyzzx∣=⟨−y,−z,−x⟩ \vec\nabla\times\vec F =\begin{vmatrix} \hat\imath & \hat\jmath & \hat k \\ \partial_x & \partial_y & \partial_z \\ xy & yz & zx \end{vmatrix} =\left\langle -y,-z,-x\right\rangle Then the curl of the curl is: ∇⃗×∇⃗×F⃗=∣ı^ȷ^k^∂x∂y∂z−y−z−x∣=⟨1,1,1⟩\begin{aligned} \vec\nabla\times\vec\nabla\times\vec F =\begin{vmatrix} \hat\imath & \hat\jmath & \hat k \\ \partial_x & \partial_y & \partial_z \\ -y & -z & -x \end{vmatrix} =\left\langle 1,1,1\right\rangle \end{aligned}

    [×]

    Remark

    Although there is an identity for the curl of the curl which we will learn on a later page, it is not very useful in computations.

    [×]

    28. d. Differential Identities

  28. If f=2xcos⁡zf=2x\cos z and g=2ysin⁡zg=2y\sin z, compute ∇⃗(fg)\vec\nabla(fg) in two ways.

    1. Find f∇⃗gf\vec\nabla g and g∇⃗fg\vec\nabla f and add them to get ∇⃗(fg)\vec\nabla(fg).

      Answer

      ∇⃗(fg)=⟨2ysin⁡(2z),2xsin⁡(2z),4xycos⁡(2z)⟩=⟨4ysin⁡zcos⁡z,4xsin⁡zcos⁡z,4xycos⁡2z−4xysin⁡2z⟩\begin{aligned} \vec\nabla(fg) &=\left\langle 2y\sin(2z),2x\sin(2z),4xy\cos(2z)\right\rangle \\ &=\left\langle 4y\sin z\cos z,4x\sin z\cos z,4xy\cos^2 z-4xy\sin^2 z\right\rangle \end{aligned}

      [×]

      Solution

      f∇⃗g=2xcos⁡z⟨0,2sin⁡z,2ycos⁡z⟩=⟨0,4xsin⁡zcos⁡z,4xycos⁡2z⟩g∇⃗f=2ysin⁡z⟨2cos⁡z,0,−2xsin⁡z⟩=⟨4ysin⁡zcos⁡z,0,−4xysin⁡2z⟩∇⃗(fg)=f∇⃗g+g∇⃗f=⟨4ysin⁡zcos⁡z,4xsin⁡zcos⁡z,4xycos⁡2z−4xysin⁡2z⟩\begin{aligned} f\vec\nabla g &=2x\cos z\left\langle 0,2\sin z,2y\cos z\right\rangle \\ &=\left\langle 0,4x\sin z\cos z,4xy\cos^2 z\right\rangle \\[6pt] g\vec\nabla f &=2y \sin z\left\langle 2\cos z,0,-2x\sin z\right\rangle \\ &=\left\langle 4y\sin z\cos z,0,-4xy\sin^2 z\right\rangle \\[6pt] \vec\nabla(fg) &=f\vec\nabla g+g\vec\nabla f \\ &=\left\langle 4y\sin z\cos z,4x\sin z\cos z,4xy\cos^2 z-4xy\sin^2 z\right\rangle \end{aligned}

      [×]

    2. Find fgfg and then ∇⃗(fg)\vec\nabla(fg).

      Answer

      ∇⃗(fg)=⟨2ysin⁡(2z),2xsin⁡(2z),4xycos⁡(2z)⟩=⟨4ysin⁡zcos⁡z,4xsin⁡zcos⁡z,4xycos⁡2z−4xysin⁡2z⟩\begin{aligned} \vec\nabla(fg) &=\left\langle 2y\sin(2z),2x\sin(2z),4xy\cos(2z)\right\rangle \\ &=\left\langle 4y\sin z\cos z,4x\sin z\cos z,4xy\cos^2 z-4xy\sin^2 z\right\rangle \end{aligned}

      [×]

      Solution

      fg=4xycos⁡zsin⁡z=2xysin⁡(2z)∇⃗(fg)=⟨2ysin⁡(2z),2xsin⁡(2z),4xycos⁡(2z)⟩=⟨4ysin⁡zcos⁡z,4xsin⁡zcos⁡z,4xycos⁡2z−4xysin⁡2z⟩\begin{aligned} fg&=4xy\cos z\sin z=2xy\sin(2z) \\[6pt] \vec\nabla(fg) &=\left\langle 2y\sin(2z),2x\sin(2z),4xy\cos(2z)\right\rangle \\ &=\left\langle 4y\sin z\cos z,4x\sin z\cos z,4xy\cos^2 z-4xy\sin^2 z\right\rangle \end{aligned}

      [×]

  29. At a point PP, we know: f=3∇⃗f=⟨−2,1,3⟩g=2∇⃗g=⟨2,−1,4⟩\begin{aligned} f&=3\qquad& \vec\nabla f&=\left\langle -2,1,3\right\rangle \\ g&=2\qquad& \vec\nabla g&=\left\langle 2,-1,4\right\rangle \end{aligned} Find ∇⃗(fg)\vec\nabla(fg) at PP.

    Answer

    ∇⃗(fg)=⟨2,−1,18⟩\vec\nabla(fg) =\left\langle 2,-1,18\right\rangle

    [×]

    Solution

    Using the formula, we have: ∇⃗(fg)=f∇⃗g+g∇⃗f=3⟨2,−1,4⟩+2⟨−2,1,3⟩=⟨6,−3,12⟩+⟨−4,2,6⟩=⟨2,−1,18⟩\begin{aligned} \vec\nabla(fg) &=f\vec\nabla g+g\vec\nabla f \\ &=3\left\langle 2,-1,4\right\rangle+2\left\langle -2,1,3\right\rangle \\ &=\left\langle 6,-3,12\right\rangle+\left\langle -4,2,6\right\rangle \\ &=\left\langle 2,-1,18\right\rangle \end{aligned}

    [×]

  30. If f=xyzf=xyz and G⃗=⟨xz,yz,z2⟩\vec G=\left\langle xz,yz,z^2\right\rangle, compute ∇⃗⋅(fG⃗)\vec\nabla\cdot (f\vec G) in two ways.

    1. Find (∇⃗f)⋅G⃗(\vec\nabla f)\cdot\vec G and f∇⃗⋅G⃗f\vec\nabla\cdot\vec G and add them to get ∇⃗⋅(fG⃗)\vec\nabla\cdot(f\vec G).

      Answer

      ∇⃗⋅(fG⃗)=7xyz2\vec\nabla\cdot(f\vec G) =7xyz^2

      [×]

      Solution

      ∇⃗f=⟨yz,xz,xy⟩(∇⃗f)⋅G⃗=xyz2+xyz2+xyz2=3xyz2∇⃗⋅G⃗=z+z+2z=4zf∇⃗⋅G⃗=4xyz2∇⃗⋅(fG⃗)=(∇⃗f)⋅G⃗+f∇⃗⋅G⃗=3xyz2+4xyz2=7xyz2\begin{aligned} \vec\nabla f &=\left\langle yz,xz,xy\right\rangle \\ (\vec\nabla f)\cdot\vec G &=xyz^2+xyz^2+xyz^2=3xyz^2 \\[6pt] \vec\nabla\cdot\vec G &=z+z+2z=4z \\ f\vec\nabla\cdot\vec G &=4xyz^2 \\[6pt] \vec\nabla\cdot(f\vec G) &=(\vec\nabla f)\cdot\vec G +f\vec\nabla\cdot\vec G \\ &=3xyz^2+4xyz^2=7xyz^2 \end{aligned}

      [×]

    2. Find fG⃗f\vec G and then ∇⃗⋅(fG⃗)\vec\nabla\cdot(f\vec G).

      Answer

      ∇⃗⋅(fG⃗)==7xyz2\vec\nabla\cdot(f\vec G) ==7xyz^2

      [×]

      Solution

      fG⃗=xyz⟨xz,yz,z2⟩=⟨x2yz2,xy2z2,xyz3⟩∇⃗⋅(fG⃗)=2xyz2+2xyz2+3xyz2=7xyz2\begin{aligned} f\vec G &=xyz\left\langle xz,yz,z^2\right\rangle \\ &=\left\langle x^2yz^2,xy^2z^2,xyz^3\right\rangle \\[6pt] \vec\nabla\cdot(f\vec G) &=2xyz^2+2xyz^2+3xyz^2=7xyz^2 \end{aligned}

      [×]

  31. At a point PP, we know: F⃗=⟨4,−2,1⟩∇⃗×F⃗=⟨2,3,−2⟩G⃗=⟨1,−3,2⟩∇⃗×G⃗=⟨−2,1,4⟩\begin{aligned} \vec F&=\left\langle 4,-2,1\right\rangle\qquad& \vec\nabla\times\vec F&=\left\langle 2,3,-2\right\rangle \\ \vec G&=\left\langle 1,-3,2\right\rangle\qquad& \vec\nabla\times\vec G&=\left\langle -2,1,4\right\rangle \end{aligned} Find ∇⃗⋅(F⃗×G⃗)\vec\nabla\cdot(\vec F\times\vec G) at PP.

    Answer

    ∇⃗⋅(F⃗×G⃗)=−5\vec\nabla\cdot(\vec F\times\vec G) =-5

    [×]

    Solution

    Using the formula, we have: ∇⃗⋅(F⃗×G⃗)=∇⃗×F⃗⋅G⃗−F⃗⋅∇⃗×G⃗=⟨2,3,−2⟩⋅⟨1,−3,2⟩−⟨4,−2,1⟩⋅⟨−2,1,4⟩=(2−9−4)−(−8−2+4)=−5\begin{aligned} \vec\nabla\cdot(\vec F\times\vec G) &=\vec\nabla\times\vec F\cdot\vec G-\vec F\cdot\vec\nabla\times\vec G \\ &=\left\langle 2,3,-2\right\rangle\cdot\left\langle 1,-3,2\right\rangle -\left\langle 4,-2,1\right\rangle\cdot\left\langle -2,1,4\right\rangle \\ &=(2-9-4)-(-8-2+4)=-5 \end{aligned}

    [×]

  32. If f=xyzf=xyz and G⃗=⟨−yz,xz,z2⟩\vec G=\left\langle -yz,xz,z^2\right\rangle, compute ∇⃗×(fG⃗)\vec\nabla\times (f\vec G) in two ways.

    1. Find (∇⃗f)×G⃗(\vec\nabla f)\times\vec G and f∇⃗×G⃗f\vec\nabla\times\vec G and add them to get ∇⃗×(fG⃗)\vec\nabla\times(f\vec G).

      Answer

      ∇⃗×(fG⃗)=⟨xz3−2x2yz,−yz3−2xy2z,4xyz2⟩\vec\nabla\times(f\vec G) =\left\langle xz^3-2x^2yz,-yz^3-2xy^2z,4xyz^2\right\rangle

      [×]

      Solution

      ∇⃗f=⟨yz,xz,xy⟩(∇⃗f)×G⃗=∣ı^ȷ^k^yzxzxy−yzxzz2∣=ı^(xz3−x2yz)−ȷ^(yz3−−xy2z)+k^(xyz2−−xyz2)=⟨xz3−x2yz,−yz3−xy2z,2xyz2⟩∇⃗×G⃗=∣ı^ȷ^k^∂x∂y∂z−yzxzz2∣=ı^(0−x)−ȷ^(0−−y)+k^(z−−z)=⟨−x,−y,2z⟩f∇⃗×G⃗=⟨−x2yz,−xy2z,2xyz2⟩∇⃗×(fG⃗)=(∇⃗f)×G⃗+f∇⃗×G⃗=⟨xz3−2x2yz,−yz3−2xy2z,4xyz2⟩\begin{aligned} \vec\nabla f &=\left\langle yz,xz,xy\right\rangle \\[6pt] (\vec\nabla f)\times\vec G &=\begin{vmatrix} \hat\imath & \hat\jmath & \hat k \\ yz & xz & xy \\ -yz & xz & z^2 \end{vmatrix} \\ &=\hat\imath(xz^3-x^2yz)-\hat\jmath(yz^3--xy^2z) \\ &\quad+\hat k(xyz^2--xyz^2) \\ &=\left\langle xz^3-x^2yz,-yz^3-xy^2z,2xyz^2\right\rangle \\[6pt] \vec\nabla\times\vec G &=\begin{vmatrix} \hat\imath & \hat\jmath & \hat k \\ \partial_x & \partial_y & \partial_z \\ -yz & xz & z^2 \end{vmatrix} \\ &=\hat\imath(0-x)-\hat\jmath(0--y) +\hat k(z--z) \\ &=\left\langle -x,-y,2z\right\rangle \\[6pt] f\vec\nabla\times\vec G &=\left\langle -x^2yz,-xy^2z,2xyz^2\right\rangle \\[6pt] \vec\nabla\times(f\vec G) &=(\vec\nabla f)\times\vec G +f\vec\nabla\times\vec G \\ &=\left\langle xz^3-2x^2yz,-yz^3-2xy^2z,4xyz^2\right\rangle \end{aligned}

      [×]

    2. Find fG⃗f\vec G and then ∇⃗×(fG⃗)\vec\nabla\times(f\vec G).

      Answer

      ∇⃗×(fG⃗)=⟨xz3−2x2yz,−yz3−2xy2z,4xyz2⟩\vec\nabla\times(f\vec G) =\left\langle xz^3-2x^2yz,-yz^3-2xy^2z,4xyz^2\right\rangle

      [×]

      Solution

      fG⃗=⟨−xy2z2,x2yz2,xyz3⟩∇⃗×(fG⃗)=∣ı^ȷ^k^∂x∂y∂z−xy2z2x2yz2xyz3∣=ı^(xz3−2x2yz)−ȷ^(yz3−−2xy2z)+k^(2xyz2−−2xyz2)=⟨xz3−2x2yz,−yz3−2xy2z,4xyz2⟩\begin{aligned} f\vec G &=\left\langle -xy^2z^2,x^2yz^2,xyz^3\right\rangle \\[6pt] \vec\nabla\times(f\vec G) &=\begin{vmatrix} \hat\imath & \hat\jmath & \hat k \\ \partial_x & \partial_y & \partial_z \\ -xy^2z^2 & x^2yz^2 & xyz^3 \end{vmatrix} \\ &=\hat\imath(xz^3-2x^2yz)-\hat\jmath(yz^3--2xy^2z) +\hat k(2xyz^2--2xyz^2) \\ &=\left\langle xz^3-2x^2yz,-yz^3-2xy^2z,4xyz^2\right\rangle \end{aligned}

      [×]

  33. At a point PP, we know: f=3∇⃗f=⟨3,4,5⟩G⃗=⟨−3,2,4⟩∇⃗×G⃗=⟨2,−1,3⟩\begin{aligned} f&=3\qquad& \vec\nabla f&=\left\langle 3,4,5\right\rangle \\ \vec G&=\left\langle -3,2,4\right\rangle\qquad& \vec\nabla\times\vec G&=\left\langle 2,-1,3\right\rangle \end{aligned} Find ∇⃗×(fG⃗)\vec\nabla\times(f\vec G) at PP.

    Answer

    ∇⃗×(fG⃗)=⟨12,−30,27⟩\vec\nabla\times(f\vec G) =\left\langle 12,-30,27\right\rangle

    [×]

    Solution

    (∇⃗f)×G⃗=∣ı^ȷ^k^345−324∣=ı^(16−10)−ȷ^(12+15)+k^(6+12)=⟨6,−27,18⟩f∇⃗×G⃗=3⟨2,−1,3⟩=⟨6,−3,9⟩∇⃗×(fG⃗)=(∇⃗f)×G⃗+f∇⃗×G⃗=⟨12,−30,27⟩\begin{aligned} (\vec\nabla f)\times\vec G &=\begin{vmatrix} \hat\imath & \hat\jmath & \hat k \\ 3 & 4 & 5 \\ -3 & 2 & 4 \end{vmatrix} \\ &=\hat\imath(16-10)-\hat\jmath(12+15)+\hat k(6+12) \\ &=\left\langle 6,-27,18\right\rangle \\[6pt] f\vec\nabla\times\vec G &=3\left\langle 2,-1,3\right\rangle =\left\langle 6,-3,9\right\rangle \\[6pt] \vec\nabla\times(f\vec G) &=(\vec\nabla f)\times\vec G +f\vec\nabla\times\vec G \\ &=\left\langle 12,-30,27\right\rangle \end{aligned}

    [×]

  34. If f=sin⁡xcos⁡yf=\sin x\cos y, compute the Laplacian ∇2f\nabla^2f and show it satisfies ∇2f=−2f\nabla^2f=-2f.

    Answer

    ∇2f=−2sin⁡xcos⁡y=−2f\nabla^2f=-2\sin x\cos y=-2f

    [×]

    Solution

    ∇2f=∂x2(sin⁡xcos⁡y)+∂y2(sin⁡xcos⁡y)=∂x(cos⁡xcos⁡y)+∂y(−sin⁡xsin⁡y)=−sin⁡xcos⁡y−sin⁡xcos⁡y=−2sin⁡xcos⁡y=−2f\begin{aligned} \nabla^2f &=\partial_x^2(\sin x\cos y)+\partial_y^2(\sin x\cos y) \\ &=\partial_x(\cos x\cos y)+\partial_y(-\sin x\sin y) \\ &=-\sin x\cos y-\sin x\cos y=-2\sin x\cos y=-2f \\ \end{aligned}

    [×]

  35. If f=x2+y2−2z2f=x^2+y^2-2z^2, compute the Laplacian ∇2f\nabla^2f and show it satisfies the Laplace equation ∇2f=0\nabla^2f=0.

    Answer

    ∇2f=0\nabla^2f=0

    [×]

    Solution

    ∇2f=∂x2(x2+y2−2z2)+∂y2(x2+y2−2z2)+∂z2(x2+y2−2z2)=∂x(2x)+∂y(2y)+∂z(−4z)=2+2−4=0\begin{aligned} \nabla^2f &=\partial_x^2(x^2+y^2-2z^2)+\partial_y^2(x^2+y^2-2z^2)+\partial_z^2(x^2+y^2-2z^2) \\ &=\partial_x(2x)+\partial_y(2y)+\partial_z(-4z) \\ &=2+2-4=0 \\ \end{aligned}

    [×]

  36. If f=zsin⁡xcos⁡yf=z\sin x\cos y, compute the curl of the gradient of ff, i.e. ∇⃗×∇⃗f\vec\nabla\times\vec\nabla f.

    Answer

    ∇⃗×∇⃗f=0⃗\vec\nabla\times\vec\nabla f=\vec0

    [×]

    Solution

    For any function: ∇⃗×∇⃗f=0⃗ \vec\nabla\times\vec\nabla f=\vec0

    [×]

  37. If F⃗=⟨ysin⁡z,xcos⁡z,xy⟩\vec F=\left\langle y\sin z,x\cos z,xy\right\rangle, compute the divergence of the curl of F⃗\vec F, i.e. ∇⃗⋅∇⃗×F⃗\vec\nabla\cdot\vec\nabla\times\vec F.

    Answer

    ∇⃗⋅∇⃗×F⃗=0\vec\nabla\cdot\vec\nabla\times\vec F=0

    [×]

    Solution

    For any vector field: ∇⃗⋅∇⃗×F⃗=0 \vec\nabla\cdot\vec\nabla\times\vec F=0

    [×]

    38. e. Generalizing Antiderivatives

  38. Verify that F=cos⁡2xF=\cos^2 x and G=−sin⁡2xG=-\sin^2 x are both antiderivatives of f=−2sin⁡xcos⁡xf=-2\sin x\cos x. Why is this possible?

    Check

    By chain rule: F′=2(cos⁡x)(−sin⁡x)=−2sin⁡xcos⁡xF'=2(\cos x)(-\sin x)=-2\sin x\cos x Similarly: G′=−2(sin⁡x)(cos⁡x)=−2sin⁡xcos⁡xG'=-2(\sin x)(\cos x)=-2\sin x\cos x This is possible because FF and GG differ by a constant: cos⁡2x=−sin⁡2x+1\cos^2 x=-\sin^2 x +1

    [×]

  39. Verify that f=xy2z3f=xy^2z^3 is a scalar potential for F⃗=⟨y2z3,2xyz3,3xy2z2⟩\vec F=\left\langle y^2z^3,2xyz^3,3xy^2z^2\right\rangle

    Check

    The gradient of f=xy2z3f=xy^2z^3 is: ∇⃗f=⟨y2z3,2xyz3,3xy2z2⟩=F⃗ \vec\nabla f=\left\langle y^2z^3,2xyz^3,3xy^2z^2\right\rangle=\vec F

    [×]

  40. Verify that f=zsin⁡xcos⁡yf=z\sin x\cos y is a scalar potential for F⃗=⟨zcos⁡xcos⁡y,−zsin⁡xsin⁡y,sin⁡xcos⁡y⟩\vec F=\left\langle z\cos x\cos y,-z\sin x\sin y,\sin x\cos y\right\rangle

    Check

    The gradient of f=zsin⁡xcos⁡yf=z\sin x\cos y is: ∇⃗f=⟨zcos⁡xcos⁡y,−zsin⁡xsin⁡y,sin⁡xcos⁡y⟩=F⃗ \vec\nabla f =\left\langle z\cos x\cos y,-z\sin x\sin y,\sin x\cos y\right\rangle =\vec F

    [×]

  41. Verify that A⃗=⟨xy,yz,zx⟩\vec A=\langle xy,yz,zx\rangle is a vector potential for F⃗=⟨−y,−z,−x⟩\vec F=\left\langle -y,-z,-x\right\rangle

    Check

    The curl of A⃗=⟨yz,xz,xy⟩\vec A=\left\langle yz,xz,xy\right\rangle is: ∇⃗×A⃗=∣ı^ȷ^k^∂x∂y∂zxyyzzx∣=⟨0−y,0−z,0−x⟩=F⃗\begin{aligned} \vec\nabla\times\vec A &=\begin{vmatrix} \hat\imath & \hat\jmath & \hat k \\ \partial_x & \partial_y & \partial_z \\ xy & yz & zx \end{vmatrix} \\ &=\left\langle 0-y,0-z,0-x\right\rangle=\vec F \end{aligned}

    [×]

  42. Verify that A⃗=⟨zsin⁡y,xsin⁡z,ysin⁡x⟩\vec A=\left\langle z\sin y,x\sin z,y\sin x\right\rangle and A⃗=⟨zsin⁡y−yzcos⁡x,xsin⁡z−zsin⁡x,0⟩\vec A=\left\langle z\sin y-yz\cos x,x\sin z-z\sin x,0\right\rangle are both vector potentials for F⃗=⟨sin⁡x−xcos⁡z,sin⁡y−ycos⁡x,sin⁡z−zcos⁡y⟩\vec F=\left\langle \sin x-x\cos z,\sin y-y\cos x,\sin z-z\cos y\right\rangle

    Check

    The curl of A⃗=⟨xsin⁡y,ysin⁡z,zsin⁡x⟩\vec A=\left\langle x\sin y,y\sin z,z\sin x\right\rangle is: ∇⃗×A⃗=∣ı^ȷ^k^∂x∂y∂zzsin⁡yxsin⁡zysin⁡x∣=⟨sin⁡x−xcos⁡z,sin⁡y−ycos⁡x,sin⁡z−zcos⁡y⟩=F⃗\begin{aligned} \vec\nabla&\times\vec A =\begin{vmatrix} \hat\imath & \hat\jmath & \hat k \\ \partial_x & \partial_y & \partial_z \\ z\sin y & x\sin z & y\sin x \end{vmatrix} \\ &=\left\langle \sin x-x\cos z,\sin y-y\cos x,\sin z-z\cos y\right\rangle \\ &=\vec F \end{aligned}

    The curl of A⃗=⟨zsin⁡y−yzcos⁡x,xsin⁡z−zsin⁡x,0⟩\vec A=\left\langle z\sin y-yz\cos x,x\sin z-z\sin x,0\right\rangle is: ∇⃗×A⃗=∣ı^ȷ^k^∂x∂y∂zzsin⁡y−yzcos⁡xxsin⁡z−zsin⁡x0∣=⟨−xcos⁡z+sin⁡x,sin⁡y−ycos⁡x,sin⁡z−zcos⁡x−zcos⁡y+zcos⁡x⟩=⟨sin⁡x−xcos⁡z,sin⁡y−ycos⁡x,sin⁡z−zcos⁡y⟩=F⃗\begin{aligned} \vec\nabla&\times\vec A =\begin{vmatrix} \hat\imath & \hat\jmath & \hat k \\ \partial_x & \partial_y & \partial_z \\ z\sin y-yz\cos x & x\sin z-z\sin x & 0 \end{vmatrix} \\ &=\left\langle -x\cos z+\sin x,\sin y-y\cos x,\right. \\ &\qquad\qquad\qquad\left.\sin z-z\cos x-z\cos y+z\cos x\right\rangle \\ &=\left\langle \sin x-x\cos z,\sin y-y\cos x,\sin z-z\cos y\right\rangle \\ &=\vec F \end{aligned}

    [×]

    43. f. Scalar Potentials

    Find a scalar potential for each vector field or show one does not exist.

  43. F⃗=⟨y+z,x+z,x+y⟩\vec F=\left\langle y+z,x+z,x+y\right\rangle

    Hint

    A scalar potential for F⃗\vec F is any function ff satisfying ∇⃗f=F⃗\vec\nabla f=\vec F. Solve the equations: ∂xf=F1∂yf=F2∂zf=F3 \partial_x f=F_1 \qquad \partial_y f=F_2 \qquad \partial_z f=F_3 for a single function ff.

    [×]

    Answer

    f=xy+xz+yzf=xy+xz+yz

    [×]

    Solution

    We solve: ∂xf=y+z∂yf=x+z∂zf=x+y \partial_x f=y+z \qquad \partial_y f=x+z \qquad \partial_z f=x+y The xx-antiderivative of the 1st1^\text{st} equation is: f=xy+xz+g(y,z) f=xy+xz+g(y,z) The yy-antiderivative of the 2nd2^\text{nd} equation is: f=xy+yz+h(x,z) f=xy+yz+h(x,z) The zz-antiderivative of the 3rd3^\text{rd} equation is: f=xz+yz+k(x,y) f=xz+yz+k(x,y) A scalar potential which satisfies all three conditions is: f=xy+xz+yz f=xy+xz+yz where g(y,z)=yzg(y,z)=yz, h(x,z)=xzh(x,z)=xz and k(x,y)=xyk(x,y)=xy.

    [×]

  44. F⃗=⟨yzcos⁡(xyz),xzcos⁡(xyz),xycos⁡(xyz)⟩\vec F=\left\langle yz\cos(xyz),xz\cos(xyz),xy\cos(xyz)\right\rangle

    Answer

    f=sin⁡(xyz)f=\sin(xyz)

    [×]

    Solution

    We solve: ∂xf=yzcos⁡(xyz)∂yf=xzcos⁡(xyz)∂zf=xycos⁡(xyz) \partial_x f=yz\cos(xyz) \qquad \partial_y f=xz\cos(xyz) \qquad \partial_z f=xy\cos(xyz) The xx-antiderivative of the 1st1^\text{st} equation, the yy-antiderivative of the 2nd2^\text{nd} equation and the zz-antiderivative of the 3rd3^\text{rd} equation are all: f=sin⁡(xyz) f=\sin(xyz) which is therefore the scalar potential.

    [×]

  45. F⃗=⟨y2+2xz,2xy−3y2z,x2−y3−2z⟩\vec F=\left\langle y^2+2xz,2xy-3y^2z,x^2-y^3-2z\right\rangle

    Hint

    Successively solve the equations: ∂xf=F1∂yf=F2∂zf=F3 \partial_x f=F_1 \qquad \partial_y f=F_2 \qquad \partial_z f=F_3 If you reach a contradiction, there is no solution.

    [×]

    Answer

    f=xy2+x2z−y3z−z2+Cf=xy^2+x^2z-y^3z-z^2+C

    [×]

    Solution

    We solve: ∂xf=y2+2xz∂yf=2xy−3y2z∂zf=x2−y3−2z \partial_x f=y^2+2xz \qquad \partial_y f=2xy-3y^2z \qquad \partial_z f=x^2-y^3-2z The xx-antiderivative of the 1st1^\text{st} equation is: f=xy2+x2z+g(y,z) f=xy^2+x^2z+g(y,z) This needs to satisfy the the 2nd2^\text{nd} equation. So we compute ∂yf\partial_y f and plug into the 2nd2^\text{nd} equation: 2xy+∂yg=2xy−3y2zor∂yg=−3y2z 2xy+\partial_y g=2xy-3y^2z \qquad \text{or} \qquad \partial_y g=-3y^2z The solution is g(y,z)=−y3z+h(z)g(y,z)=-y^3z+h(z). So: f=xy2+x2z−y3z+h(z) f=xy^2+x^2z-y^3z+h(z) This needs to satisfy the the 3rd3^\text{rd} equation. So we compute ∂zf\partial_z f and plug into the 3rd3^\text{rd} equation: x2−y3+h′(z)=x2−y3−2zorh′(z)=−2z x^2-y^3+h'(z)=x^2-y^3-2z \qquad \text{or} \qquad h'(z)=-2z A solution is h(z)=−z2+Ch(z)=-z^2+C. So a scalar potential is: f=xy2+x2z−y3z−z2+C f=xy^2+x^2z-y^3z-z^2+C

    [×]

  46. F⃗=⟨y2+2xz,2xy+3y2z,x2−y3−2z⟩\vec F=\left\langle y^2+2xz,2xy+3y^2z,x^2-y^3-2z\right\rangle

    Answer

    There is no scalar potential. The contradiction is shown in the solution and in the remark.

    [×]

    Solution

    We solve: ∂xf=y2+2xz∂yf=2xy+3y2z∂zf=x2−y3−2z \partial_x f=y^2+2xz \qquad \partial_y f=2xy+3y^2z \qquad \partial_z f=x^2-y^3-2z The xx-antiderivative of the 1st1^\text{st} equation is: f=xy2+x2z+g(y,z) f=xy^2+x^2z+g(y,z) This needs to satisfy the the 2nd2^\text{nd} equation. So we compute ∂yf\partial_y f and plug into the 2nd2^\text{nd} equation: 2xy+∂yg=2xy+3y2zor∂yg=3y2z 2xy+\partial_y g=2xy+3y^2z \qquad \text{or} \qquad \partial_y g=3y^2z The solution is g(y,z)=y3z+h(z)g(y,z)=y^3z+h(z). So: f=xy2+x2z+y3z+h(z) f=xy^2+x^2z+y^3z+h(z) This needs to satisfy the the 3rd3^\text{rd} equation. So we compute ∂zf\partial_z f and plug into the 3rd3^\text{rd} equation: x2+y3+h′(z)=x2−y3−2zorh′(z)=−2y3−2z x^2+y^3+h'(z)=x^2-y^3-2z \qquad \text{or} \qquad h'(z)=-2y^3-2z This is a contradiction because hh cannot be a function of yy. So there is no scalar potential.

    [×]

    Check

    To verify there is no scalar potential, we compute the curl: ∇⃗×F⃗=∣ı^ȷ^k^∂x∂y∂zy2+2xz2xy+3y2zx2−y3−2z∣=ı^(−3y2−3y2)−ȷ^(2x−2x)+k^(2y−2y)=⟨−6y2,0,0⟩\begin{aligned} \vec\nabla\times\vec F &=\begin{vmatrix} \hat\imath & \hat\jmath & \hat k \\ \partial_x & \partial_y & \partial_z \\ y^2+2xz & 2xy+3y^2z & x^2-y^3-2z \end{vmatrix} \\ &=\hat\imath(-3y^2-3y^2) -\hat\jmath(2x-2x) +\hat k (2y-2y) \\ &=\left\langle -6y^2,0,0\right\rangle \end{aligned} Since this is not 0⃗\vec0, there cannot be a scalar potential.

    [×]

  47. F⃗=⟨cos⁡xcos⁡y,sin⁡xsin⁡y,sec⁡2z⟩\vec F=\left\langle \cos x\cos y,\sin x\sin y,\sec^2 z\right\rangle

    Answer

    There is no scalar potential, since the curl of F⃗\vec F is not 0⃗\vec0.

    [×]

    Solution

    We first compute the curl: ∇⃗×F⃗=∣ı^ȷ^k^∂x∂y∂zcos⁡xcos⁡ysin⁡xsin⁡ysec⁡2z∣=ı^(0−0)−ȷ^(0−0)+k^(cos⁡xsin⁡y+cos⁡xsin⁡y)=⟨0,0,2cos⁡xsin⁡y⟩\begin{aligned} \vec\nabla\times\vec F &=\begin{vmatrix} \hat\imath & \hat\jmath & \hat k \\ \partial_x & \partial_y & \partial_z \\ \cos x\cos y & \sin x\sin y & \sec^2 z \end{vmatrix} \\ &=\hat\imath(0-0) -\hat\jmath(0-0) +\hat k (\cos x\sin y+\cos x\sin y)\\ &=\left\langle 0,0,2\cos x\sin y\right\rangle \end{aligned} Since this is not 0⃗\vec0, there is no scalar potential.

    [×]

  48. F⃗=⟨cos⁡xcos⁡y,−sin⁡xsin⁡y,sec⁡2z⟩\vec F=\left\langle \cos x\cos y,-\sin x\sin y,\sec^2 z\right\rangle

    Answer

    f=sin⁡xcos⁡y+tan⁡zf=\sin x\cos y+\tan z

    [×]

    Solution

    We first compute the curl: ∇⃗×F⃗=∣ı^ȷ^k^∂x∂y∂zcos⁡xcos⁡y−sin⁡xsin⁡ysec⁡2z∣=ı^(0−0)−ȷ^(0−0)+k^(−cos⁡xsin⁡y+cos⁡xsin⁡y)=0⃗\begin{aligned} \vec\nabla\times\vec F &=\begin{vmatrix} \hat\imath & \hat\jmath & \hat k \\ \partial_x & \partial_y & \partial_z \\ \cos x\cos y & -\sin x\sin y & \sec^2 z \end{vmatrix} \\ &=\hat\imath(0-0) -\hat\jmath(0-0) +\hat k (-\cos x\sin y+\cos x\sin y) =\vec0 \end{aligned} Since this is 0⃗\vec0, we expect there is a potential and solve: ∂xf=cos⁡xcos⁡y∂yf=−sin⁡xsin⁡y∂zf=sec⁡2z \partial_x f=\cos x\cos y \qquad \partial_y f=-\sin x\sin y \qquad \partial_z f=\sec^2 z The xx-antiderivative of the 1st1^\text{st} equation is: f=sin⁡xcos⁡y+g(y,z) f=\sin x\cos y+g(y,z) This needs to satisfy the the 2nd2^\text{nd} equation. So we compute ∂yf\partial_y f and plug into the 2nd2^\text{nd} equation: −sin⁡xsin⁡y+∂yg=−sin⁡xsin⁡yor∂yg=0 -\sin x\sin y+\partial_y g=-\sin x\sin y \qquad \text{or} \qquad \partial_y g=0 The solution is g(y,z)=h(z)g(y,z)=h(z). (Don't say g=0g=0 or you will not get a solution!) So: f=sin⁡xcos⁡y+h(z) f=\sin x\cos y+h(z) This needs to satisfy the the 3rd3^\text{rd} equation. So we compute ∂zf\partial_z f and plug into the 3rd3^\text{rd} equation: h′(z)=sec⁡2zorh(z)=tan⁡z h'(z)=\sec^2 z \qquad \text{or} \qquad h(z)=\tan z Therefore, a scalar potential is: f=sin⁡xcos⁡y+tan⁡z f=\sin x\cos y+\tan z

    [×]

    Remark

    In this case, computing the curl was a waste of time. But we could not know this in advance!

    [×]

    49. g. Vector Potentials

    Find a vector potential for each vector field or show one does not exist.

  49. F⃗=⟨−x,y,y−x⟩\vec F=\left\langle -x,y,y-x\right\rangle

    Hint

    A vector potential for F⃗=⟨F1,F2,F3⟩\vec F=\left\langle F_1,F_2,F_3\right\rangle is any vector field A⃗=⟨A1,A2,A3⟩\vec A=\left\langle A_1,A_2,A_3\right\rangle satisfying ∇⃗×A⃗=F⃗\vec\nabla\times\vec A=\vec F. So we need to solve the equations: ∂yA3−∂zA2=F1∂zA1−∂xA3=F2∂xA2−∂yA1=F3\begin{aligned} \partial_y A_3-\partial_z A_2&=F_1 \\ \partial_z A_1-\partial_x A_3&=F_2 \\ \partial_x A_2-\partial_y A_1&=F_3 \end{aligned} for A1A_1, A2A_2 and A3A_3. There is always a solution with A3=0A_3=0. So we actually need to solve: ∂zA2=−F1∂zA1=F2∂xA2−∂yA1=F3\begin{aligned} \partial_z A_2&=-F_1 \\ \partial_z A_1&=F_2 \\ \partial_x A_2-\partial_y A_1&=F_3 \end{aligned}

    [×]

    Answer

    A⃗=⟨yz+xy,xz+xy,0⟩\vec A=\left\langle yz+xy,xz+xy,0\right\rangle
    There are many other solutions.

    [×]

    Solution

    We assume A3=0A_3=0. So we need to solve: ∂zA2=−F1=x∂zA1=F2=y∂xA2−∂yA1=F3=y−x\begin{aligned} \partial_z A_2&=-F_1=x \\ \partial_z A_1&=F_2=y \\ \partial_x A_2-\partial_y A_1&=F_3=y-x \end{aligned} The first 22 equations say: A2=xz+f(x,y)A1=yz+g(x,y) A_2=xz+f(x,y) \qquad A_1=yz+g(x,y) We substitute into the 3rd3^\text{rd} equation: z+∂xf−z−∂yg=y−x z+\partial_x f-z-\partial_y g=y-x One solution is f=xyf=xy and g=xyg=xy. So A⃗=⟨yz+xy,xz+xy,0⟩ \vec A=\left\langle yz+xy,xz+xy,0\right\rangle There are many other solutions.

    [×]

  50. F⃗=⟨xcos⁡z,ycos⁡z,2x−2sin⁡z⟩\vec F=\left\langle x\cos z,y\cos z,2x-2\sin z\right\rangle

    Hint

    Successively solve the equations: ∂zA2=−F1∂zA1=F2∂xA2−∂yA1=F3\begin{aligned} \partial_z A_2&=-F_1 \\ \partial_z A_1&=F_2 \\ \partial_x A_2-\partial_y A_1&=F_3 \end{aligned} If you reach a contradiction, there is no solution.

    [×]

    Answer

    A⃗=⟨ysin⁡z,−xsin⁡z+x2,0⟩\vec A=\left\langle y\sin z,-x\sin z+x^2,0\right\rangle
    There are many other solutions.

    [×]

    Solution

    We assume A3=0A_3=0. So we need to solve: ∂zA2=−F1=−xcos⁡z∂zA1=F2=ycos⁡z∂xA2−∂yA1=F3=2x−2sin⁡z\begin{aligned} \partial_z A_2&=-F_1=-x\cos z \\ \partial_z A_1&=F_2=y\cos z \\ \partial_x A_2-\partial_y A_1&=F_3=2x-2\sin z \end{aligned} The first 22 equations say: A2=−xsin⁡z+f(x,y)A1=ysin⁡z+g(x,y) A_2=-x\sin z+f(x,y) \qquad A_1=y\sin z+g(x,y) We substitute into the 3rd3^\text{rd} equation: −sin⁡z+∂xf−sin⁡z−∂yg=2x−2sin⁡zor∂xf−∂yg=2x -\sin z+\partial_x f-\sin z-\partial_y g=2x-2\sin z \qquad \text{or} \qquad \partial_x f-\partial_y g=2x One solution is f=x2f=x^2 and g=0g=0. So A⃗=⟨ysin⁡z,−xsin⁡z+x2,0⟩ \vec A=\left\langle y\sin z,-x\sin z+x^2,0\right\rangle There are many other solutions.

    [×]

  51. F⃗=⟨xcos⁡z,ycos⁡z,2x−sin⁡z⟩\vec F=\left\langle x\cos z,y\cos z,2x-\sin z\right\rangle

    Answer

    There is no vector potential. The contradiction is shown in the solution and in the remark.

    [×]

    Solution

    We assume A3=0A_3=0. So we need to solve: ∂zA2=−F1=−xcos⁡z∂zA1=F2=ycos⁡z∂xA2−∂yA1=F3=2x−sin⁡z\begin{aligned} \partial_z A_2&=-F_1=-x\cos z \\ \partial_z A_1&=F_2=y\cos z \\ \partial_x A_2-\partial_y A_1&=F_3=2x-\sin z \end{aligned} The first 22 equations say: A2=−xsin⁡z+f(x,y)A1=ysin⁡z+g(x,y) A_2=-x\sin z+f(x,y) \qquad A_1=y\sin z+g(x,y) We substitute into the 3rd3^\text{rd} equation: −sin⁡z+∂xf−sin⁡z−∂yg=2x−sin⁡zor∂xf−∂yg=2x+sin⁡z -\sin z+\partial_x f-\sin z-\partial_y g=2x-\sin z \qquad \text{or} \qquad \partial_x f-\partial_y g=2x+\sin z This is a contradiction because ff and gg cannot be functions of zz. So there is no vector potential.

    [×]

    Check

    To verify there is no vector potential, we compute the divergence: ∇⃗⋅F⃗=∂x(xcos⁡z)+∂y(ycos⁡z)+∂z(2x−sin⁡z)=cos⁡z+cos⁡z−cos⁡z=cos⁡z≠0\begin{aligned} \vec\nabla\cdot\vec F &=\partial_x(x\cos z)+\partial_y(y\cos z)+\partial_z(2x-\sin z) \\ &=\cos z+\cos z-\cos z=\cos z \ne 0 \end{aligned} Since this is not 00, there cannot be a vector potential.

    [×]

  52. F⃗=⟨yz,xz,xy⟩\vec F=\left\langle yz,xz,xy\right\rangle

    Answer

    A⃗=⟨xz22,−yz22+x2y2,0⟩\vec A =\left\langle \dfrac{xz^2}{2},-\,\dfrac{yz^2}{2}+\dfrac{x^2y}{2},0\right\rangle

    [×]

    Solution

    We first compute the divergence: ∇⃗⋅F⃗=∂x(yz)+∂y(xz)+∂z(xy)=0+0+0=0 \vec\nabla\cdot\vec F =\partial_x(yz)+\partial_y(xz)+\partial_z(xy) =0+0+0=0 Since this is 0, we expect there is a potential and solve: ∂zA2=−F1=−yz∂zA1=F2=xz∂xA2−∂yA1=F3=xy\begin{aligned} \partial_z A_2&=-F_1=-yz \\ \partial_z A_1&=F_2=xz \\ \partial_x A_2-\partial_y A_1&=F_3=xy \end{aligned} The first 22 equations say: A2=−yz22+f(x,y)A1=xz22+g(x,y) A_2=-\,\dfrac{yz^2}{2}+f(x,y) \qquad A_1=\dfrac{xz^2}{2}+g(x,y) We substitute into the 3rd3^\text{rd} equation: ∂xf−∂yg=xy \partial_x f-\partial_y g=xy A solution is f=x2y2f=\dfrac{x^2y}{2} and g=0g=0. So a vector potential is: A⃗=⟨xz22,−yz22+x2y2,0⟩ \vec A =\left\langle \dfrac{xz^2}{2},-\,\dfrac{yz^2}{2}+\dfrac{x^2y}{2},0\right\rangle

    [×]

  53. F⃗=⟨x,y,z⟩\vec F=\left\langle x,y,z\right\rangle

    Answer

    There is no vector potential because ∇⃗⋅F⃗=3≠0\vec\nabla\cdot\vec F=3\ne0.

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    Solution

    We first compute the divergence: ∇⃗⋅F⃗=∂x(x)+∂y(y)+∂z(z)=1+1+1=3 \vec\nabla\cdot\vec F =\partial_x(x)+\partial_y(y)+\partial_z(z) =1+1+1=3 Since this is not 00, there is no vector potential.

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    54. PY: Checked to here.

    Review Exercises

  54. PY: Add the last problem from Edfinity Stokes.>

    Hint

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    Answer

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    Solution

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  55. Q1

    Hint

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    Answer

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    Solution

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