26. Stokes' Theorem

Since we are told to use Stokes' Theorem, we convert the integral into the line integral \(\displaystyle \oint_{\partial S} \vec{F}\cdot d\vec{s}\) around the boundary. The boundary curve may be obtained by setting \(r=2\) in the surface \(\vec{R}(r,\theta)\). So: \[ \vec r(\theta)=\vec{R}(2,\theta) =\langle 2\cos\theta,2\sin\theta,2\sin(5\theta)\rangle \] The velocity is: \[ \vec v=\langle -2\sin\theta,2\cos\theta,10\cos(5\theta)\rangle \] Since the surface is oriented upward, the boundary must be traversed counterclockwise, which is the direction \(\vec v\) points. On the curve, the vector field becomes: \[ \vec{F}=\langle-y,x,x^2+y^2\rangle =\langle-2\sin\theta,2\cos\theta,4\rangle \] So their dot product is: \[ \vec{F}\cdot\vec v =4\sin^2\theta+4\cos^2\theta+40\cos(5\theta) =4+40\cos(5\theta) \] So the line integral is: \[ \oint_{\partial S} \vec{F}\cdot d\vec{s} =\int_0^{2\pi} [4+40\cos(5\theta)]\,d\theta =\left[\rule{0pt}{10pt}4\theta+8\sin(5\theta)\right]_0^{2\pi} =8\pi \]

To see if \(\vec F=\langle 2yz,2xz,2xy\rangle\) has a vector potential, we compute \(\vec\nabla\cdot\vec F=0+0+0=0\). So \(\vec F\) has a vector potential. We write \(\vec F=\vec\nabla\times\vec A\) and assume \(A_3=0\). So we need to solve \[\begin{aligned} -\partial_z A_2&=F_1=2yz \\ \partial_z A_1&=F_2=2xz \\ \partial_x A_2-\partial_y A_1&=F_3=2xy \\ \end{aligned}\] The first two equations say \(A_2=-yz^2+f(x,y)\) and \(A_1=xz^2+g(x,y)\). Then the third equation says \(\partial_xf-\partial_yg=2xy\). This is satisfied by \(f=x^2y\) and \(g=0\). So \[ \vec A=\langle xz^2,-yz^2+x^2y,0\rangle \] So \[ \iint_H \vec{F}\cdot d\vec{S} =\iint_H \vec\nabla\times\vec A\cdot d\vec{S} =\oint_{\partial H} \vec A\cdot d\vec{s} \] The boundary is the circle \(x^2+y^2=4\) with \(z=0\) traversed counterclockwise. It may be parametrized by \(\vec r=(2\cos\theta,2\sin\theta,0)\) and has velocity \(\vec v=\langle-2\sin\theta,2\cos\theta,0\rangle\). On the curve the vector field is \(\vec A=\langle 0,4\cos^2\theta2\sin\theta,0\rangle\) and its dot product with the velocity is \(\vec A\cdot\vec v=0+4\cos^2\theta2\sin\theta2\cos\theta+0 =16\cos^3\theta\sin\theta\). So the integral is \[ \oint_{\partial H} \vec A\cdot d\vec{s} =\int_0^{2\pi} 16\cos^3\theta\sin\theta\,d\theta =\left[\rule{0pt}{10pt}-4\cos^4\theta\right]_0^{2\pi}=0 \]

To see if \(\vec F=\langle 2xz,2yz,2z^2\rangle\) has a vector potential, we compute \(\vec\nabla\cdot\vec F=2z+2z+4z\ne0\). Since the divergence is non-zero, \(\vec F\) does not have a vector potential and we need to compute the surface integral directly. The hemisphere may be parametrized in spherical coordinates as: \[ \vec R(\phi,\theta)=\langle2\sin\phi\cos\theta,2\sin\phi\sin\theta,2\cos\phi\rangle \] for \(0\le\phi\le\pi\) and \(0\le\theta\le2\pi\). The normal is: \[\begin{aligned} \vec{N}&=\vec{e}_\phi\times\vec{e}_\theta =\begin{vmatrix} \hat{\imath} & \hat{\jmath} & \hat{k} \\ (2\cos\phi\cos\theta&2\cos\phi\sin\theta&-2\sin\phi) \\ (-2\sin\phi\sin\theta&2\sin\phi\cos\theta&\quad0\quad) \end{vmatrix} \\ &=\langle4\sin^2\phi\cos\theta,4\sin^2\phi\sin\theta,4\sin\phi\cos\phi\rangle \end{aligned}\] Next, we evaluate \(\vec F=\langle 2xz,2yz,2z^2\rangle\) on the surface and compute its dot product with the normal: \[\begin{aligned} \vec F(\vec R(\phi,\theta)) &=\langle 8\sin\phi\cos\phi\cos\theta,8\sin\phi\cos\phi\sin\theta,8\cos^2\phi\rangle \\ \vec F\cdot\vec N &=32\sin^3\phi\cos\phi\cos^2\theta +32\sin^3\phi\cos\phi\sin^2\theta +32\cos^3\phi\sin\phi \\ &=32\sin^3\phi\cos\phi+32\sin\phi\cos^3\phi =32\sin\phi\cos\phi \end{aligned}\] So the integral is: \[\begin{aligned} \iint_H \vec{F}\cdot d\vec{S} &=\int_0^{2\pi}\int_0^{\pi/2} \vec{F}\cdot\vec N\,d\phi\,d\theta \\ &=\int_0^{2\pi}\int_0^{\pi/2} 32\sin\phi\cos\phi\,d\phi\,d\theta \\ &=64\pi\left[\dfrac{\sin^2\phi}{2}\right]_0^{\pi/2} =32\pi \end{aligned}\]