9. Line Integrals

The arc length integral can be generalized to the integral of any function.

b. Line Integrals of Scalars

Line Integral of a Scalar Function
The line integral of a function, f(x,y,z)f(x,y,z), along a parametric curve, r(t)=(x(t),y(t),z(t))\vec r(t)=\left(x(t),y(t),z(t)\right), between A=r(a)A=\vec r(a) and B=r(b)B=\vec r(b) is given by ABfds=abf(r(t))vdt \int_A^B f\,ds =\int_a^b f(\vec r(t))|\vec v|\,dt where f(r(t))f(\vec r(t)) is the composition of ff and r\vec r, i.e. the function f(x,y,z)f(x,y,z) with xx, yy and zz replaced by x(t)x(t), y(t)y(t) and z(t)z(t). The composition f(r(t))f(\vec r(t)) is called the value of the function along the curve or the restriction of the function to the curve.

We sometimes write f(t)f(t) to mean f(r(t))f(\vec r(t)), the value of ff at time tt, even though this is not mathematically correct. Consequently, this is sometimes called abuse of notation.

Notice that we again switch the limits on the integral from the abstract points AA and BB to the values aa and bb of the parameter tt when we switch from the integral with the general differential dsds to the integral with the parameter differential dtdt.

Compute (0,0,0)(1,1,2/3)(3xz+y2)ds\displaystyle \int_{(0,0,0)}^{(1,1,2/3)} (3xz+y^2)\,ds along the twisted cubic r(t)=(t,t2,23t3)\vec r(t)=\left(t,t^2,\dfrac{2}{3}t^3\right).

The value of the function along the curve is f(r(t))=3x(t)z(t)+y(t)2=3(t)(23t3)+(t2)2=3t4 f(\vec r(t)) =3x(t)z(t)+y(t)^2 =3(t)\left(\dfrac{2}{3}t^3\right)+(t^2)^2=3t^4 The velocity is v=(1,2t,2t2)\vec v=(1,2t,2t^2) and its length is v=1+4t2+4t4=(1+2t2)2=1+2t2|\vec v|=\sqrt{1+4t^2+4t^4}=\sqrt{(1+2t^2)^2}=1+2t^2. Finally, the endpoints are: (0,0,0)=r(0)(1,1,23)=r(1) (0,0,0)=\vec r(0) \qquad \left(1,1,\dfrac{2}{3}\right)=\vec r(1) So (0,0,0)(1,1,2/3)(3xz+y2)ds=01f(r(t))vdt=013t4(1+2t2)dt=[35t5+67t7]01=35+67=5135\begin{aligned} \int_{(0,0,0)}^{(1,1,2/3)} (3xz+y^2)\,ds &=\int_0^1 f(\vec r(t))|\vec v|\,dt =\int_0^1 3t^4(1+2t^2)\,dt \\ &=\left[\dfrac{3}{5}t^5+\dfrac{6}{7}t^7\right]_0^1 =\dfrac{3}{5}+\dfrac{6}{7}=\dfrac{51}{35} \end{aligned}

Compute x2ds\displaystyle \int x^2\,ds along the helix x=4cosz3,y=4sinz3x=4\cos\dfrac{z}{3},\quad y=4\sin\dfrac{z}{3} between z=0z=0 and z=6πz=6\pi. You can rotate this plot with your mouse.

1

Hint

For functions of zz, the differential of arc length is ds=(dxdz)2+(dydz)2+1dz ds=\sqrt{\left(\dfrac{dx}{dz}\right)^2+\left(\dfrac{dy}{dz}\right)^2+1}\,dz

[×]

Answer

x2ds=80π\displaystyle \int x^2\,ds=80\pi

[×]

Solution

The derivatives are: dxdz=43sinz3anddydz=43cosz3 \dfrac{dx}{dz}=-\,\dfrac{4}{3}\sin\dfrac{z}{3} \qquad \text{and} \qquad \dfrac{dy}{dz}=\dfrac{4}{3}\cos\dfrac{z}{3} So the differential of arc length is ds=(dxdz)2+(dydz)2+1dz=(43sinz3)2+(43cosz3)2+1dz=169+1dz=53dz\begin{aligned} ds&=\sqrt{\left(\dfrac{dx}{dz}\right)^2+\left(\dfrac{dy}{dz}\right)^2+1}\,dz \\ &=\sqrt{\left(-\,\dfrac{4}{3}\sin\dfrac{z}{3}\right)^2 +\left(\dfrac{4}{3}\cos\dfrac{z}{3}\right)^2+1}\,dz \\ &=\sqrt{\dfrac{16}{9}+1}\,dz=\dfrac{5}{3}\,dz \end{aligned} Thus, using the identity cos2A=1+cos2A2\cos^2A=\dfrac{1+\cos2A}{2}: x2ds=06π16cos2(z3)53dz=80306π1+cos2z32dz=403[z+32sin2z3]06π=80π\begin{aligned} \int x^2\,ds &=\int_0^{6\pi} 16\cos^2\left(\dfrac{z}{3}\right)\dfrac{5}{3}\,dz =\dfrac{80}{3}\int_0^{6\pi} \dfrac{1+\cos\dfrac{2z}{3}}{2}\,dz \\ &=\dfrac{40}{3}\left[z+\dfrac{3}{2}\sin\dfrac{2z}{3}\right]_0^{6\pi} =80\pi \end{aligned}

[×]

Compute (4,0,0)(4,0,6π)(x2+y2+z2)ds\displaystyle \int_{(4,0,0)}^{(4,0,6\pi)} (x^2+y^2+z^2)\,ds along the helix r(t)=(4cost,4sint,3t)\vec r(t)=(4\cos t,4\sin t,3t).


Notice this is the same helix as in the previous exercise, except written in parametric form.

eg_LineIntScal_helix

Answer

(4,0,0)(4,0,6π)(x2+y2+z2)ds=160π+120π3\displaystyle \int_{(4,0,0)}^{(4,0,6\pi)} (x^2+y^2+z^2)\,ds =160\pi+120\pi^3

[×]

Solution

We first identify the integrand and evaluate it on the curve: f=x2+y2+z2f(r(t))=16cos2t+16sin2t+9t2=16+9t2\begin{aligned} f&=x^2+y^2+z^2 \\ f(\vec r(t))&=16\cos^2t+16\sin^2t+9t^2=16+9t^2 \end{aligned} The velocity is v(t)=(4sint,4cost,3)\vec v(t)=(-4\sin t,4\cos t,3) and its length is v=5|\vec v|=5. Also, the point (4,0,0)(4,0,0) occurs at t=0t=0 and the point (4,0,6π)(4,0,6\pi) occurs at t=2πt=2\pi. So the integral becomes: (4,0,0)(4,0,6π)(x2+y2+z2)ds=02π(16+9t2)5dt=5[16t+3t3]02π=160π+120π3\begin{aligned} \int_{(4,0,0)}^{(4,0,6\pi)} &(x^2+y^2+z^2)\,ds =\int_0^{2\pi} (16+9t^2)5\,dt \\ &=5\left[\rule{0pt}{10pt}16t+3t^3\right]_0^{2\pi}=160\pi+120\pi^3 \end{aligned}

[×]

We look at applications starting on the next page.

© 2025 MYMathApps

Supported in part by NSF Grant #1123255