Math Insight

Scalar surface integral examples

Example 1

For the cylindrical surface of radius 3 and height 5 given by $x^2+y^2=3^2$ and $0 \le z \le 5$, let the charge density be proportional to the distance from the $xy$-plane. Find the total charge on the surface.

Solution: Since the distance from the point $(x,y,z)$ to the $xy$-plane is $z$, let the charge density be $f(x,y,z) = kz$ for some constant $k$. The total charge on the cylindrical surface $\dls$ is the surface integral of $f$ over $\dls$: $\ssint{\dls}{f}$.

Parameterize the cylindrical surface by \begin{align*} \dlsp(\theta,t) = (3 \cos\theta, 3\sin\theta, t) \end{align*} for $0 \le \theta \le 2\pi$ and $0 \le t \le 5$. The total charge on the cylindrical surface is \begin{gather*} \ssint{\dls}{f} \goodbreak = \pssint{0}{5}{0}{2\pi}{f}{\dlsp}{\theta}{t} \end{gather*}

Calculate the components of the above integral as follows. \begin{align*} f(\dlsp(\theta,t)) &= kt\\ \pdiff{\dlsp}{\theta}(\theta,t) &= (-3\sin\theta, 3\cos\theta,0)\\ \pdiff{\dlsp}{t}(\theta,t) &= (0,0,1)\\ \pdiff{\dlsp}{\theta}(\theta,t) \times \pdiff{\dlsp}{t}(\theta,t) &= \left| \begin{array}{ccc} \vc{i} & \vc{j} & \vc{k}\\ -3\sin\theta & 3\cos\theta & 0\\ 0 & 0 & 1 \end{array} \right|\\ &= \vc{i} 3 \cos\theta - \vc{j} (-3\sin \theta)\\ &= (3\cos\theta, 3\sin\theta,0)\\ \left\|\pdiff{\dlsp}{\theta}(\theta,t) \times \pdiff{\dlsp}{t}(\theta,t)\right\| &= \sqrt{9 \cos^2\theta + 9 \sin^2\theta}\\ &= 3 \end{align*} Therefore, the total charge is \begin{align*} \ssint{\dls}{f} &=\int_0^5 \!\!\!\int_0^{2\pi} kt 3 \,d\theta\, dt\\ &= \int_0^5 \left( \left.3t \theta \right|_{\theta=0}^{\theta=2\pi} \right) dt\\ &= \int_0^5 k6\pi t \,dt\\ &= \left.k3 \pi t^2 \right|_0^5 = 3\pi (25)k = 75\pi k \end{align*}