Multivariable calculus and vector analysis

Surface integral examples

Example 1

For the cylinder 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 cylinder.

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

Parameterize the cylinder by

$$\begin{array}{lll}\hfill \mathbf{\Phi }\left(\theta ,t\right)=\left(3cos\theta ,3sin\theta ,t\right)& & \hfill \end{array}$$

for $0\le \theta \le 2\pi$ and $0\le t\le 5$. The total charge on the cylinder is

$$\begin{array}{lll}\hfill {\iint }_{S}fdS={\int \; <!--nolimits\; -->}_0^5{\int \; <!--nolimits\; -->}_0^{2\pi }f\left(\mathbf{\Phi }\left(\theta ,t\right)\right)∥\frac{\partial \mathbf{\Phi }}{\partial \theta }\left(\theta ,t\right)\times \frac{\partial \mathbf{\Phi }}{\partial t}\left(\theta ,t\right)∥d\theta dt& & \hfill \end{array}$$

Calculate the components of the above integral as follows.

$$\begin{array}{llll}\hfill f\left(\mathbf{\Phi }\left(\theta ,t\right)\right)& =kt& \hfill & \\ \hfill \frac{\partial \mathbf{\Phi }}{\partial \theta }\left(\theta ,t\right)& =\left(-3sin\theta ,3cos\theta ,0\right)& \hfill & \\ \hfill \frac{\partial \mathbf{\Phi }}{\partial t}\left(\theta ,t\right)& =\left(0,0,1\right)& \hfill & \\ \hfill \frac{\partial \mathbf{\Phi }}{\partial \theta }\left(\theta ,t\right)\times \frac{\partial \mathbf{\Phi }}{\partial t}\left(\theta ,t\right)& =\left|\begin{array}{ccc}\hfill \mathbf{i}\hfill & \hfill \mathbf{j}\hfill & \hfill \mathbf{k}\hfill \\ \hfill -3sin\theta \hfill & \hfill 3cos\theta \hfill & \hfill 0\hfill \\ \hfill 0\hfill & \hfill 0\hfill & \hfill 1\hfill \end{array}\right|& \hfill & \\ \hfill & =\mathbf{i}3cos\theta -\mathbf{j}\left(-3sin\theta \right)& \hfill & \\ \hfill & =\left(3cos\theta ,3sin\theta ,0\right)& \hfill & \\ \hfill ∥\frac{\partial \mathbf{\Phi }}{\partial \theta }\left(\theta ,t\right)\times \frac{\partial \mathbf{\Phi }}{\partial t}\left(\theta ,t\right)∥& =\sqrt{9{cos}^2\theta +9{sin}^2\theta }& \hfill & \\ \hfill & =3& \hfill & \end{array}$$

Therefore, the total charge is

$$\begin{array}{llll}\hfill {\iint }_{S}fdS& ={\int \; <!--nolimits\; -->}_0^5{\int \; <!--nolimits\; -->}_0^{2\pi }kt3d\theta dt& \hfill & \\ \hfill & ={\int \; <!--nolimits\; -->}_0^5\left({\left.3t\theta \right|}_{\theta =0}^{\theta =2\pi }\right)dt& \hfill & \\ \hfill & ={\int \; <!--nolimits\; -->}_0^{5}k6\pi tdt& \hfill & \\ \hfill & ={\left.k3\pi t^2\right|}_0^5=3\pi \left(25\right)k=75\pi k& \hfill & \end{array}$$

Example 2

Let $S$ be the same cylinder as above ($x^2+y^2=9$ for $0\le z\le 5$).

Let $\mathbf{F}$ be the vector field $\mathbf{F}\left(x,y,z\right)=\left(2x,2y,2z\right)$.

Find the integral of $\mathbf{F}$ over $S$. Use outward pointing normal.

What is the sign of integral? Since the vector field and unit normal point outward, the integral better be positive.

As above, parameterize the cylinder by

$$\begin{array}{lll}\hfill \mathbf{\Phi }\left(\theta ,t\right)=\left(3cos\theta ,3sin\theta ,t\right)& & \hfill \end{array}$$

for $0\le \theta \le 2\pi$ and $0\le t\le 5$. We calculated above that

$$\begin{array}{llll}\hfill \frac{\partial \mathbf{\Phi }}{\partial \theta }\left(\theta ,t\right)\times \frac{\partial \mathbf{\Phi }}{\partial t}\left(\theta ,t\right)& =\left(3cos\theta ,3sin\theta ,0\right)& \hfill & \end{array}$$

Is $\left(3cos\theta ,3sin\theta ,0\right)$ an outward pointing normal? It is a normal at the point $\mathbf{\Phi }\left(\theta ,t\right)=\left(3cos\theta ,3sin\theta ,t\right)$. As shown in the below figure, it is an outward pointing normal.

Calculate total flux:

$$\begin{array}{llll}\hfill {\iint }_S\mathbf{F}\cdot d\mathbf{S}& ={\int \; <!--nolimits\; -->}_0^5{\int \; <!--nolimits\; -->}_0^{2\pi }\mathbf{F}\left(\mathbf{\Phi }\left(\theta ,t\right)\right)\cdot \left(\frac{\partial \mathbf{\Phi }}{\partial \theta }\left(\theta ,t\right)\times \frac{\partial \mathbf{\Phi }}{\partial t}\left(\theta ,t\right)\right)d\theta dt& \hfill & \\ \hfill & ={\int \; <!--nolimits\; -->}_0^5{\int \; <!--nolimits\; -->}_0^{2\pi }\mathbf{F}\left(3cos\theta ,3sin\theta ,t\right)\cdot \left(3cos\theta ,3sin\theta ,0\right)d\theta dt& \hfill & \\ \hfill & ={\int \; <!--nolimits\; -->}_0^5{\int \; <!--nolimits\; -->}_0^{2\pi }\left(6cos\theta ,6sin\theta ,2t\right)\cdot \left(3cos\theta ,3sin\theta ,0\right)d\theta dt& \hfill & \\ \hfill & ={\int \; <!--nolimits\; -->}_0^5{\int \; <!--nolimits\; -->}_0^{2\pi }18{cos}^2\theta +18{sin}^2\theta d\theta dt& \hfill & \\ \hfill & ={\int \; <!--nolimits\; -->}_0^5{\int \; <!--nolimits\; -->}_0^{2\pi }18d\theta dt& \hfill & \\ \hfill & ={\int \; <!--nolimits\; -->}_0^{5}18\left(2\pi \right)dt& \hfill & \\ \hfill & =18\left(2\pi \right)\left(5\right)=180\pi & \hfill & \end{array}$$

There are two orientations for the cylinder. If we had instead chose the orientation given by the inward point normal, what must ${\iint }_S\mathbf{F}\cdot d\mathbf{S}$ be? $-180\pi$

Contents by topic

1. Geometry in higher dimensions

2. Differential Calculus

3. Integral calculus

4. Vector calculus

4a. Vector field basics

4b. Parametrized curves

4c. Parametrized surfaces

4d. Path integrals and line integrals

4e. Surface integrals

• Surface integrals
• Surface integral examples

5. The fundamental theorems of vector calculus

6. Review material