Scalar Surface Integrals

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Objectives:

1.: Understand the procedure to find the surface area of a bounded surface.
2.: Understand what the scaling factor $\|\textbf {r}_u \times \textbf {r}_v \|$ represents for a parametrized surface $\textbf {r}(u,v)$ .
3.: Be able to compute a surface integral $\iint _S f(x,y,z)dS$ .
4.: Be comfortable parametrizing surfaces in the process of computing scalar surface integrals.

Recap Video

Here is a video highlights the main points of the section.

To summarize:

To compute a scalar surface integral of $f(x,y,z)$ over a surface $S$ :

Parametrize the surface as $\textbf {r}(u,v)$ , where the domain of $(u,v)$ is $D$ .
Compute $\iint _S f(x,y,z)dS = \iint _D f(\textbf {r}(u,v))\| \textbf {r}_u \times \textbf {r}_v \| du\,dv.$

Example Video

The following video shows a worked example using two different parameterizations (it is worth looking at both approaches).

If $S$ is the portion of $z = 4-x^2-y^2$ in the first octant, evaluate $\iint _S \sqrt {4x^2+4y^2+1}dS.$

Approach 1:
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Approach 2:
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Problems

If $S$ is the surface given by $\textbf {r}(u,v) = \left \langle u\cos (v),u\sin (v),v \right \rangle$ , where $0 \leq u \leq 1$ and $0\leq v \leq \pi$ , compute $\iint _S \sqrt {1+x^2+y^2}dS$ .

We already have the parametrization, so step (1) is done for us. We need to compute $\|\textbf {r}_u \times \textbf {r}_v\|$ . We take the partials: $\textbf {r}_u = \left \langle \answer {\cos (v)},\answer {\sin (v)},\answer {0} \right \rangle ,$ $\textbf {r}_v = \left \langle \answer {-u\sin (v)},\answer {u\cos (v)},\answer {1} \right \rangle .$ The cross product is $\textbf {r}_u \times \textbf {r}_v = \left \langle \answer {\sin (v)}, \answer {-\cos (v)}, \answer {u} \right \rangle ,$ which has magnitude $\|\textbf {r}_u \times \textbf {r}_v\| = \sqrt {\answer {1+u^2}}.$ We now need $f(\textbf {r}(u,v))$ , which in this case is $f(\textbf {r}(u,v)) = \sqrt {\answer {1+u^2}}.$ Therefore

$\begin{eqnarray*} \iint_S \sqrt{1+x^2+y^2}dS &=& \iint_D f(\textbf{r}(u,v))\| \textbf{r}_u \times \textbf{r}_v\| dudv \\ &=& \int_0^{\pi} \int_0^1 \answer{1+u^2}dudv \\ &=& \answer{4\pi/3}. \end{eqnarray*}$

If $f(x,y,z) = x$ and $S$ is the portion of $x^2+y^2=4$ with $y \geq 0$ and $0 \leq z \leq 1$ , evaluate $\displaystyle \iint _S f(x,y,z)dS$ .

We can parametrize the cylinder as $\textbf {r}(\theta ,z) = (2\cos (\theta ),2\sin (\theta ),z).$ The bounds for the variables are $\answer {0} \leq \theta \leq \answer {\pi },\quad \answer {0} \leq z \leq \answer {1}.$ With this parametrization, we have $\textbf {r}_{\theta } = \left \langle \answer {-2\sin (\theta )},\answer {2\cos (\theta )},\answer {0} \right \rangle ,$ $\textbf {r}_z = \left \langle \answer {0},\answer {0},\answer {1} \right \rangle .$ The cross product is $\textbf {r}_{\theta } \times \textbf {r}_z = \left \langle \answer {2\cos (\theta )},\answer {2\sin (\theta )},\answer {0} \right \rangle ,$ which has magnitude $\| \textbf {r}_{\theta } \times \textbf {r}_z \| = \answer {2}.$ The composition $f(\textbf {r}(\theta ,z)) = \answer {2\cos (\theta )}.$ The surface integral is therefore

$\begin{eqnarray*} \iint_S xdS &=& \iint_D f(\textbf{r}(\theta,z))\| \textbf{r}_{\theta} \times \textbf{r}_z\| d\theta dz \\ &=& \int_0^{1} \int_0^{\pi} \answer{4\cos(\theta)}d\theta dz\\ &=& \answer{0}. \end{eqnarray*}$

To get the surface area of $S$ , we can compute a scalar surface integral over $S$ with $f(x,y,z) = \answer {1}$ .

If $f(x,y,z) = z$ and $S$ is the portion of $y = 9-z^2$ where $0 \leq x \leq 3$ and $0 \leq z \leq 3$ , then $\displaystyle \iint _S zdS = \answer {\frac {37\sqrt {37}-1}{4}}$ .

The surface can be parametrized as $\textbf {r}(x,z) = \left \langle x,9-z^2,z \right \rangle$ , which has $\textbf {r}_x \times \textbf {r}_z = \left \langle \answer {0},\answer {-1},\answer {-2z} \right \rangle .$ This has magnitude $\answer {\sqrt {1+4z^2}}$ . We also have $f(\textbf {r}(x,z)) = \answer {z}$ . Therefore, the surface integral is $\int _0^3 \int _0^3 z\sqrt {1+4z^2}dz\,dx.$

If $f(x,y,z) = x$ and $S$ is the portion of $x^2+y^2+z^2=4$ where $z \geq 0$ and $x \geq |y|$ , then $\iint _S f(x,y,z)dS = \answer { 2\pi \sqrt {2}}.$

The parametrization is $\textbf {r}(\theta ,\phi ) = \left \langle 2\cos (\theta )\sin (\phi ),2\sin (\theta )\sin (\phi ),2\cos (\phi ) \right \rangle ,$ and the bounds are $-\pi /4 \leq \theta \leq \answer {\pi /4}$ and $0 \leq \phi \leq \answer {\pi /2}$ (the bounds for $\theta$ come from looking at the $xy$ -plane). The magnitude of the cross product should be $\|\textbf {r}_{\theta } \times \textbf {r}_{\phi } \| = 4\sin (\phi ).$ The integral turns out to be $\int _{-\pi /4}^{\pi /4} \int _0^{\pi /2} 8\cos (\theta )\sin ^2(\phi ) d\phi d\theta .$

True/False

Determine whether the following statements are true/false.

If $f(x,y,z)$ is a continuous function on a surface $S$ with $f(x,y,z) > 0$ , then $\displaystyle \iint _S f(x,y,z)dS > 0$ .

True False

If $f(x,y,z)$ is a continuous function on a surface $S$ , then $\displaystyle \iint _S f(x,y,z)dS$ does not depend on how we parametrize $S$ .

True False

If $\textbf {r}(u,v)$ , where the domain of $(u,v)$ is $D$ , parametrizes a surface with surface area $5$ , then $2\textbf {r}(u,v)$ , where the domain of $(u,v)$ is still $D$ , parametrizes a surface with surface area $10$ .

True False

This is false, as you can check that the scaling factor is increased by a factor of $4$ . This should make sense because increasing both sides of a parallelogram by a factor of $2$ increases the area by a factor of $4$ .