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## Multivariable calculus

### Course: Multivariable calculus>Unit 4

Lesson 12: Surface integrals (articles)

# Surface integral example

Practice computing a surface integral over a sphere.

## The task at hand: Surface integral on a sphere.

In the last article, I talked about what surface integrals do and how you can interpret them. Here, you can walk through the full details of an example. If you prefer videos you can also watch Sal go through a different example.
Consider the sphere of radius $2$, centered at the origin.
Your task will be to integrate the following function over the surface of this sphere:
$f\left(x,y,z\right)=\left(x-1{\right)}^{2}+{y}^{2}+{z}^{2}$

#### Step 1: Take advantage of the sphere's symmetry

The sphere with radius $2$ is, by definition, all points in three-dimensional space satisfying the following property:
${x}^{2}+{y}^{2}+{z}^{2}={2}^{2}$
This expression is very similar to the function:
$f\left(x,y,z\right)=\left(x-1{\right)}^{2}+{y}^{2}+{z}^{2}$
In fact, we can use this to our advantage...
Concept check: When you evaluate $f\left(x,y,z\right)=\left(x-1{\right)}^{2}+{y}^{2}+{z}^{2}$ on points that happen to be on the sphere with radius $2$, what simpler expression do you get?
Keep in mind, $f\left(x,y,z\right)$ does not equal this simpler expression everywhere, but only on the points where ${x}^{2}+{y}^{2}+{z}^{2}=4$. Since we will only integrate over points on this sphere, though, we can justifiably replace the function $f$ in the integral with this value.
$\begin{array}{r}{\iint }_{\text{Sphere}}\left(\left(x-1{\right)}^{2}+{y}^{2}+{z}^{2}\right)\phantom{\rule{0.167em}{0ex}}d\mathrm{\Sigma }={\iint }_{\text{Sphere}}\left(-2x+5\right)\phantom{\rule{0.167em}{0ex}}d\mathrm{\Sigma }\end{array}$
Of course, this is not something you can do for every surface integral, but it's a good lesson to take advantage of symmetry when you can to make these integrals easier.

#### Step 2: Parameterize the sphere

To relate this surface integral to a double integral on a flat plane, we need to first find a function which parameterizes the sphere.
Concept check: Which of the following functions parameterizes the sphere with radius $2$?

Great! Now we have a formula for the parameterization $\stackrel{\to }{\mathbf{\text{v}}}\left(t,s\right)$ of the sphere, along with a corresponding region on the $ts$-plane. We can start expanding out surface integral like this:
$\begin{array}{rl}& \phantom{\rule{1em}{0ex}}{\iint }_{\text{Sphere}}\left(-2x+5\right)\phantom{\rule{0.167em}{0ex}}d\mathrm{\Sigma }\\ \\ & ={\int }_{0}^{\pi }{\int }_{0}^{2\pi }\left(-2\underset{x\text{-value of parameterization}}{\underset{⏟}{\left(2\mathrm{cos}\left(t\right)\mathrm{sin}\left(s\right)\right)}}+5\right)\phantom{\rule{0.167em}{0ex}}\underset{\text{We need work this out}}{\underset{⏟}{|\frac{\partial \stackrel{\to }{\mathbf{\text{v}}}}{\partial t}×\frac{\partial \stackrel{\to }{\mathbf{\text{v}}}}{\partial s}|}}\phantom{\rule{-0.167em}{0ex}}\phantom{\rule{-0.167em}{0ex}}\phantom{\rule{-0.167em}{0ex}}\phantom{\rule{-0.167em}{0ex}}\phantom{\rule{-0.167em}{0ex}}\phantom{\rule{-0.167em}{0ex}}\phantom{\rule{0.167em}{0ex}}dt\phantom{\rule{0.167em}{0ex}}ds\end{array}$

#### Step 3: Compute both partial derivatives

The main beast to wrangle with in any surface integral is this little guy:
$\begin{array}{r}|\frac{\partial \stackrel{\to }{\mathbf{\text{v}}}}{\partial t}×\frac{\partial \stackrel{\to }{\mathbf{\text{v}}}}{\partial s}|\end{array}$
Concept check: To start, compute both partial derivatives of our parametric function:
$\begin{array}{r}\stackrel{\to }{\mathbf{\text{v}}}\left(t,s\right)=\left[\begin{array}{c}2\mathrm{cos}\left(t\right)\mathrm{sin}\left(s\right)\\ 2\mathrm{sin}\left(t\right)\mathrm{sin}\left(s\right)\\ 2\mathrm{cos}\left(s\right)\end{array}\right]\end{array}$
$\frac{\partial \stackrel{\to }{\mathbf{\text{v}}}}{\partial t}\left(t,s\right)=$
$\stackrel{^}{\mathbf{\text{i}}}+$
$\stackrel{^}{\mathbf{\text{j}}}+$
$\stackrel{^}{\mathbf{\text{k}}}$

$\frac{\partial \stackrel{\to }{\mathbf{\text{v}}}}{\partial s}\left(t,s\right)=$
$\stackrel{^}{\mathbf{\text{i}}}+$
$\stackrel{^}{\mathbf{\text{j}}}+$
$\stackrel{^}{\mathbf{\text{k}}}$

#### Step 4: Compute the cross product

Compute the cross product of the two partial derivative vectors that you just found.
$\frac{\partial \stackrel{\to }{\mathbf{\text{v}}}}{\partial t}×\frac{\partial \stackrel{\to }{\mathbf{\text{v}}}}{\partial s}=$
$\stackrel{^}{\mathbf{\text{i}}}+$
$\stackrel{^}{\mathbf{\text{j}}}+$
$\stackrel{^}{\mathbf{\text{k}}}$

#### Step 5: Find the magnitude of the cross product.

Find the magnitude of the cross product that you just found.
$|\frac{\partial \stackrel{\to }{\mathbf{\text{v}}}}{\partial t}×\frac{\partial \stackrel{\to }{\mathbf{\text{v}}}}{\partial s}|=$

Notice, technically the answer should have an absolute value sign in it. However, because our parameterization only applies to the region where $0\le s\le \pi$, the value of $\mathrm{sin}\left(s\right)$ will always be positive anyway, so we are free to leave that out.

#### Step 6: Compute the integral

Taking everything we've done so far, here's what the surface integral has turned into:
$\begin{array}{rl}& \phantom{\rule{1em}{0ex}}{\iint }_{\text{Sphere}}f\left(x,y,z\right)\phantom{\rule{0.167em}{0ex}}d\mathrm{\Sigma }\\ \\ & ={\iint }_{\text{Sphere}}\left(-2x+5\right)\phantom{\rule{0.167em}{0ex}}d\mathrm{\Sigma }\phantom{\rule{1em}{0ex}}←\text{Step 1}\\ \\ & ={\int }_{0}^{\pi }{\int }_{0}^{2\pi }\left(-2\left(2\mathrm{cos}\left(t\right)\mathrm{sin}\left(s\right)\right)+5\right)\phantom{\rule{0.167em}{0ex}}|\frac{\partial \stackrel{\to }{\mathbf{\text{v}}}}{\partial t}×\frac{\partial \stackrel{\to }{\mathbf{\text{v}}}}{\partial s}|\phantom{\rule{0.167em}{0ex}}dt\phantom{\rule{0.167em}{0ex}}ds\phantom{\rule{1em}{0ex}}←\text{Step 2}\\ \\ & ={\int }_{0}^{\pi }{\int }_{0}^{2\pi }\left(-2\left(2\mathrm{cos}\left(t\right)\mathrm{sin}\left(s\right)\right)+5\right)\phantom{\rule{0.167em}{0ex}}\left(4\mathrm{sin}\left(s\right)\right)\phantom{\rule{0.167em}{0ex}}dt\phantom{\rule{0.167em}{0ex}}ds\phantom{\rule{1em}{0ex}}←\text{Steps 3, 4, 5}\\ \\ & ={\int }_{0}^{\pi }{\int }_{0}^{2\pi }\left(-16\mathrm{cos}\left(t\right){\mathrm{sin}}^{2}\left(s\right)+20\mathrm{sin}\left(s\right)\right)\phantom{\rule{0.167em}{0ex}}dt\phantom{\rule{0.167em}{0ex}}ds\end{array}$
As a the final step, compute this double integral.
$\begin{array}{r}{\int }_{0}^{\pi }{\int }_{0}^{2\pi }\left(-16\mathrm{cos}\left(t\right){\mathrm{sin}}^{2}\left(s\right)+20\mathrm{sin}\left(s\right)\right)\phantom{\rule{0.167em}{0ex}}dt\phantom{\rule{0.167em}{0ex}}ds=\end{array}$