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### 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.
$\begin{array}{r}{\int }_{0}^{\pi }{\int }_{0}^{2\pi }\left(-16\mathrm{cos}\left(t\right){\mathrm{sin}}^{2}\left(s\right)+20\mathrm{sin}\left(s\right)\right)\phantom{\rule{0.167em}{0ex}}dt\phantom{\rule{0.167em}{0ex}}ds=\end{array}$

## Want to join the conversation?

• I could only understand the parameterization of ( x = r cos 𝜽 ; y = r sin 𝜽 ; z = 0 ) in the xy-plane from (0 to 2 π ) around z-axis.
Could you please explain the parameterization of the varying radius from (0 to π). How do you get ( x y z ) as ( r cos sin; r sin sin; r cos) or (r cos cos; r sin sin; r sin) ?
• can someone explain why we have taken s only from 0 to pie and t 0 to 2pie
• s tells us how far up we need to go in the z-direction, and once you know your z-coordinate you trace out a circle parallel to the xy-plane on top of the sphreres surface. Since you can reach every point between -2 and 2 by letting s range from 0 to pi (remember, your z-component is 2cos(s) and cos reaches every value between -1 and 1 if you use every s between 0 and pi) any other value of s would only let you trace out a circle you already traced out

For visuals on what I just said open the answer for the first concept check in step 2
• Can you please help me solving this surface area integral
Find the surface area of the portion of the sphere of radius 4,centre at origin that lies inside the cylinder x^2+y^2=12 and above the xy-plane.
• Here in this Sal parameterized the sphere as:
[ cos(t)cos(s) , cos(t)sin(s) , sin(t) ]
Why is it so different? Does it give the same result?
(1 vote)
• Yes, both approaches will yield that same final result, assuming the dimensions of the sphere are consistent and the functions that are being integrated over, are identical. It all hinges on how you define your s and t angles and their corresponding datums. This in turn affects the parametrization of the system.

Sal's method defines his base circle in the x-y plane in terms of angle 0 < s < 2pi rotating about the z-axis. He then defines the varying radius of that circle in terms of the angle -pi/2 < t < pi/2 rotating about an arbitrary axis in the x-y plane. Sal seems to have set the x-y plane as his datum for angle t, such that t=0 in this plane.

The above method seems to define the base circle in the x-y plane as well, but in terms of angle 0 < t < 2pi rotating about the z-axis. The varying radius of that circle is then defined in terms of the angel 0 < s < pi rotating about an arbitrary axis in the x-y plane. This method however has defined the positive z-axis as a datum, such that s = 0 in the positive vertical orientation.

This is my understanding of the methods in these two solutions. I hope my explanation is clear enough for the distinction to be apparent to you. Remember that each mothod will have slightly different integral boundaries due to the choice of datum. Perhaps an experienced mathematician can do a better job of verballising this xD