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### Course: Multivariable calculus>Unit 4

Lesson 12: Surface integrals (articles)

# Surface area example

Here you have the chance to practice computing surface area, using the example of a torus.

## Who is this for?

This article is meant for anyone who read the last article on computing the surface area of parametric surfaces using a certain double integral, and who wants to practice this concept. You will compute the surface area of a torus (a doughnut shape) using this method, which requires no small amount of computation.
If you neither want or need practice with this computation, and you feel comfortable with the general concept of how these surface area integrals work, feel free to skip ahead to the next article.

## Quick recap of the surface area integral

Before diving into the example, let's quickly remind ourselves of the method described for finding the area of a surface discussed in the last article.
• Parameterize the surface. In other words, find a vector-valued function $\stackrel{\to }{\mathbf{\text{v}}}\left(t,s\right)$ which maps some region $T$ of the two-dimensional $ts$-plane onto your surface in three dimensions. Sometimes this parameterization will be given to you, if that's how your surface is defined. Other times the surface is defined some other way, and you have to find it yourself.
• Imagine chopping up the parameter space with horizontal and vertical lines, thus dividing your region $T$ into little rectangles. Each of these rectangles gets mapped onto a little piece of your surface which is well-approximated by a parallelogram. If your little rectangle sits at the point $\left(t,s\right)$, and it has width $dt$ and height $ds$, you can approximate its area with the following expression:
$\begin{array}{r}|\frac{\partial \stackrel{\to }{\mathbf{\text{v}}}}{\partial t}×\frac{\partial \stackrel{\to }{\mathbf{\text{v}}}}{\partial s}|\phantom{\rule{0.167em}{0ex}}dt\phantom{\rule{0.167em}{0ex}}ds\end{array}$
The smaller your initial rectangles, the more closely the corresponding piece of your surface resembles an actual flat parallelogram, and the closer this expression is to giving the true area of that piece.
• Add up the areas of these pieces with a double integral:
$\begin{array}{r}{\iint }_{T}|\frac{\partial \stackrel{\to }{\mathbf{\text{v}}}}{\partial t}×\frac{\partial \stackrel{\to }{\mathbf{\text{v}}}}{\partial s}|\phantom{\rule{0.167em}{0ex}}dt\phantom{\rule{0.167em}{0ex}}ds\end{array}$

## Surface area of a torus

English makes it hard to describe the dimensions of this torus, but I'll give it a shot with the help of some doughnut terminology. Imagine this torus as the glaze on a doughnut.
• Let's say the distance between the origin and the innermost part of this jelly filling is $3$. Call this the "outer radius".
• Let's also say the distance between the innermost part of the jelly filling and the glaze itself is $1$. Call this the "inner radius"
With these dimensions, the torus (i.e. glaze) can be parameterized with the following function:
$\begin{array}{r}\stackrel{\to }{\mathbf{\text{v}}}\left(t,s\right)=\left[\begin{array}{c}3\mathrm{cos}\left(t\right)+\mathrm{cos}\left(s\right)\mathrm{cos}\left(t\right)\\ \\ 3\mathrm{sin}\left(t\right)+\mathrm{cos}\left(s\right)\mathrm{sin}\left(t\right)\\ \\ \mathrm{sin}\left(s\right)\end{array}\right]\end{array}$
For this parameterization to cover the torus once and only once, apply it to the region of the $ts$-plane where
$\begin{array}{rl}0\le & t\le 2\pi \\ 0\le & s\le 2\pi \end{array}$
For a description of where this parameterization comes from, check out the last example in this article.

## Step 1: Compute each partial derivative

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

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

Remember, you should think of these vectors as representing the edges of little parallelograms, which piece together to make the torus as a whole. More accurately, you must multiply the first one by $dt$ and the second one by $ds$ to scale them down to the infinitesimal size of one of these parallelograms.
As it so happens, these vectors are perpendicular to each other (you can check by taking their dot product). This implies all the little parallelograms making up the torus happen to be rectangles, at least when we use this particular parameterization. You can see this in the picture of the torus above.

## Step 2: Compute the cross product

To find the area of a parallelogram spanned by the two vectors you just found, the first step is to take their cross product. (Warning: This one gets hairy)
$\frac{\partial \stackrel{\to }{\mathbf{\text{v}}}}{\partial t}\left(t,s\right)×\frac{\partial \stackrel{\to }{\mathbf{\text{v}}}}{\partial s}\left(t,s\right)=$
$\stackrel{^}{\mathbf{\text{i}}}+$
$\stackrel{^}{\mathbf{\text{j}}}+$
$\stackrel{^}{\mathbf{\text{k}}}$

## Step 3: Find the magnitude of this cross product

The cross product you just computed is a vector. In order to find the area of a paralellogram spanned by the two partial derivative vectors, we must find its magnitude. (Warning: This one gets even more hairy).
$\begin{array}{r}|\frac{\partial \stackrel{\to }{\mathbf{\text{v}}}}{\partial t}\left(t,s\right)×\frac{\partial \stackrel{\to }{\mathbf{\text{v}}}}{\partial s}\left(t,s\right)|=\end{array}$

Once you scale this down by $dt\phantom{\rule{0.167em}{0ex}}ds$, this tells you the area of each of the little parallelograms making up the torus, as a function of $s$ and $t$. Well, in this case, the answer is just a function of $s$, which means the area of these parallelograms doesn't change as you let $t$ vary.

## Step 4: Set up the appropriate double integral

Which of the following represents the right bounds to place on the double integral representing surface area for this torus?

## Step 5: Compute the double integral

Surface area of this torus:

## Congratulations

These integrals are a lot of work, so pat yourself on the back for working all the way through this!

## Want to join the conversation?

• Should the sign of the j component in step two not be positive?
• I also came to the same conclusion and yes, I did include the sign of the checker board pattern.
• shouldn't the outer radius be 4 and inner radius 2 for this specific torus?
• It can get confusing when we are talking about "outer radius" and "inner radius." In this case, the "inner radius" of 3 refers to the radius from the origin to the middle of the ring. (Sort of like half-way between the inner and outer radius you are thinking about). The "inner radius" of 1 refers to the radius of a cross-section of the ring.

Another way of thinking about it is that the torus is created by taking a circle of radius 1 and rotating that circle 360 degrees (2pi radians) around an axis. When performing this rotation, the distance from the axis to the center of the circle is 3.
• What if we compute integrals in an arbitrary dimension?
Can we define the cross product in R^4, R^5, and so on, too?
• unfortunately not, we can only talk about a cross-product in R^3
• For step 2, it asks us to calculate the cross product. I performed the calculations and got the following as my answer: (3+cos(s))*cos(s)*cos(t) î + (3+cos(s))*sin(t)*cos(s)*j + sin(s)(3+cos(s))*k. When I plugged it in to check my answer, the system tells me that my answer is incorrect, so I recalculated it and still got the same answer. Then, I thought I was doing something wrong and went ahead to check the answer key. However, I look at the answer key and the system has the exact same answer as me. Could someone tell me if I'm doing something wrong, my computer is broken or they have the same issue too? Thanks in advance for any help!
(1 vote)
• You are probably not doing anything wrong. I had this same problem. I checked several times that my answers were typed with exactly the same characters in the hint, but it still marked my answer as incorrect.
(1 vote)
• Dear Instructor, is there a way by which we can determine that it will be rectangles and not other parallelogram?

আমরা কিভাবে বুঝতে পারি যে সারফেস এরিয়ার ক্ষুদ্রতম অংশগুলি আয়ত হবে?? এটা বোঝার কি কোন উপায় আছে?
(1 vote)
• yeah, there is if you take the dot product and if it's equal to 0 then it is a rectangle.
(1 vote)
• Jelly filled doughnuts are not torus shaped. The resemble short cylinders with rounded corners. Did you mean a glazed yeast-raised ring doughnut?
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