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### Course: Multivariable calculus > Unit 4

Lesson 12: Surface integrals (articles)# Surface area integrals

How do you find the surface area of a parametric surface? This will lead to the more general idea of a surface integral.

## Background

- Partial derivatives of parametric surfaces

In particular, make sure you have a strong intuition for the partial derivatives of a function parameterizing a surface, and what they represent. - Double integrals
- Cross product (video)

## What we are building to

Setup:

is some surface in three-dimensional space.$S$ is a vector-valued function parameterizing$\overrightarrow{\mathbf{\text{v}}}({t},{s})$ .${S}$ is the region of the$T$ -plane (also known as the parameter space) that corresponds with$ts$ .$S$

The surface area of $S$ can be computed with the following double integral:

These integrals can be very labor intensive to compute.

## Surface area

From geometry, you might be familiar with the surface areas of a few specific shapes. For example, the surface area of a sphere with radius $r$ is $4\pi {r}^{2}$ .

But what if someone gives you an arbitrary surface, defined using some parametric function that maps a region of two-dimensional parameter space into three-dimensional space? How do you find its surface area?

The answer is to use a certain integral, or rather a certain

*double*integral, which you are about to learn. This is analogous to how you can find the arc length of an arbitrary curve using a certain single integral, or the volume of a strangely shaped solid using the appropriate triple integral.## Example: Breaking down surface area

Define a parametric surface with the following function:

Let's name this surface $S$ .

Of course, with parametric surfaces, it is not enough just to specify the function which parameterizes it. We also need to know the region of the parameter space that gets mapped onto the surface. "Parameter space" is a fancy word for where the point $({t},{s})$ lives, also known as the "domain". In this case, let's say it is the rectangle defined by

Let's call this rectangle $T$ . Here is what it looks like for $\overrightarrow{\mathbf{\text{v}}}$ to transform the rectangle $T$ in the parameter space into the surface $S$ in three-dimensional space.

Our strategy for computing this surface area involves three broad steps:

**Step 1**: Chop up the surface into little pieces.**Step 2**: Compute the area of each piece.**Step 3**: Add up these areas.

After studying line integrals, double integrals and triple integrals, you may recognize this idea of chopping something up and adding all its pieces as a more general pattern in how integration can be used to solve problems. As with those examples, our final computation will not

*actually*involve chopping up the surface into a specific number of pieces and adding them up; we let an integral take care of that for us.## Step 1: Chopping up the surface

To start, think of chopping up the rectangle $T$ in the parameter space into many tiny little rectangles. In the drawing, I'll only chop it into a few rectangles so that we can see and reference each one, but in principle you should think of very many, really small rectangles.

For one of these tiny rectangles, you can think its width as being ${dt}$ , a tiny change to the parameter ${t}$ . Similarly, think of its height as being ${ds}$ , a tiny change to the parameter ${s}$ .

Now consider how the function $\overrightarrow{\mathbf{\text{v}}}({t},{s})$ maps one of these tiny rectangles onto the surface $S$ . In the following animation, I'll make most of the surface a faded grey, and leave just one of the tiny rectangles colored as we watch $T$ transform into $S$ .

Strictly speaking, the rectangle will become slightly curved as it is pasted onto $S$ . However, as you consider smaller and smaller rectangles, that curvature becomes more and more negligible, and we can basically treat this tiny piece as if it was flat.

In fact, as we consider smaller and smaller rectangles in the parameter space, the portions of the surface $S$ that these rectangles map to will look more and more like parallelograms.

Our first task, then, will be to find a formula giving the area of these parallelograms.

## Step 2: Seeking the area of a parallelogram piece

For one of these tiny rectangles that we chopped $T$ into, let $({{t}_{A}},{{s}_{A}})$ represent its lower left corner, and $({{t}_{B}},{{s}_{B}})$ represent its lower right corner.

Now consider the vector pointing from $\overrightarrow{\mathbf{\text{v}}}({{t}_{A}},{{s}_{A}})$ to $\overrightarrow{\mathbf{\text{v}}}({{t}_{B}},{{s}_{B}})$ on the surface. Let's name that vector $\overrightarrow{\mathbf{\text{a}}}$ .

**Concept check**: Take the same setup as the previous problem, but let

Okay, here's where we are so far: We are thinking about a tiny rectangle in the parameter space with the following properties

- Bottom left corner:
$({{t}_{A}},{{s}_{A}})$ - Width:
${dt}$ - Height:
${ds}$

When you apply the function $\overrightarrow{\mathbf{\text{v}}}$ to this rectangle, you end up with what is basically a parallelogram on the surface $S$ . Based on the previous two questions, the sides of this parallelogram are determined by the vectors

and

## Where this gets labor intensive

Boy is this a complicated expression. It involves two partial derivatives of a vector-valued function, taking their cross product, then taking the magnitude. It's as if someone was

*trying*to create the most complicated expression they could imagine.Right now we have a purely theoretical expression for the area of one of these little parallelograms:

However, if you want to get a feel for what this actually entails, I encourage you to work through it.

**Work it out**: Given the definition of

evaluate the expression found in the previous question to get a function in terms of ${t}$ , ${s}$ , ${dt}$ and ${ds}$ .

## Step 3: Integrating everything together

**Here's where we are so far**. After breaking up the rectangle

Then, through many computations, you found an expression for the area of one of these parallelograms:

Where

describes the position of the initial little rectangle.$({t},{s})$ is its width.${dt}$ is its height.${ds}$

To add up the areas of all these little parallelograms, we take a double integral of this quantity over the region $T$ . As a reminder, $T$ was defined as the region where

Using those bounds, here is the double integral representing the surface area of ${S}$ :

Working this out by hand seems tricky, given that finding the antiderivative of $\sqrt{{{s}^{2}}+4{{t}^{2}}+4{{t}^{4}}}$ will be difficult. But using a calculator (or Wolfram Alpha), we can find the answer:

The important thing to remember here is how to construct the appropriate double integral, and to think about adding up many tiny pieces of area on the surface itself.

## Summary: This ain't easy

Generalizing everything we did in the previous example, the surface area of our parametric surface $S$ is expressed using the integral

where $S$ is described using a parametric function $\overrightarrow{\mathbf{\text{v}}}({t},{s})$ applied to a region $T$ of the ${t}{s}$ -plane.

You've already had a glimpse of this, but it's worth pointing out that this can be a really complicated thing to compute.

- First you have to take two partial derivatives of vector-valued functions, which if you count each component includes
partial derivatives in total.$6$ - You then have to take the cross product of these two partial derivative vectors, which itself requires taking a determinant whose components are vectors and functions.
- Then you have to compute the norm of that cross product.
- After all that, there is still a double integral ahead of you. And remember, just setting up a double integral isn't always easy, especially if the region you are integrating over is not rectangular.
- And all this is assuming you already know the function
and the region$\overrightarrow{\mathbf{\text{v}}}({t},{s})$ . Sometimes you are just given a surface which is implicitly defined, like a sphere defined by$T$ . In that case you need to find a function which parameterizes this surface, as well as which specific region of the parameter space corresponds to the surface.${x}^{2}+{y}^{2}+{z}^{2}=1$

The key when going through all of this is to stay organized, and be patient. One way to think about it is that setting up and computing just one of these surface area integrals is akin to doing $10$ practice problems in single-variable calculus.

The thought process that goes into all of this is actually very useful for thinking about surfaces and three-dimensional geometry in general, beyond the specific case of computing surface area. For example, how do you think computer graphics works? Quite often, displaying a three-dimensional figure involves subdividing a surface into polygons, and getting the computer to display those polygons. Even if this never involves performing a surface area integral, per se, the reasoning associated with how to do this is remarkably similar, using cross products of partial derivatives, etc.

If you want to practice this more, the next article walks through another full example. If you do choose to work through it, prepare to mark up a lot of paper.

## Want to join the conversation?

- In step 2:Why should we multiply dt with partial derivative in approximating a.

I think the partial derivative may posses the magnitude of vector a and there may not be any need to multiply the partial derivative with dt.(5 votes)- Remember that we are taking the integral, which means that we are summing infinitesimally small pieces together. In parameter space, these pieces are of size Ds and Dt, which in the limit becomes ds and dt. The vector function v maps from parameter space to the surface S in "result"-space. dv/dt gives the rise of the surface S in result space, but this is not yet a distance. Only when we multiply it with a distance in parameter space do we get a reasonably approximation in result space. This distance in parameter space is Dt, or in the limit: dt. And the distance in the "result"-space is dv/dt*Dt.

Imagine how you would do the same for 1-dimensional functions. Imagine a function v(x)=x^2. We want to approximate v(1.01)-v(1) = 1.01^2-1^2. The distance in the domain: Dx=0.01. The derivative dv/dx=2x. dv/dx(1)=2. But dv/dx(1) alone is not an approximation for v(1.01)-v(1.0), we need to multiply the derivative by the distance Dx in the domain which is 0.01. dv/dx*Dx=2*0.01 = 0.02. By the way: the real answer is 0.0201(8 votes)

- Why must we parameterize this problem? What box do we get into if instead of parameterizing with say s & t, we use x & y throughout?

A second question is more general: in general, we parameterize to express something as say a function of time, t; we parameterize to aid in the solution of a problem. What can you say generally as to why we parameterize?(4 votes)- Because we want to simplify the problem to do a simpler operation on it. Since surfaces are flat (have no thickness), surfaces in 3D space can be converted to 2D (and back) without losing information. So if we want, say, the surface area of some surface in real-life 3D like a curved sheet of paper, we can factor out the "curve" of the paper and pretend it's a flat, 2D rectangle on the xy-plane (or ts-plane). Now what's the area? simple - width times height. It's much easier to do plain old area when in 2D, and to answer your first question, sometimes it's the only way. So to the question "why we parameterize" your answer of "to aid in solution" is still correct in the general case. In the cases of functions of only time, you were reducing a 2D problem (say projectile motion, in the xy-plane) downward by one dimension (going from 2 variables to one) since lines are one-dimensional. In each case, parameterizing lets us simplify the problem.(5 votes)

- I think there was a mistake in the example when we're finding the area of the little parallelograms. When using the determinant trick to find the cross product, Grant wrote a plus sign for the j-hat term instead of negative. It didn't affect the answer though, since we squared it while finding the magnitude.(3 votes)
- In my multivariable calculus class, we were taught a method that doesn't involve paramaterization of the surface. we simply prove that:

dS=|∇g|*dA=sqrt(1+(fx)^2+(fy)^2)*dxdy

Do you cover this method on KhanAcademy?(3 votes) - In step 2: Why do we multiply the distance, which could also depend on s, with the partial derivative of t? What if that distance depends only on s? wouldn't our approximation be totally wrong?(2 votes)
- In step 2:While approximating the vector a from ( tA, sA ) to ( tB , sB) why did you consider the change of vector with respect to t only why didn't you consider the change due to s.You have considered dV/dt but you did not consider dV/ds.(1 vote)
- I know this is an old forum, but I am very confused by the explanation of why the vector a is given by the partial of v w.r.t t but not s? In the diagram it shows a as the vector connecting v(t_A, s_A) to v(t_B, s_B) but if only t is changing should it not be from v(t_A, s_A) to (t_B, s_A)? I feel like I am missing something(1 vote)
- Step 1: ...In fact, as we consider smaller and smaller rectangles in the parameter space, the portions of the surface S that these rectangles map to will look more and more like parallelograms.

Instructor, what does this mean? How do they end up transforming into a parallelogram?

Are you saying that a tiny rectangle in T is stretched into a parallelogram in S in the conversion from T to S?(1 vote)