# Inspiration for Calculus Students: The Things We Take for Granted

Updated: Jan 12, 2020

*“Alright class, let’s take a look at a really interesting example today. This question is much too hard to be on the test, but I think it’s really fascinating, and may transform the way you think about our classwork”.*

Chances are, you’ve heard a professor say something like this before. What did you do? I think most people would tune out their professor for the next thirty minutes. Hopefully I’m wrong about that.

But if you’re interested, let’s take a trip down to thought-experiment-lane, where the examples truly are interesting, the questions are truly too hard to be on the test, and maybe—just maybe—they could legitimately transform the way you think about your classwork. Here is today’s question:

“In the third grade, you ‘learned’ that the area of a circle is (pi r^{2}). No one really explained *why* this is the case, but you memorized it anyways, and have since used the formula to solve countless math problems. However, unlike you, mathematicians and philosophers who lived thousands of years ago were fascinated by circles and their properties—partly because they had no idea what the area of a circle was. **Using integrals and cartesian coordinates, can you prove that the area of a circle is (pi r^{2})?”**

Hundreds of years from now, when I become a math professor for integral calculus, this question will be the extra credit question on my final exam. A formula so simple and universally-memorized, that we all take it for granted. Can you prove it? Let’s get started.

**Step 1: The Setup**

At the risk of blowing your mind, I think the first step in solving this problem is drawing a circle with radius (r). So, let’s do that right away:

Already on the first step, I have performed a sneaky trick: in order to find the area of the whole circle, I will only calculate the area of one quarter of the circle. Multiplying that area by 4 at the end of my calculations will give me the final answer. Technically, you don’t have to use this trick, but I like it because it keeps all my numbers positive, since we are only dealing with quadrant 1.

The problem statement asks me to use integrals and cartesian coordinates, so I already know what direction I’m heading in. Let’s see if we can set up an integral using our quarter-circle.

I often describe an integral as “a sum of an infinite number of tiny things”. In our case, we will use a sum of an infinite number of rectangles with length (y) and width (dx). Thus, our integral is written as (int y d x). We know that the minimum (x) value in our quarter-circle is zero, and the maximum (x) alue is equal to the radius (r). So now we can input our limits of integration:

$$int_{0}^{r} y d x$$

Would you believe me if I told you we were almost done? There isn’t much left to do in the setup, we just have to write (y) in terms of (x). Here we call on the formula for a circle: (x^{2}+y^{2}=r^{2}), and solve the equation for (y):

$$y=sqrt{r^{2}-x^{2}}$$

Now we have our completed integral ready to solve:

$$int_{0}^{r} sqrt{r^{2}-x^{2}} d x$$

And let’s not forget that we need to multiply this integral by 4 at the end.

**Step 2: The Calculations (so literally everything else)**

If you made it this far, you’d get a good amount of partial credit in my class. Any computer can tell you the answer to that integral above. But we’re not going to settle for anything less than a hand-written solution! For all we know, the establishment is tricking us into believing that the area of a circle is (pi r^{2})—and all the computers are in on the ruse! Time to solve this thing.

### 2.1: The Trig Substitution

$$int_{0}^{r} sqrt{r^{2}-x^{2}} d x$$

You might immediately feel stuck when looking at this integral, especially since everything is under a radical. But fortunately, there’s a strategy specifically designed to remove radicals from integrals: **Trig Substitutions**. A trig substitution is kind of like a glorified u-substitution. One of the fundamental ideas governing this type of substitution is the trig identity: (sin ^{2}(x)+cos ^{2}(x)=1). If we can get the inside of our radical to look like (1-sin ^{2}(x)), then we can set it equal to (cos ^{2}(x)), and take the square root of (cos ^{2}(x)), which is simply (cos(x)).

Thus, the trig substitution we make here is: (x=r sin (theta)); (d x=r cos (theta) d theta).

$$int sqrt{r^{2}-r^{2} sin ^{2}(theta)} * r cos (theta) d theta$$

Sometimes things get messy before they get cleaner. The good news is we can get rid of the radical. Also, notice that I didn’t bother messing with the limits of integration here. Once I substitute (x) back everywhere, then I’ll reintroduce the limits of integration to the problem.

$$begin{array}{c}{int sqrt{r^{2} *left(1-sin ^{2}(theta)right)} * r cos (theta) d theta} \ {int sqrt{r^{2} cos ^{2}(theta) *} r cos (theta) d theta} \ {int r cos (theta) * rcos(theta) d theta} \ {int r^{2} cos ^{2}(theta) d theta}end{array}$$

As it turns out, we managed to greatly simplify our integral with that trig substitution. Unfortunately, it looks like we might be stuck. Integrating (cos ^{2}(theta)) is no easy feat. Got any ideas?

### 2.2: The Double Angle Formulas

A quick review of your notes might reveal that (cos ^{2}(theta)) can be rewritten as (frac{1+cos (2 theta)}{2}), thanks to a useful trig identity. Let’s rewrite our integral with this in mind, and pull some constants out front:

$$int r^{2} cos ^{2}(theta) d theta=frac{r^{2}}{2} int 1+cos (2 theta) d theta$$

What are you waiting for? We can integrate this thing now!

$$frac{r^{2}}{2} int d theta+frac{r^{2}}{2} int cos (2 theta) d theta$$

$$frac{r^{2}}{2}left(theta+frac{1}{2} sin (2 theta)right)$$

This (sin (2 theta)) term looks fine now, but it might confuse us later if we don’t take care of it now. So, I’ll do us both a favor and remind you of another trig identity that you *definitely* remember: (sin (2 theta)=2 sin (theta) cos (theta)). Don’t worry, I had to check my notes for that one too.

$$frac{r^{2}}{2}(theta+sin (theta) cos (theta))$$

### 2.3: The Road Back to x

If you’ve ever done a trig substitution before, you’ll know that converting back from (theta) to (x) requires high amounts of attention and brainpower. Let’s recall our earlier equation that we used to get (theta) in the first place, and solve it for (theta).

$$x=r sin (theta)$$

$$theta=sin ^{-1}left(frac{x}{r}right)$$

You already know the next step: let’s plug our expression for (theta) into the equation above:

$$frac{r^{2}}{2}(theta+sin (theta) cos (theta))$$

$$frac{r^{2}}{2}left[sin ^{-1}left(frac{x}{r}right)+sin left(sin ^{-1}left(frac{x}{r}right)right) cos left(sin ^{-1}left(frac{x}{r}right)right)right] left[ begin{array}{c}{x=r} \ {x=0}end{array}right.$$

Things just got very messy. There are a couple pieces of good news though; our limits of integration have returned now that everything is in terms of (x), and some of the trig functions above are solvable right now. For example, we know how to solve (left[sin ^{-1}left(frac{x}{r}right)right]), so we don’t need to worry about that term. Also, we can simplify the term (left[sin left(sin ^{-1}left(frac{x}{r}right)right)right]) to just read (left[frac{x}{r}right]).

$$frac{r^{2}}{2}left[sin ^{-1}left(frac{x}{r}right)+frac{x}{r} cos left(sin ^{-1}left(frac{x}{r}right)right)right] left[ begin{array}{l}{x=r} \ {x=0}end{array}right.$$

The only thing left to worry about is this final term: (cos left(sin ^{-1}left(frac{x}{r}right)right)).

How do we deal with that?

Now we’re about to see why trig substitutions are more nuanced than u-substitutions. You might have wondered why we used the variable (theta) earlier instead of using (u). The reason we used (theta) here is because we can graphically represent our trig substitution as a right triangle. Here’s what I mean:

I’ll give you a moment to convince yourself that the above image makes sense.

According to the Pythagorean theorem, the third side of that triangle must be equal to (sqrt{r^{2}-x^{2}}). And that means we know one more thing about this triangle:

$$cos (theta)=frac{sqrt{r^{2}-x^{2}}}{r}$$

Let’s plug that information back into our main equation:

$$frac{r^{2}}{2}left[sin ^{-1}left(frac{x}{r}right)+frac{x}{r} * frac{sqrt{r^{2}-x^{2}}}{r}right] left[ begin{array}{l}{x=r} \ {x=0}end{array}right]$$

### 2.4: The Final Answer

Remember that we don’t have to solve this equation for every value of (x)—we only have to evaluate it at the two limits of integration: at (x=r) and (x =0 ). First, I’ll evaluate the equation at (x = 0).

$$frac{r^{2}}{2}left[sin ^{-1}left(frac{0}{r}right)+frac{0}{r} * frac{sqrt{r^{2}-0^{2}}}{r}right]=frac{r^{2}}{2}left[sin ^{-1}(0)+(0 * 1)right]=0$$

Well that was anticlimactic. Let’s try evaluating at (x=r)

$$frac{r^{2}}{2}left[sin ^{-1}left(frac{r}{r}right)+frac{r}{r} * frac{sqrt{r^{2}-r^{2}}}{r}right]=frac{r^{2}}{2}left[sin ^{-1}(1)+(1 * 0)right]=frac{r^{2}}{2} * frac{pi}{2}$$

You don’t need to memorize the fact that (sin ^{-1}(1)=frac{pi}{2}). If you take a long look at the triangle we drew above, you might be able to convince yourself of its truth.

Anyway, we found the solution to our original integral:

$$int_{0}^{r} sqrt{r^{2}-x^{2}} d x=frac{r^{2}}{2} * frac{pi}{2}=frac{pi r^{2}}{4}$$

At the beginning of all this, we agreed that to find the area of the entire circle, we would multiply our integral solution by 4. And thus, we get (A=pi r^{2}).

**Why Did We Do all of This Again?**

The point of this exercise was not to fill a bunch of pages with math calculations. The real point is to send this message: __Mathematics is a results-oriented discipline. Math techniques (like integrals) are always invented to solve problems__, not just to bore students. And if we have an appreciation for *why* certain math techniques exist—*what problems they were invented to solve*—we can better understand when to make use of them.

The first proof of the area of a circle was made by Archimedes over two thousand years ago. But his method sucked. It involved a bunch of triangles, and his explanations were probably longer than the calculations we just did. Archimedes could only have dreamed of the amazing tools—like integral calculus—that we use today. So be glad. Enjoy the fact that we spent 30 minutes making a better proof than Archimedes did using his entire life. Don’t take this stuff for granted. And appreciate that folks like Archimedes worked really hard on this stuff so that we don’t have to.