So I when I woke up today, I decided to learn U-sub and trig-sub integration. About 30 mins after I started, I set myself to proving the area of a circle by integration. Here's what I got.

WLOG, assume the circle's center is located at \((0,0)\)

Thus the equation of the circle is \(y=\sqrt{r^2-x^2}\)

To find the area, we will take the indefinite integral for now.

\(\displaystyle \int \sqrt{r^2-x^2} \mathbb{d}x\)

From here we substitute \(x=r\sin(u)\). Now, we take the derivative of each side and we get

\(\dfrac{dx}{du}=r\cos(u)\)

\(dx=r\cos(u)du\)

Substituting for x and dx we have

\(\displaystyle \int \sqrt{r^2-x^2}~ \mathbb{d}x=\int \sqrt{r^2-r^2\sin^2(u)} (r\cos (u)) ~du\)

Focusing on the square root, we will use the identity \(1-\sin^2(a)=\cos^2(a)\)

\(\sqrt{r^2(1-\sin^2(u))}=r\cos(u)\)

Now we have

\(\displaystyle \int r^2\cos^2(u)~\mathbb{d}u\)

\(r^2\displaystyle \int \cos^2(u)~\mathbb{d}u\)

Now, to integrate \(\cos^2(u)\) we must use \(\cos^2(a)=\frac{\cos{2a}}{2}+\frac{1}{2}\) aka the half angle identity (Spent 15 mins trying u-sub on this step, then went to wolfram, glanced at first step to integrate \(\cos^2(u)\) and slapped my self on the forehead). Now focusing on the cos part of the eq above

\(r^2\displaystyle \int \frac{\cos{2u}}{2}+\frac{1}{2} ~du\)

Splitting terms

\(r^2\left(\displaystyle \int \frac{\cos{2u}}{2} ~du+\int \frac{1}{2} du\right)\)

\(r^2\left(\displaystyle \int \frac{\cos{2u}}{2}~du+\frac{u}{2}\right)\)

Doing a w-sub for u

\(w=2u\)

\(\frac{dw}{dx}=2\)

\(dw=2du\)

Substituting back

\(\displaystyle \int \frac{\cos(w)}{4}~dw+\frac{u}{2}\)

\(\frac{\sin(w)}{4}+\frac{u}{2}\)

Re-substituting for w

\(\frac{\sin(2u)}{4}+\frac{u}{2}\)

Using double angle formula \(\sin(2a)=2\sin(a)\cos(a)\)

\(\frac{\sin(2u)}{4}+\frac{u}{2}=\frac{\sin(u)\cos(u)}{2}+\frac{u}{2}\)

Going back to our original eq

\(r^2\displaystyle \int \cos^2(u)~\mathbb{d}u=r^2\left(\frac{\sin(u)\cos(u)}{2}+\frac{u}{2}\right)\)

Now comes the slightly hard part, we must resubstitute (jeez, I've said "Resubstitute/substitute back" like 10 times already) \(x=r\sin(u)\). Looking at the sin/cos half of our eq we have

\((r\sin(u))(r\cos(u))\)

Using \(\cos(a)=\sqrt{1-sin^2(a)}\)

\(\frac{1}{2}(r\sin(u))(r\sqrt{1-sin^2(u)})=\frac{1}{2}(r\sin(u))(\sqrt{r^2-r^2sin^2(u)})\)

Substituting

\(\frac{1}{2}(x)(\sqrt{r^2-x^2})\)

We now have

\(r^2\left(\frac{\sin(u)\cos(u)}{2}+\frac{u}{2}\right)=\frac{1}{2}(x)(\sqrt{r^2-x^2})+\frac{r^2u}{2}\)

We can rearrange \(x=r\sin(u)\) to \(u=\sin^{-1}\left(\frac{x}{r}\right)\). Substituting (said it again)

\(\frac{1}{2}x\sqrt{r^2-x^2}+\frac{r^2u}{2}=\frac{1}{2}x\sqrt{r^2-x^2}+\frac{r^2}{2}\sin^{-1}\left(\frac{x}{r}\right)\)

You can skip down to the part that says "skip to here" if you don't want to read the next part, I just give an explanation of why wolfram's equation is slightly different than this one.

Now we are done. If you plug it into wolfram, they change the above equation slightly to \(\frac{1}{2}x\sqrt{r^2-x^2}+\frac{r^2}{2}\tan^{-1}\left(\dfrac{x}{\sqrt{r^2-x^2}}\right)\)

Let me explain, if we have an angle u, and \(\sin(u)=\frac{x}{r}\) then r is the hypotenuse and x is the leg opposite of angle u. By Pythagorean theorem, the other leg is \(\sqrt{r^2-x^2}\). This is the adjacent to angle u.

\(\therefore \sin^{-1}\left(\dfrac{opposite}{hypotenuse}\right)=\tan^{-1}\left(\dfrac{opposite}{adjacent}\right)=\tan^{-1}\left(\dfrac{opposite}{\sqrt{hypotenuse^2-opposite^2}}\right)=u\)

\(\therefore \sin^{-1}\left(\dfrac{x}{r}\right)=\tan^{-1}\left(\dfrac{x}{y}\right)=\tan^{-1}\left(\dfrac{x}{\sqrt{r^2-x^2}}\right)=u\)

*skip here*

Now, to find the area, we need the definite integral from \(x=-r\) to \(x=r\). Thus we have

\(\displaystyle \int_{-r}^{r} \sqrt{r^2-x^2} \mathbb{d}x=\frac{1}{2}x\sqrt{r^2-x^2}+\frac{r^2}{2}\sin^{-1}\left(\frac{x}{r}\right)\)

Equating

\(\left(\frac{1}{2}(r)\sqrt{r^2-(r)^2}+\frac{r^2}{2}\sin^{-1}\left(\frac{(r)}{r}\right)\right)-\left(\frac{1}{2}(-r)\sqrt{r^2-(-r)^2}+\frac{r^2}{2}\sin^{-1}\left(\frac{(-r)}{r}\right)\right)\)

Simplifying.

\(\frac{r^2}{2}\sin^{-1}(1)-\frac{r^2}{2}\sin^{-1}(-1)=\frac{r^2}{2}\cdot\frac{\pi}{2}-\frac{r^2}{2}\cdot(-\frac{\pi}{2})\)

\(\Rightarrow \frac{r^2\pi}{2}\)

Now, you may ask, "well isn't the formula \(r^2\pi\) what's up with the two in the denominator?"

Well, the graph of this equation is a semi-circle, not a full circle. So when we multiply our equation by two since we have the area of half the circle, we finally get our desired result of \(\pi r^2\)

Fin.

## Comments

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TopNewestSign #23 you're a math nerd: you wake up one day and decide to teach yourself U-sub and trig-sub integration. :) I like how, after 30 minutes, you decided the best way to learn was to go about solving a familiar but beautiful problem and deal with each challenge as it came along. You've covered a lot of ground in a short period of time; congrats. :)

And now you know why God invented double integrals and polar coordinates: the area of a radius \(R\) circle is

\(\displaystyle\int_{0}^{2\pi} \int_{0}^{R} r dr d\theta = \int_{0}^{2\pi} \dfrac{R^{2}}{2} d\theta = 2\pi * \dfrac{R^{2}}{2} = \pi R^{2}\).

P.S.. I just had a look at your "Into the Woods" question, and there are some nuances here that are open to interpretation. The first to consider is the use of the word "into" in this context; does this mean "through" or "to a point where you are further into the forest than you are from getting out of it". A second issue to consider is the use of the word "can" in this context; does it mean "the maximum possible distance" or "the distance along the random direction you've chosen". Which brings us to a third issue, namely, is the direction of travel chosen randomly, or is it some preordained vector that guides you to the center of the arboreal labyrinth? Could I possibly be guilty of overanalyzing this? :) – Brian Charlesworth · 2 years, 1 month ago

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Sorry about the ambiguity of the problem, but I changed the wording a short while ago to say "in a straight line". That should fix your second problem.

For the first issue, I'm not too sure how to fix it. The word "into" is meant to be the main part of the riddle. Where there is a difference between into and out of. – Trevor Arashiro · 2 years, 1 month ago

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– Brian Charlesworth · 2 years, 1 month ago

Thanks, I survived my journey into the woods and found everlasting peace and serenity. :)Log in to reply

– Daniel Liu · 2 years, 1 month ago

How does that double integral work?Log in to reply

here. Depending on the symmetries of the problem you're dealing with, one coordinate system will be preferable to another. (There are cylindrical and spherical coordinate systems at your disposal when dealing with triple integrals.) – Brian Charlesworth · 2 years, 1 month ago

For this specific integral, you first evaluate the "interior" integral, i.e., the \(dr\) one, holding \(\theta\) constant, and then once that is done you evaluate the "outer" integral, i.e., the \(d\theta\) one to get the final result. In rectangular coordinates the area increment is \(dA = dx dy\), and in polar coordinates the area increment is \(dA = r dr d\theta\). A good summary of the theory behind this is givenLog in to reply

Here's something neat. It's not a complete proof, just something to give you a little intuition as to why the area of a circle should be \(\pi r^2\). – Ryan Tamburrino · 2 years, 1 month ago

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– Trevor Arashiro · 2 years, 1 month ago

WOW, that is really neat.Log in to reply

I fixed some slight latex errors. Also, at one point you called the property \[\int a+b\text{ d}x=\int a\text{ d}x+\int b\text{ d}x\] integrating by parts. It's actually just "splitting the integral" I suppose you can call it, integrating by parts is a whole other idea. – Daniel Liu · 2 years, 1 month ago

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– Bogdan Simeonov · 2 years, 1 month ago

You can even make higher dimension integrals, which give you volumes of multidimensional objects.In fact, you can even calculate the mass of an object with an n-dimensional integral, where the mass of point \((x_1,x_2,...,x_n)\) is defined by \(\delta(x_1,x_2,...,x_n)\) .The normal integral gives you the mass of a string.Log in to reply

– Trevor Arashiro · 2 years, 1 month ago

Yes, good point. Thanks for that!Log in to reply

I think there's an easier way. Find the circumference of the circle (2pir) and multiply by radius(r) then divide by 2. You can have infinitely many triangles with base infinitely small. the sums of the bases of these triangles would be the circumference, the area would be the circumference times radius divided by 2. Sorry that I didn't use LATEX – Brian Wang · 1 year, 5 months ago

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Shouldn't the equation be \(y=\pm\sqrt{r^2-x^2}\) and not just \(\sqrt{r^2-x^2}\). I think an equation must have an equal sign... :p – Pranjal Jain · 2 years, 1 month ago

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– Trevor Arashiro · 2 years, 1 month ago

Well, I'm finding the area of just half a circle" the integral becomes more complicated with the \(\pm\).Log in to reply

– Pranjal Jain · 2 years, 1 month ago

Equation without = sign? It should be \(y=\sqrt{r^2-x^2}\)Log in to reply

– Trevor Arashiro · 2 years, 1 month ago

Oh, Ya, you're right.Log in to reply

Yay! You can start revising with the Integration by substitution quizzes, and also add to the wiki pages! – Calvin Lin Staff · 2 years, 1 month ago

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– Trevor Arashiro · 2 years, 1 month ago

I added this to the trig sub wiki already, should I also add it to that wiki?Log in to reply

Given U-substitution and Ln |f| that require lots of love. – Calvin Lin Staff · 2 years, 1 month ago

Oh, I was referring to "I decided to learn U-sub and trig-sub integration". There are other wikis likeLog in to reply

– Trevor Arashiro · 2 years, 1 month ago

Hahah, "requireclots of love," ill be sure to check those out.Log in to reply

great typng – Incredible Mind · 2 years, 1 month ago

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