Proof for area of a circle

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)(0,0)

Thus the equation of the circle is y=r2x2y=\sqrt{r^2-x^2}

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

r2x2dx\displaystyle \int \sqrt{r^2-x^2} \mathbb{d}x

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



Substituting for x and dx we have

r2x2 dx=r2r2sin2(u)(rcos(u)) du\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 1sin2(a)=cos2(a)1-\sin^2(a)=\cos^2(a)


Now we have

r2cos2(u) du\displaystyle \int r^2\cos^2(u)~\mathbb{d}u

r2cos2(u) dur^2\displaystyle \int \cos^2(u)~\mathbb{d}u

Now, to integrate cos2(u)\cos^2(u) we must use cos2(a)=cos2a2+12\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 cos2(u)\cos^2(u) and slapped my self on the forehead). Now focusing on the cos part of the eq above

r2cos2u2+12 dur^2\displaystyle \int \frac{\cos{2u}}{2}+\frac{1}{2} ~du

Splitting terms

r2(cos2u2 du+12du)r^2\left(\displaystyle \int \frac{\cos{2u}}{2} ~du+\int \frac{1}{2} du\right)

r2(cos2u2 du+u2)r^2\left(\displaystyle \int \frac{\cos{2u}}{2}~du+\frac{u}{2}\right)

Doing a w-sub for u




Substituting back

cos(w)4 dw+u2\displaystyle \int \frac{\cos(w)}{4}~dw+\frac{u}{2}


Re-substituting for w


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


Going back to our original eq

r2cos2(u) du=r2(sin(u)cos(u)2+u2)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=rsin(u)x=r\sin(u). Looking at the sin/cos half of our eq we have


Using cos(a)=1sin2(a)\cos(a)=\sqrt{1-sin^2(a)}




We now have


We can rearrange x=rsin(u)x=r\sin(u) to u=sin1(xr)u=\sin^{-1}\left(\frac{x}{r}\right). Substituting (said it again)


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 12xr2x2+r22tan1(xr2x2)\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)=xr\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 r2x2\sqrt{r^2-x^2}. This is the adjacent to angle u.

sin1(oppositehypotenuse)=tan1(oppositeadjacent)=tan1(oppositehypotenuse2opposite2)=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

sin1(xr)=tan1(xy)=tan1(xr2x2)=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=rx=-r to x=rx=r. Thus we have

rrr2x2dx=12xr2x2+r22sin1(xr)\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)





r2π2\Rightarrow \frac{r^2\pi}{2}

Now, you may ask, "well isn't the formula r2πr^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 πr2\pi r^2


Note by Trevor Arashiro
6 years, 5 months ago

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Sign #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 RR circle is

02π0Rrdrdθ=02πR22dθ=2πR22=πR2\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 - 6 years, 5 months ago

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Haha, yes #mathnerd, I never took the time to learn these techniques (I also just learned the chain rule) simply because I thought they would take way too long to learn. Turns out you can learn them relatively quickly.

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 - 6 years, 5 months ago

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

Brian Charlesworth - 6 years, 5 months ago

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How does that double integral work?

Daniel Liu - 6 years, 5 months ago

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For this specific integral, you first evaluate the "interior" integral, i.e., the drdr one, holding θ\theta constant, and then once that is done you evaluate the "outer" integral, i.e., the dθd\theta one to get the final result. In rectangular coordinates the area increment is dA=dxdydA = dx dy, and in polar coordinates the area increment is dA=rdrdθdA = r dr d\theta. A good summary of the theory behind this is given 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 - 6 years, 5 months ago

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I fixed some slight latex errors. Also, at one point you called the property a+b dx=a dx+b dx\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 - 6 years, 5 months ago

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Yes, good point. Thanks for that!

Trevor Arashiro - 6 years, 5 months ago

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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 (x1,x2,...,xn)(x_1,x_2,...,x_n) is defined by δ(x1,x2,...,xn)\delta(x_1,x_2,...,x_n) .The normal integral gives you the mass of a string.

Bogdan Simeonov - 6 years, 5 months ago

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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 πr2\pi r^2.

Ryan Tamburrino - 6 years, 5 months ago

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WOW, that is really neat.

Trevor Arashiro - 6 years, 5 months ago

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Yay! You can start revising with the Integration by substitution quizzes, and also add to the wiki pages!

Calvin Lin Staff - 6 years, 5 months ago

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

Trevor Arashiro - 6 years, 5 months ago

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Oh, I was referring to "I decided to learn U-sub and trig-sub integration". There are other wikis like Given U-substitution and Ln |f| that require lots of love.

Calvin Lin Staff - 6 years, 5 months ago

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@Calvin Lin Hahah, "requireclots of love," ill be sure to check those out.

Trevor Arashiro - 6 years, 5 months ago

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Shouldn't the equation be y=±r2x2y=\pm\sqrt{r^2-x^2} and not just r2x2\sqrt{r^2-x^2}. I think an equation must have an equal sign... :p

Pranjal Jain - 6 years, 5 months ago

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Well, I'm finding the area of just half a circle" the integral becomes more complicated with the ±\pm.

Trevor Arashiro - 6 years, 5 months ago

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Equation without = sign? It should be y=r2x2y=\sqrt{r^2-x^2}

Pranjal Jain - 6 years, 5 months ago

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@Pranjal Jain Oh, Ya, you're right.

Trevor Arashiro - 6 years, 5 months ago

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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 - 5 years, 9 months ago

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great typng

incredible mind - 6 years, 5 months ago

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