How do you find the formula for an area of the circle through integration?

I got this question on my Maths exam today, and the Department of Education has stated it is on the syllabus, but none of the three textbooks I could get my hands mention anything about it. One mentions the volume of a cone, but none the area of a circle.

Use integration methods to establish the formula $A = \pi r^2$ for the area of a disc with radius $r$

How would I do this?

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How many types of integration do you know? – Jack Jun 11 '11 at 0:44

You would write an equation for a (piece of a) circle of radius $r$ and integrate to find the area. For example, one quarter of the circle is $x^2+y^2=r^2, x,y \ge 0$. So the area of the circle is $$4\int_0^r y \; dx=4\int_0^r \sqrt{r^2-x^2} \; dx$$

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Probably the intended answer. Depending on how early in the course the question occurs, the students may not know how to do areas in polar coordinates, or in double integrals. – GEdgar Aug 9 '11 at 16:26

The limit of the Riemann sum of annulae will give you $$\int_0^r 2\pi r \; dr$$

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Sure, but that just establishes that $\pi$ is half of $2\pi$. – Michael Lugo Aug 12 '11 at 15:15
@Michael, surely that's all you need to establish, given that dimensional analysis gives you the rest? Or have I missed your point? – Peter Taylor Aug 12 '11 at 16:21
My point is that it would be nice to have some way to calculate &pi;, for example some integral that it's equal to. – Michael Lugo Aug 12 '11 at 23:44

I think the best answer is the one given here: http://www.nick-santos.com/jokes/2007/05/mathematician-and-plumber.html

EDIT (at Aryabhatta's suggestion): The area is $$\int\int_{\rm circle}1\,dy\,dx=\int_0^{2\pi}\int_0^Rr\,dr\,d\theta={\rm etc.}$$ but this loses all the flavor of the link.

MORE EDIT: The link died, so I updated in the comments. But in case that link dies, too, here's what's actually there. Nick Santos got it from the geometer Peter Doyle.

Once upon a time there was a mathematician. His toilet was clogged. So he called the plumber. The plumber arrived later that evening, unclogged the toilet in 15 minutes, and handed the mathematician the bill. The mathematician looked at the bill and shouted: "Great scott! What a bill! You plumbers must make a fortune charging people this much. Do you mind if I ask how much you make?"

The plumber doesn't mind. He grins and names his salary.

The Mathematician: "That's more than I make. And it looks a lot easier than theoretical mathematics."

The Plumber: "Well, friend, you sound eager, so I'll give you a tip. The plumbing business is swell. And my foreman needs more men for the job. Tell him I sent you, and you can see for yourself what the work is like. One thing though: don't tell him you're a mathematician. He hates elitists, and he won't hire anyone with higher than an 8th grade education."

Well the mathematician was as serious as he claimed. The next day he did indeed go to see the plumber's foreman. By pretending to be a middle-school drop-out, he got the job easily. And lo! It was better than he had hoped. He found himself with better pay, fewer hours, and more respect than he ever had as a theoretical mathematician.

The mathematician worked many years as a happy plumber. The plumbing industry flourished, and one day the foreman decided that his plumbers really had to be at a 9th-grade level in order to remain competitive in this rapidly-growing field. He hired several private tutors, and required his plumbers to attend night classes.

On the first day of class, the math teacher was trying to gauge what his students knew. He singled out one plumber and asked, "Do you know the formula for the area of the circle?" Of course, the tutor had picked our mathematician, and like any good mathematician, he responded, "No. But I know how to derive it." "Derive it for me then," challenged the teacher.

So the mathematician went to the blackboard and began to compute the area of the circle as any good mathematician would. He wrote out the double integral with respect to x and y, computed the Jacobian with respect to r and theta so that he could perform a change of basis in terms of polar coordinates, then evaluated the double integral. And came to the solution $-\pi r^2$.

"Wait!" said the mathematician right before he announced his answer. "That can't be right!" He began to check his work, looking for where the negative got introduced. But he couldn't find his mistake. At last he threw up his hands, erased his computation, and started over again. He quickly set up the double integral, did the change of basis to polar coordinates, and simplified. And again he arrived at $-\pi r^2$!

By now the mathematician was frantic. How could he have gotten such a ridiculous answer twice? Had his math skills really gotten so bad? He wildly looked to the teacher, but the teacher knew nothing of multi-variable calculus, and had no idea what was going on. Then he looked to the class...and noticed something quite odd.

The eyes of every plumber were fixed upon him! The class was anxiously, quietly trying to get his attention without alerting the teacher. Each one had his hands cupped around his mouth, and they were all whispering the same words over and over again in unison. Slowly, he leaned in to hear better. And this is what he heard:

"You forgot to take the absolute value of the Jacobian!"

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Funny, but better suited as a comment :-) – Aryabhata Jun 10 '11 at 23:52
@Aryabhatta, you're probably right, but there is an answer there, and it's correct, and it differs significantly from the other three answers currently posted in that it uses a double integral. – Gerry Myerson Jun 11 '11 at 0:07
Why don't you edit the answer to mention the exact details? Currently it seems too generic. – Aryabhata Jun 11 '11 at 0:08
The equation of a circle of radius $r$ centered at the origin is $$x^2+y^2=r^2.$$ You can take the portion of the circle that lies on the first quadrant, which is exactly one quarter of the entire circle, and express it as a function of $x$ by solving for $y$: $$y = \sqrt{r^2-x^2},\qquad 0\leq x\leq r.$$ The area of the circle is four times the area of the region between the $x$-axis, the $y$-axis, and the graph above.