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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
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4 Answers

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
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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 π, for example some integral that it's equal to. –  Michael Lugo Aug 12 '11 at 23:44
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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.

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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.

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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
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