Integrating a circle with rectangular elements

**maximade** · August 3rd 2010, 08:24 PM

Imagine a circle where its center is at the origin with radius r. Now suppose I make infinitesimally thin rectangles only at the first quadrant, the rectangles length>width, from that one can see that I'm just trying to find a fourth of the total area of that circle. In terms of why, I have set up my integration problem to be: the integral from 0 to 4 of sqrt(r^2-x^2)dx times 4(since I'm trying to find the whole area, not just a fourth of the area). From that I'm just obviously trying to make the answer end up being (pi)r^2, the equation of a circle, but I can't figure out how to evaluate the integral from 0 to 4 of sqrt(r^2-x^2)dx times 4, unless its wrong... Anyways if it were right, how would I even do the integral, please show me your work, thank you.

**TheEmptySet** · August 3rd 2010, 08:58 PM

Originally Posted by maximade

Imagine a circle where its center is at the origin with radius r. Now suppose I make infinitesimally thin rectangles only at the first quadrant, the rectangles length>width, from that one can see that I'm just trying to find a fourth of the total area of that circle. In terms of why, I have set up my integration problem to be: the integral from 0 to 4 of sqrt(r^2-x^2)dx times 4(since I'm trying to find the whole area, not just a fourth of the area). From that I'm just obviously trying to make the answer end up being (pi)r^2, the equation of a circle, but I can't figure out how to evaluate the integral from 0 to 4 of sqrt(r^2-x^2)dx times 4, unless its wrong... Anyways if it were right, how would I even do the integral, please show me your work, thank you.

$\int_{0}^{r}\sqrt{r^2-x^2}dx$

let $x=r\sin(t) \implies dx=r\cos(t)dt$

$\int_{0}^{\frac{\pi}{2}}\sqrt{r^2-r^2\sin^{2}t}(r\cos(t))dt=r^{2}\int_{0}^{\frac{\pi }{2}}\cos^2(t)dt$

$r^{2}\int_{0}^{\frac{\pi}{2}}\cos^2(t)dt=r^{2}\int _{0}^{\frac{\pi}{2}}\frac{1}{2}(1+\cos(2t))dt=\fra c{\pi r^2}{4}$

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