double integral over 2 circles with 1 common point

Question

Find the value of the integral $\int\int \sqrt{x^2+y^2}$ over region $D={(x,y): x\le x^2+y^2 \le 2x}$.

after drawing a sketch of the region.and converting x and y into polar co-ortdinates $x= r\cos\theta$,$y=r\sin\theta$.The region became $\cos\theta\le r$$ \le 2\cos\theta$. $-\pi/2\le \theta\le\pi/2$. after integration the value became $28/9$. i dont know if its correct or not.

mathlove · Accepted Answer · 2014-01-02T06:57:52.443

2

You are correct. Your integral will be $$\int_{-\pi/2}^{\pi/2}\int_{\cos\theta}^{2\cos\theta}r\cdot rdrd\theta=\int_{-\pi/2}^{\pi/2}\left(\frac 13(2\cos\theta)^3-\frac 13\cos^3\theta\right)d\theta$$ $$=\frac 73\left(\frac{3\sin\theta}{4}+\frac{\sin(3\theta)}{12}\right)\large|_{-\pi/2}^{\pi/2}=\frac{28}{9}.$$

edited Jan 02 '14 at 06:57

answered Jan 02 '14 at 06:52

mathlove

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how are u determining the theta?? by drawing the diagram or can u arrive at it by the equations only?? – Prayagdeep Parija Jan 03 '14 at 05:05
the equations only. I think you've already written it... – mathlove Jan 03 '14 at 05:07
@Prayagdeep: oh, you are asking if $-\pi/2\le \theta \pi/2$ is correct? I think $0\le \theta\le \pi/2$ is better. We usually use this. – mathlove Jan 03 '14 at 05:26

double integral over 2 circles with 1 common point

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