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I have a function $f(x,y) = x$, and I want to find the double integral over the circular region $(x-2)^2 + y^2 =1 $ using polar coordinates.

Converting the region to polar, we get

$r^2 -4cos\theta r + 3 = 0$

Solving for $r$ gives me the limits for $r$ in the integral:

$r = 2 cos\theta + \sqrt{4cos^2\theta -3}$

$r = 2 cos\theta - \sqrt{4cos^2\theta -3}$

My question is how do I find the limits for $\theta$?

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  • $\begingroup$ If you're integrating over the whole circular region, then $\theta \in [0,2\pi]$ $\endgroup$
    – Alexei0709
    Commented Sep 8, 2015 at 22:54
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    $\begingroup$ It might be easier to shift the whole thing to the origin (depends on your function). $\endgroup$
    – Feyre
    Commented Sep 8, 2015 at 22:57

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hint: Let $x-2 = u, y = v \to \left|\dfrac{\partial(x,y)}{\partial(u,v)}\right|=1\to \displaystyle \int\int_{A} f(x,y)dxdy=\displaystyle \int\int_{B} 1\cdot f(x(u,v),y(u,v))dudv=\displaystyle \int\int_{B} (2+u)dudv=\displaystyle \int_{0}^{2\pi} \int_{0}^1 r(2+r\cos \theta)drd\theta$, whereas $A = \{(x,y): (x-2)^2+y^2 \le 1\}$, and $B = \{(u,v): u^2+v^2\le1\}$

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  • $\begingroup$ (I changed the equals signs in the last line to inequalities, but please change them back if this is incorrect.) $\endgroup$
    – user84413
    Commented Sep 8, 2015 at 23:20

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