Using polar coordinates, evaluate the integral $$ \int\int_R\sin(x^2 + y^2)dA $$ where R is the region $1\le x^2 + y^2\le 64$
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2What is the shape of $1 \leq x^2 + y^2$? What is the shape of $x^2 + y^2 \leq 64?$ – dfan Mar 07 '14 at 00:30
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I do not know what do you mean – user131040 Mar 07 '14 at 00:37
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2If you draw the region $x^2 + y^2 \leq 64$ on the x-y plane, what does it look like? – dfan Mar 07 '14 at 00:39
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I think that it would look like a circle but i am not sure – user131040 Mar 07 '14 at 00:40
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Make the change of variable $(x,y)=(r\cos t, r\sin t)$ into the integral, using $dA = dxdy = rdrdt$:
$$ I = \int\int_R \sin(x^2 + y^2) dA \\= \int_{t=0}^{2\pi}\int_1^{\sqrt{64}=8} \sin(r^2) rdrdt \\= {2\pi}\int_1^{8} \sin(r^2) rdr \\= {\pi}[-\cos(r^2)]_1^{8} = {\pi}(\cos 1 - \cos 64) $$
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