Double integral

Submitted by Structure on Sat, 06/02/2012 - 05:11.

$M=\{(x,y)|x^2+y^2\le a^2 \}$

$\int\int_{M}f(x,y)dxdy=\int_{-a}^{a}(\int_{-\sqrt{a^2-x^2}}^{\sqrt{a^2-x^2}}f(x,y)dy)dx=\int_{0}^{2\pi}(\int_0^af(r\cos t,r\sin t)rdr)dt$

$M=\{(x,y)|b^2\le x^2+y^2\le a^2 \}$

$\int\int_{M}f(x,y)dxdy=\int_{-a}^{-b}(\int_{-\sqrt{a^2-x^2}}^{\sqrt{a^2-x^2}}f(x,y)dy)dx+\int_{-b}^{b}(\int_{-\sqrt{a^2-x^2}}^{-\sqrt{b^2-x^2}}f(x,y)dy)dx+$

$\int_{-b}^{b}(\int_{\sqrt{b^2-x^2}}^{\sqrt{a^2-x^2}}f(x,y)dy)dx+\int_{b}^{a}(\int_{-\sqrt{a^2-x^2}}^{\sqrt{a^2-x^2}}f(x,y)dy)dx=\int_{0}^{2\pi}(\int_b^af(r\cos t,r\sin t)rdr)dt$

$\int\int_{R^2}e^{-x^2-y^2}dx dy=\int_0^{2\pi}\int_0^{+ \infty}r e^{-r^2}dr dt$

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Double integral