<s>[INST] Double integration including e? I would like to calculate the following expression:
$\iint_D x^2e^\sqrt{x^2+y^2} dxdy$
for D = ${ a^2\le x^2+y^2 \le b^2} $ and $ 0 \le a \le b$
How would you do it, in a step-by-step, please?
 [/INST] $$\iint_D x^2e^\sqrt{x^2+y^2} dxdy=\iint_D (rcos\phi)^2e^r\cdot r drd\phi= \iint_D r^3cos^2\phi e^r drd\phi=$$
$$=\int^{2\pi}_0d\phi \int^b_a r^3cos^2\phi e^r dr=\int^{2\pi}_0cos^2\phi d\phi \int^b_a r^3 e^r dr=$$
$$=\int^{2\pi}_0cos^2\phi d\phi ( r^3-3r^2+6r-6) e^r|^b_a =\int^{2\pi}_0(( b^3-3b^2+6b-6) e^b-(a^3-3a^2+6a-6) e^a) cos^2\phi d\phi= $$
$$=(( b^3-3b^2+6b-6) e^b-(a^3-3a^2+6a-6) e^a)\int^{2\pi}_0 cos^2\phi d\phi $$
$$=(( b^3-3b^2+6b-6) e^b-(a^3-3a^2+6a-6) e^a)(\frac{1}{2}\phi +\frac{1}{4}sin2\phi)|^{2\pi}_0= $$
$$=(( b^3-3b^2+6b-6) e^b-(a^3-3a^2+6a-6) e^a)\pi $$
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