Let $ D_a =$ {$~(x,y) ; x^2 + y^2 \leq a~$} circule .
the double integral is on $D_a$
$f(a) = \frac{a^6}{8} - \int\int~\frac{x^2y^2}{e^{(x^2 + y^2)}} ~dy~dx$
Find $a$ such that $f(a)$ is minimum.
assure that is is indeed the minimum
$(1)~$Can one calculate the integral ?
What I did :
going to polar coordinates :
$f(a) = \frac{a^6}{8} -\frac{\pi}{4}\int_{0}^{a} r^5e^{(-r^4)}~dr$
now $f'(a) = \frac{6a^5}{8} - \frac{\pi}{4}a^5e^{-a^4}$ Fundumental theorem of calc.
and i get
$a_{\min} = \ln(\frac{\pi}{3}~)^{\tfrac{1}{4}}$
The problem is I dont know how to find the minimum since I don't know how to calculate the integral.
You already have it. Now just differentiate once more:
$$f''(a)=\frac{15}4a^4-\frac{5\pi}4a^4e^{-a^4}+4a^8\pi e^{-a^4}\implies$$
$$f''\left(\log\left(\frac\pi3\right)^{1/4}\right)=\frac{15}4\,\log\frac\pi3-\frac{5\pi}4\,\log\frac\pi3\cdot\frac\pi3+8\,\log\frac\pi3\cdot\pi\cdot\frac\pi3=$$
$$=\log\frac\pi3\left[\frac{15}4-\frac{5\pi^2}{12}+\frac{8\pi^2}3\right]=\left[\frac{15}4+\frac{9\pi^2}4\right]\log\frac\pi3>0$$
and thus that point indeed is a minimum...Of course, it is possible to shorten the above. Up to you.