Proving that $\iint_{B_2(0)}e^{(x^2+y^2)^2}dA\leq e(1+3e^{15})\pi$ 
Show that $$\iint_{B_2(0)}e^{(x^2+y^2)^2}dA\leq e(1+3e^{15})\pi$$ [Hint: If you cannot get the desired estimate directly, try using domain decomposition.]

I am having trouble with this problem. I understand that using polar coordinates,$$\iint_D{e^{x^2+y^2}dA}=\pi(e-1)$$ where $D$ is the unit disk of radius one, because it becomes $$\int_0^{2\pi}\int_0^1{re^{r^2}dr \ d\theta}=\pi(e-1)$$
So, in the problem above, does this simply become $$\int_0^{2\pi}\int_0^1{re^{{r^2}^2}dr \ d\theta}=\pi(e-1)$$
$$\Rightarrow\int_0^{2\pi}\int_0^1{re^{r^4}dr \ d\theta}=\pi(e-1)?$$
If not, what is it? How would one then prove the inequality?
 A: The disk has radius 2... Following your calculations the integral is given by
$$
\int_{0}^{2 \pi}\int_0^2 r e^{r^4} dr d \theta = 2 \pi \int_0^2 r e^{r^4} dr= 2\pi\left(\int_0^1 r e^{r^4} dr+\int_1^2 r e^{r^4} dr \right)
$$
Now,
$$
\int_0^1 r e^{r^4} dr \leq \int_0^1 r e^{r^2} dr = \frac 12 (e-1)
$$
$$
\int_1^2 r e^{r^4} dr \leq \int_1^2 r^3 e^{r^4} dr = \frac 14 e (e^{15}-1)
$$
So the initial integral is bounded by
$$
2 \pi (\frac 12 (e-1) + \frac 14 e (e^{15}-1)) = \frac{\pi}{2}(-2+e+e^{16})\leq \frac{\pi}{2}(1+3 e^{15})
$$
This is smaller than the proposed bound.
A: In the example you showed, we have
$$\iint_De^{x^2+y^2}~\mathrm dA=\int_0^{2\pi}\int_0^1re^{r^2}~\mathrm dr~\mathrm d\theta=\pi\int_0^12re^{r^2}~\mathrm dr$$
which is computed with the substitution $u=r^2$. For your problem, we get
$$\iint_{B_2(0)}e^{(x^2+y^2)^2}~\mathrm dA=\int_0^{2\pi}\int_0^2re^{r^4}~\mathrm dr~\mathrm d\theta=\pi\int_0^22re^{r^4}~\mathrm dr$$
Note that $r\in[0,2]$. Again using $u=r^2$ will get us
$$\iint_{B_2(0)}e^{(x^2+y^2)^2}~\mathrm dA=\pi\int_0^4e^{u^2}~\mathrm du$$
which, by convexity, gives us
$$\iint_{B_2(0)}e^{(x^2+y^2)^2}~\mathrm dA\le2\pi(e^{16}+1)$$
which is smaller than the provided upper bound.
A: We have $$\begin{split}
\int_{B_2(0)}e^{(x^2+y^2)^2}dA&=2\pi\int_0^2re^{r^4}dr
\\&=2\pi\left(\int_0^1re^{r^4}dr+\int_1^2re^{r^4}d\right)
\\&\leq2\pi\left(\int_0^1e\cdot rdr+\int_1^22e^{r^4}dr\right).\end{split}$$ 
