How can I solve these two tough integrals?

Question

\begin{equation*} J_{1} = \int_{0}^{\sqrt{{\pi}/{6}}} \int_{y}^{\sqrt{{\pi}/{6}}} \cos{(x^2)}\,dx\,dy \end{equation*}

\begin{equation*} J_{2} = \int\int_{E}\int z e^{(x^2+y^2)} + xe^{x^8}\,dV, \end{equation*}

where E is the region of space defined by $x^2+y^2 \le \pi^2$, $0 \le z \le 2$.

I'm stuck, so any tip will be helpful

Thanks in advance!

Answer 1

We have: $$ J_1 = \int_{D}\cos(x^2)\,dx\,dy $$ where $D$ is the region given by $\{(x,y)\in [0,\sqrt{\pi/6}]^2: x\geq y\}$, so: $$ J_1 = \int_{0}^{\sqrt{\pi/6}} x\cos x^2\,dx = \frac{1}{2}\left.\sin(x^2)\right|_{0}^{\sqrt{\pi/6}} = \frac{1}{4},$$ while in $J_2$ the integral of $x e^{x^8}$ is zero by symmetry ($xe^{x^8}$ is an odd function and $E$ is symmetric with respect to the $x=y=0$ line), so: $$ J_2 = \left(\int_{0}^{2}z\,dz\right)\cdot\left(\int_{x^2+y^2\leq \pi^2}e^{x^2+y^2}\,dx\,dy\right)$$ and by switching to polar coordinates we get: $$ J_2 = 2\int_{0}^{\pi}\int_{0}^{2\pi}\rho e^{\rho^2}d\theta d\rho = 4\pi\int_{0}^{\pi}\rho e^{\rho^2}\,d\rho= 2\pi\left(e^{\pi^2}-1\right). $$

Answer 2

Change the order of integration in $J_1:$ $$ J_{1} = \int_{0}^{\sqrt{\frac{\pi}{6}}} \int_{y}^{\sqrt{\frac{\pi}{6}}} \cos{(x^2)}\,dx\,dy = \int_{0}^{\sqrt{\frac{\pi}{6}}} \int_{0}^{x} \cos{(x^2)}\,dy\,dx \\ = \int_{0}^{\sqrt{\frac{\pi}{6}}} \cos{(x^2)}\left(\int_{0}^{x} \,dx \right)\,dy=\int_{0}^{\sqrt{\frac{\pi}{6}}} x \cos{(x^2)} \,dx. $$