Use cylindrical coordinates to evaluate:
$$\int_0^4\ \int_0^{\frac{\sqrt2}2} \int_x^{\sqrt{1-x^2}} e^{-x^2-y^2} dydxdz$$
I have tried to solve for $\theta$, $r$ and $z$.
I have obtained $0\leq\theta\leq\pi/4$, $0\leq r\leq\sec(\theta)/\sqrt{2}$ and $0\leq z \leq 4$.
But I am not obtaining the correct solution. Are my integrals correct or is it wrong?
On
Since $0\leqslant x\leqslant\frac{\sqrt2}2$, and since $x\leqslant y\leqslant\sqrt{1-x^2}$, you have $x^2\leqslant y^2\leqslant1-x^2$. In cylindrical coordinates, this means that$$r^2\cos^2\theta\leqslant r^2\sin^2\theta\leqslant1-r^2\cos^2\theta.$$Now, note that all of those pairs $(x,y)$ belong to the first quadrant, and that therefore $\theta\in\left[0,\frac\pi2\right]$. Here, $\cos^2\theta\leqslant\sin^2\theta\iff\theta\in\left[\frac\pi4,\frac\pi2\right]$. On the other hand,$$r^2\sin^2\theta\leqslant1-r^2\cos^2\theta\iff r^2\leqslant1.$$So, take$$\int_0^4\int_{\pi/4}^{\pi/2}\int_0^1re^{-r^2}\,\mathrm dr\,\mathrm d\theta\,\mathrm dz.$$
The lower bound of $y$ is $x$
$y = x \implies \tan \theta = 1 \text{, i.e } \theta = \frac{\pi}{4}$
$x \geq 0$ means we are in first quadrant so $\theta \leq \frac{\pi}{2}$.
The upper bound of $y$ is $\sqrt{1-x^2} \implies x^2+y^2 \leq 1 \implies r \leq 1$.
So it should be,
$\displaystyle \int_{\pi/4}^{\pi/2} \int_0^4 \int_0^1 r \ e^{-r^2} \ dr \ dz \ d\theta = \frac{(e-1) \pi}{2e}$