Write triple integral as cylindrical coordinate of given region, confused in determining lower and upper bound.

Question

Given $E$ is a region as follows: $$E=\left\{0\leq x\leq 1, 0\leq y\leq \sqrt{1-x^2}, \sqrt{x^2+y^2}\leq z\leq \sqrt{2-x^2-y^2}\right\}.$$ Write triple integral $$\iiint_\limits{E}xydzdydx$$ as triple integral in cylindrical coordinates.

First I try to plot the region as follows.

I try to convert triple integral in cylindrical coordinates as below.

\begin{align} \int\limits_{0}^{\frac{\pi}{2}}\int\limits_{0}^{1}\int\limits_{\sqrt{(r\cos\theta)^2+(r\sin\theta)^2}}^{\sqrt{2-(r\cos\theta)^2-(r\sin\theta)^2}} (r\cos\theta)(r\sin\theta)r dzdrd\theta \end{align}

and simplify as

\begin{align} \int\limits_{0}^{\frac{\pi}{2}}\int\limits_{0}^{1}\int\limits_{r}^{\sqrt{2-r^2}} r^3\cos\theta \sin\theta dzdrd\theta. \end{align}

I'm not sure to determining the lower and upper bound of triple integral in this problem. I try as above. Is it right?

Also, I confused if the order of integral is $drd\theta dz$, so I try $dzdrd\theta$.

Answer 1

The upper and lower bound of triple integral are right. You can explicit the variable $r$ or $\theta$ or $z$ as the following $$ \iiint_E xy \, dz \, dy \, dx = \int_{\theta=0}^{\theta=\frac{\pi}{2}} \int_{r=0}^{r=1} \int_{z=r}^{z=\sqrt{2 - r^2}} r^3 \cos \theta \sin \theta \, dz \, dr \, d\theta= $$

$$=\left\{\int_{\theta=0}^{\theta=\frac{\pi}{2}} \cos \theta \sin \theta \ d\theta\left[\int_{r=0}^{r=1} r^3\, dr\left(\int_{z=r}^{z=\sqrt{2 - r^2}} dz \right)\right]\right\} $$

and solve the triple integral solving the integral into the rounded brackets, ecc.