31-32 evaluate the integral by changing to cylindrical coordinates. 31. ∫-2,2 ∫-√(4 - y²),√(4 - y²) ∫√(x² +…

Explanation:

Step1: Recall cylindrical - coordinate conversions

In cylindrical coordinates, $x = r\cos\theta$, $y = r\sin\theta$, $z = z$, and $dV=dzr\ dr\ d\theta$, and $x^{2}+y^{2}=r^{2}$.

Step2: Determine the limits of integration

The limits for $y$ are $- 2\leqslant y\leqslant2$, and for $x$ are $-\sqrt{4 - y^{2}}\leqslant x\leqslant\sqrt{4 - y^{2}}$. This is a circle of radius $r = 2$ centered at the origin in the $xy$-plane, so $0\leqslant r\leqslant2$ and $0\leqslant\theta\leqslant2\pi$. The limits for $z$ are $\sqrt{x^{2}+y^{2}}\leqslant z\leqslant2$, which in cylindrical coordinates is $r\leqslant z\leqslant2$.

Step3: Rewrite the integrand

The integrand $xz$ becomes $(r\cos\theta)z$.

Step4: Set up the triple - integral in cylindrical coordinates

The triple - integral $\int_{-2}^{2}\int_{-\sqrt{4 - y^{2}}}^{\sqrt{4 - y^{2}}}\int_{\sqrt{x^{2}+y^{2}}}^{2}xz\ dz\ dx\ dy$ becomes $\int_{0}^{2\pi}\int_{0}^{2}\int_{r}^{2}(r\cos\theta)z\cdot r\ dz\ dr\ d\theta$.

Step5: Integrate with respect to $z$

First, integrate $\int_{r}^{2}(r\cos\theta)z\cdot r\ dz$. We have $r^{2}\cos\theta\int_{r}^{2}z\ dz=r^{2}\cos\theta\left[\frac{z^{2}}{2}\right]_{r}^{2}=r^{2}\cos\theta\left(\frac{4}{2}-\frac{r^{2}}{2}\right)=\frac{r^{2}\cos\theta(4 - r^{2})}{2}$.

Step6: Integrate with respect to $r$

Next, integrate $\int_{0}^{2}\frac{r^{2}\cos\theta(4 - r^{2})}{2}dr=\frac{\cos\theta}{2}\int_{0}^{2}(4r^{2}-r^{4})dr$. $\frac{\cos\theta}{2}\left[\frac{4r^{3}}{3}-\frac{r^{5}}{5}\right]_{0}^{2}=\frac{\cos\theta}{2}\left(\frac{4\times2^{3}}{3}-\frac{2^{5}}{5}\right)=\frac{\cos\theta}{2}\left(\frac{32}{3}-\frac{32}{5}\right)=\frac{\cos\theta}{2}\times\frac{32\times(5 - 3)}{15}=\frac{32\cos\theta}{15}$.

Step7: Integrate with respect to $\theta$

Finally, integrate $\int_{0}^{2\pi}\frac{32\cos\theta}{15}d\theta=\frac{32}{15}[\sin\theta]_{0}^{2\pi}=0$.

Answer:

$0$