integrate $x^{2}+y^{2}+z^{2}$ over the cylinder.\n$x^{2}+y^{2}leq4$ and $-2leq zleq3$\n(express numbers in…

Explanation:

Step1: Convert to cylindrical coordinates

In cylindrical coordinates, $x = r\cos\theta$, $y = r\sin\theta$, $z=z$, and $dV=dxdy dz=r\ dz\ dr\ d\theta$. Also, $x^{2}+y^{2}=r^{2}$. The region $x^{2}+y^{2}\leq4$ and $- 2\leq z\leq3$ becomes $0\leq r\leq2$, $0\leq\theta\leq2\pi$, $-2\leq z\leq3$. The integrand $x^{2}+y^{2}+z^{2}$ becomes $r^{2}+z^{2}$.

Step2: Set up the triple - integral

The triple - integral $\iiint_{W}(x^{2}+y^{2}+z^{2})dxdy dz$ in cylindrical coordinates is $\int_{0}^{2\pi}\int_{0}^{2}\int_{-2}^{3}(r^{2}+z^{2})r\ dz\ dr\ d\theta=\int_{0}^{2\pi}\int_{0}^{2}\int_{-2}^{3}(r^{3}+rz^{2})dz\ dr\ d\theta$.

Step3: Integrate with respect to $z$

First, integrate $\int_{-2}^{3}(r^{3}+rz^{2})dz=\left[r^{3}z+\frac{rz^{3}}{3}\right]_{z = - 2}^{z = 3}=r^{3}(3-( - 2))+\frac{r}{3}(3^{3}-(-2)^{3})=5r^{3}+\frac{r}{3}(27 + 8)=5r^{3}+\frac{35r}{3}$.

Step4: Integrate with respect to $r$

Then, integrate $\int_{0}^{2}(5r^{3}+\frac{35r}{3})dr=\left[\frac{5r^{4}}{4}+\frac{35r^{2}}{6}\right]_{0}^{2}=\frac{5\times2^{4}}{4}+\frac{35\times2^{2}}{6}=20+\frac{140}{6}=20+\frac{70}{3}=\frac{60 + 70}{3}=\frac{130}{3}$.

Step5: Integrate with respect to $\theta$

Finally, integrate $\int_{0}^{2\pi}\frac{130}{3}d\theta=\frac{130}{3}[\theta]_{0}^{2\pi}=\frac{260\pi}{3}$.

Answer:

$\frac{260\pi}{3}$