15.7 triple integrals in cylindrical coordinates: problem 3 (1 point) use cylindrical coordinates to…

Explanation:

Step1: Convert to cylindrical coordinates

In cylindrical coordinates, $x = r\cos\theta$, $y = r\sin\theta$, $z=z$ and $dV=r\ dz\ dr\ d\theta$. Also, $x^{2}+y^{2}=r^{2}$. The region is defined by $r^{2}\leq z\leq9$, and for the intersection of $z = r^{2}$ and $z = 9$, we have $r^{2}=9$ or $r = 3$. The limits for $\theta$ are $0\leq\theta\leq2\pi$, $0\leq r\leq3$, and $r^{2}\leq z\leq9$. The function $f(x,y,z)=z$ becomes $f(r,\theta,z)=z$.

Step2: Set up the triple - integral

The triple - integral $\iiint_{W}f(x,y,z)dV$ in cylindrical coordinates is $\int_{0}^{2\pi}\int_{0}^{3}\int_{r^{2}}^{9}z\cdot r\ dz\ dr\ d\theta$.

Step3: Integrate with respect to $z$

First, integrate $\int_{r^{2}}^{9}zr\ dz$. Using the power rule $\int z^{n}dz=\frac{z^{n + 1}}{n+1}+C$ ($n\neq - 1$), we have $r\int_{r^{2}}^{9}z\ dz=r\left[\frac{z^{2}}{2}\right]_{z = r^{2}}^{z = 9}=\frac{r}{2}(81 - r^{4})$.

Step4: Integrate with respect to $r$

Next, integrate $\int_{0}^{3}\frac{r}{2}(81 - r^{4})dr=\frac{1}{2}\int_{0}^{3}(81r-r^{5})dr$. Using the power rule again, $\frac{1}{2}\left[\frac{81r^{2}}{2}-\frac{r^{6}}{6}\right]_{0}^{3}=\frac{1}{2}\left(\frac{81\times9}{2}-\frac{729}{6}\right)=\frac{1}{2}\left(\frac{729}{2}-\frac{729}{6}\right)=\frac{1}{2}\times\frac{729\times(3 - 1)}{6}=\frac{729}{6}$.

Step5: Integrate with respect to $\theta$

Finally, integrate $\int_{0}^{2\pi}\frac{729}{6}d\theta$. Since $\int_{0}^{2\pi}d\theta=2\pi$, the result is $\frac{729}{6}\times2\pi=\frac{729\pi}{3}=243\pi$.

Answer:

$243\pi$