15.7 triple integrals in cylindrical coordinates: problem 2 (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 $W$ is described by $0\leq r\leq7$ (since $x^{2}+y^{2}=r^{2}\leq49$), $0\leq\theta\leq 2\pi$ and $0\leq z\leq4$.

Step2: Set up the triple - integral

The integral $\iiint_{W}(x^{2}+y^{2})dV$ becomes $\int_{0}^{2\pi}\int_{0}^{7}\int_{0}^{4}r^{2}\cdot r\ dz\ dr\ d\theta=\int_{0}^{2\pi}\int_{0}^{7}\int_{0}^{4}r^{3}\ dz\ dr\ d\theta$.

Step3: Integrate with respect to $z$

$\int_{0}^{2\pi}\int_{0}^{7}r^{3}\left[z\right]{0}^{4}dr\ d\theta=\int{0}^{2\pi}\int_{0}^{7}4r^{3}dr\ d\theta$.

Step4: Integrate with respect to $r$

$\int_{0}^{2\pi}\left[4\times\frac{r^{4}}{4}\right]{0}^{7}d\theta=\int{0}^{2\pi}7^{4}d\theta$.

Step5: Integrate with respect to $\theta$

$7^{4}\left[\theta\right]_{0}^{2\pi}=2401\times2\pi = 4802\pi$.

Answer:

$4802\pi$