section 15.6 : triple integrals in cylindrical coordinates - practice problems\n1. evaluate $iiint_e 4xy dv$…

Explanation:

Step1: Convert to cylindrical coordinates

In cylindrical coordinates, $x = r\cos\theta$, $y = r\sin\theta$, $dV=r\ dz\ dr\ d\theta$, and $z$ ranges from $z = 2r^{2}-7$ to $z = 1$. To find the bounds for $r$ and $\theta$, set $2r^{2}-7=1$, then $2r^{2}=8$, $r^{2} = 4$, so $r$ ranges from $0$ to $2$, and $\theta$ ranges from $0$ to $2\pi$. The integrand $4xy$ becomes $4r\cos\theta\times r\sin\theta=4r^{2}\cos\theta\sin\theta$.

Step2: Set up the triple - integral

The triple - integral $\iiint_E4xy\ dV$ in cylindrical coordinates is $\int_{0}^{2\pi}\int_{0}^{2}\int_{2r^{2}-7}^{1}(4r^{2}\cos\theta\sin\theta)r\ dz\ dr\ d\theta=\int_{0}^{2\pi}\int_{0}^{2}\int_{2r^{2}-7}^{1}4r^{3}\cos\theta\sin\theta\ dz\ dr\ d\theta$.

Step3: Integrate with respect to $z$

$\int_{0}^{2\pi}\int_{0}^{2}4r^{3}\cos\theta\sin\theta\left[z\right]{z = 2r^{2}-7}^{z = 1}dr\ d\theta=\int{0}^{2\pi}\int_{0}^{2}4r^{3}\cos\theta\sin\theta(1-(2r^{2}-7))dr\ d\theta=\int_{0}^{2\pi}\int_{0}^{2}4r^{3}\cos\theta\sin\theta(8 - 2r^{2})dr\ d\theta=\int_{0}^{2\pi}\int_{0}^{2}(32r^{3}-8r^{5})\cos\theta\sin\theta\ dr\ d\theta$.

Step4: Integrate with respect to $r$

$\int_{0}^{2\pi}\cos\theta\sin\theta\left[\frac{32r^{4}}{4}-\frac{8r^{6}}{6}\right]{0}^{2}d\theta=\int{0}^{2\pi}\cos\theta\sin\theta\left(8\times16-\frac{4}{3}\times64\right)d\theta=\int_{0}^{2\pi}\cos\theta\sin\theta\left(128-\frac{256}{3}\right)d\theta=\int_{0}^{2\pi}\cos\theta\sin\theta\times\frac{128}{3}d\theta$.

Step5: Integrate with respect to $\theta$

Let $u=\sin\theta$, then $du=\cos\theta\ d\theta$. When $\theta = 0$, $u = 0$; when $\theta=2\pi$, $u = 0$. So $\int_{0}^{2\pi}\frac{128}{3}\cos\theta\sin\theta\ d\theta=\frac{128}{3}\int_{0}^{0}u\ du = 0$.

Answer:

$0$