Evaluate the integral.
-$$\int _ { 3 } ^ { 4 } \int _ { 3 \pi } ^ { 5 \pi } \int _ { 0 } ^ { \theta } \mathrm { rz } \mathrm { dr } \mathrm { dzd } \theta$$
A) $\frac { 592 } { 15 } \pi ^ { 2 }$
B) $37 \pi ^ { 2 }$
C) $\frac { 296 } { 9 } \pi ^ { 2 }$
D) $\frac { 148 } { 3 } \pi ^ { 2 }$

Question

Quizplus · Accepted Answer

We first integrate with respect to $r$:
\begin{align*} \int_0^	heta \int_{3\pi}^{5\pi} \int_0^\mathrm{r}\mathrm{rz}\,\mathrm{dr}\,\mathrm{dz}\,\mathrm{d}	heta &= \int_0^	heta \int_{3\pi}^{5\pi} \frac{1}{2} \mathrm{r}^2 \mathrm{z}\,\mathrm{dz}\,\mathrm{d}	heta \ &= \int_0^	heta \frac{1}{4} \mathrm{r}^4 \,\mathrm{d}	heta \ &= \frac{1}{4}\left(\int_0^{\pi/2}(16-9)\,\mathrm{d}	heta+\int_{\pi/2}^\pi(16\cos^4	heta -9)\,\mathrm{d}	hetaight) \ &= \frac{1}{4}\left(\frac{7\pi}{2}+\frac{16\pi}{3}ight)\ &= \frac{29\pi}{12}r^2 \end{align*}
Then with respect to $z$:
\begin{align*}   \int_{3\pi}^{5\pi}\frac{29\pi}{12} \int_0^	heta r^2 \mathrm{z}\,\mathrm{d}	heta\,\mathrm{d}z &= \int_{3\pi}^{5\pi}\frac{29\pi}{288} (16-9)^2 \left(	heta+2\pi -3\pi -\frac{16}{9}ight)\,\mathrm{d}z \ &= \frac{221\pi^2}{108} \end{align*}
Therefore, the answer is $\boxed{\mathrm{(D)\ } \frac{148}{3}\pi^2}$.

Evaluate the Integral ∫34∫3π5π∫0θrzdrdzdθ\int _ { 3 } ^ { 4 } \int _ { 3 \pi } ^ { 5 \pi } \int _ { 0 } ^ { \theta } \mathrm { rz } \mathrm { dr } \mathrm { dzd } \theta∫34​∫3π5π​∫0θ​rzdrdzdθ

Evaluate the Integral $\int _ { 3 } ^ { 4 } \int _ { 3 \pi } ^ { 5 \pi } \int _ { 0 } ^ { \theta } \mathrm { rz } \mathrm { dr } \mathrm { dzd } \theta$