The iterated integral $\int _ { 0 } ^ { \frac { \pi } { 2 } } \int _ { \frac { \pi } { 6 } } ^ { \frac { \pi } { 2 } } \int _ { 0 } ^ { 2 } \rho ^ { 3 } \sin ( \varphi ) \cos \theta d \rho d \varphi d \theta$ is

A) $4 \sqrt { 3 }$
B) $- 3 \sqrt { 3 }$
C) $3 \sqrt { 3 }$
D) $- 2 \sqrt { 3 }$
E) $2 \sqrt { 3 }$

Question

The iterated integral

\int _ { 0 } ^ { \frac { \pi } { 2 } } \int _ { \frac { \pi } { 6 } } ^ { \frac { \pi } { 2 } } \int _ { 0 } ^ { 2 } \rho ^ { 3 } \sin ( \varphi ) \cos \theta d \rho d \varphi d \theta

is

A) $4 \sqrt { 3 }$
B) $- 3 \sqrt { 3 }$
C) $3 \sqrt { 3 }$
D) $- 2 \sqrt { 3 }$
E) $2 \sqrt { 3 }$

Quizplus · Accepted Answer

The integral is evaluated by first integrating with respect to $\rho$, then $\varphi$, and finally $\theta$. The integral of $\rho^3$ from 0 to 2 with respect to $\rho$ gives $\frac{\rho^4}{4}$ evaluated from 0 to 2, which is $4$. The integral of $\sin(\varphi)$ from $\frac{\pi}{6}$ to $\frac{\pi}{2}$ gives $1 - \frac{\sqrt{3}}{2}$. The integral of $\cos(\theta)$ from 0 to $\frac{\pi}{2}$ is 1. Multiplying these results gives $4 \times (1 - \frac{\sqrt{3}}{2}) \times 1 = 4 - 2\sqrt{3}$, which simplifies to $2\sqrt{3}$.

The Iterated Integral Is A) 434 \sqrt { 3 }43​

The Iterated Integral Is

A) $4 \sqrt { 3 }$