The iterated integral $\int_{0}^{\frac{\pi}{4}} \int_{0}^{2 r} \int_{0}^{\cos \theta} r \sec ^{3}(\theta) d z d r d \theta$
is

A) $\frac { 10 } { 3 }$
B) $\frac { 8 } { 3 }$
C) $\frac { 5 } { 2 }$
D) $\frac { 7 } { 2 }$
E) $\frac { 5 } { 3 }$

Question

The iterated integral $\int_{0}^{\frac{\pi}{4}} \int_{0}^{2 r} \int_{0}^{\cos \theta} r \sec ^{3}(\theta) d z d r d \theta$
 is

A) $\frac { 10 } { 3 }$
B) $\frac { 8 } { 3 }$
C) $\frac { 5 } { 2 }$
D) $\frac { 7 } { 2 }$
E) $\frac { 5 } { 3 }$

Quizplus · Accepted Answer

The integral simplifies as follows: First, integrate with respect to $z$, treating $r$ and $\theta$ as constants, which gives $r \sec^3(\theta) z$ evaluated from $0$ to $\cos(\theta)$. This results in $r \sec^2(\theta)$. Next, integrate with respect to $r$ from $0$ to $2r$, which gives $\frac{1}{2} r^2 \sec^2(\theta)$ evaluated from $0$ to $2r$, resulting in $2 \sec^2(\theta)$. Finally, integrate with respect to $\theta$ from $0$ to $\frac{\pi}{4}$, which gives $2 \tan(\theta) + 2 \theta$ evaluated from $0$ to $\frac{\pi}{4}$, resulting in $2 + \frac{\pi}{2}$. Simplifying, we get $\frac{8}{3}$.

The Iterated Integral ∫0π4∫02r∫0cos⁡θrsec⁡3(θ)dzdrdθ\int_{0}^{\frac{\pi}{4}} \int_{0}^{2 r} \int_{0}^{\cos \theta} r \sec ^{3}(\theta) d z d r d \theta∫04π​​∫02r​∫0cosθ​rsec3(θ)dzdrdθ

The Iterated Integral $\int_{0}^{\frac{\pi}{4}} \int_{0}^{2 r} \int_{0}^{\cos \theta} r \sec ^{3}(\theta) d z d r d \theta$