Evaluate the double integral below. $\int _ { 0 } ^ { \pi / 2 } \int _ { 0 } ^ { 5 + 5 \sin \theta } \theta r d r d \theta$
A) $\frac { 225 } { 8 } + \frac { 5 } { 16 } \pi ^ { 2 }$
B) $\frac { 225 } { 16 } + \frac { 75 } { 32 } \pi ^ { 2 }$
C) $\frac { 225 } { 8 } + \frac { 75 } { 32 } \pi ^ { 2 }$
D) $\frac { 225 } { 8 } + \frac { 75 } { 32 } \pi$
E) $\frac { 5 } { 8 } + \frac { 5 } { 32 } \pi$

Question

Quizplus · Accepted Answer

To solve the double integral, first integrate with respect to $r$, treating $\theta$ as a constant. The integral of $\theta r$ with respect to $r$ from $0$ to $5 + 5\sin\theta$ is $\frac{1}{2}\theta r^2$ evaluated from $0$ to $5 + 5\sin\theta$, which simplifies to $\frac{1}{2}\theta(5 + 5\sin\theta)^2$. Expanding and simplifying gives $\frac{1}{2}\theta(25 + 50\sin\theta + 25\sin^2\theta)$. Integrating this expression with respect to $\theta$ from $0$ to $\pi/2$ involves integrating $\theta$, $\theta\sin\theta$, and $\theta\sin^2\theta$. The integral of $\theta$ from $0$ to $\pi/2$ is $\frac{\pi^2}{8}$, the integral of $\theta\sin\theta$ from $0$ to $\pi/2$ is $1$, and the integral of $\theta\sin^2\theta$ requires using integration by parts or a trigonometric identity, leading to a value that contributes to the final answer. Combining these results and multiplying by the coefficients and constants from the original integral gives the final answer of $\frac{225}{8} + \frac{75}{32}\pi^2$.

Evaluate the Double Integral Below ∫0π/2∫05+5sin⁡θθrdrdθ\int _ { 0 } ^ { \pi / 2 } \int _ { 0 } ^ { 5 + 5 \sin \theta } \theta r d r d \theta∫0π/2​∫05+5sinθ​θrdrdθ

Evaluate the Double Integral Below $\int _ { 0 } ^ { \pi / 2 } \int _ { 0 } ^ { 5 + 5 \sin \theta } \theta r d r d \theta$