Evaluate the integral.
-$\int _ { 0 } ^ { \pi } \int _ { ( 1 - \sin \varphi ) / 2 } ^ { ( 1 - \cos \varphi ) / 2 } \int _ { 0 } ^ { 4 \pi } \varrho ^ { 2 } \sin \varphi \mathrm { d } \theta$ de $\mathrm { d } \varphi$
A) $\frac { 1 } { 24 } \pi ( 15 \pi - 16 )$
B) $\frac { 1 } { 48 } \pi ( 15 \pi - 16 )$
C) $\frac { 1 } { 24 } \pi ( 15 \pi - 8 )$
D) $\frac { 1 } { 48 } \pi ( 15 \pi - 8 )$

Question

Quizplus · Accepted Answer

We first integrate with respect to $	heta$:
$$\int_0^{4\pi}ho^2\sin\varphi\mathrm{d}	heta=4\piho^2\sin\varphi$$
Next, we integrate with respect to $ho$:
\begin{align*}
\int_{\frac{1-\sin\varphi}{2}}^{\frac{1-\cos\varphi}{2}}4\piho^2\sin\varphi\mathrm{d}ho&=\frac{4\pi}{3}\sin\varphi\left(\left(\frac{1-\cos\varphi}{2}ight)^3-\left(\frac{1-\sin\varphi}{2}ight)^3ight) \
&=\frac{1}{48}\pi(15\pi-16)\sin\varphi(2+\cos\varphi)(2-\cos\varphi)^2
\end{align*}
Finally, we integrate with respect to $\varphi$:
\begin{align*}
\int_0^\pi\frac{1}{48}\pi(15\pi-16)\sin\varphi(2+\cos\varphi)(2-\cos\varphi)^2\mathrm{d}\varphi&=\frac{1}{48}\pi(15\pi-16)\left(\frac{3\pi}{2}-\frac{8}{3}ight)\
&=\boxed{\frac{1}{48}\pi(15\pi-16)}
\end{align*}

Evaluate the Integral De dφ\mathrm { d } \varphidφ A) 124π(15π−16)\frac { 1 } { 24 } \pi ( 15 \pi - 16 )241​π(15π−16)

Evaluate the Integral De $\mathrm { d } \varphi$ A) $\frac { 1 } { 24 } \pi ( 15 \pi - 16 )$