Solve the problem.
-Let $\mathrm { D }$ be the region that is bounded below by the cone $\varphi = \frac { \pi } { 4 }$ and above by the sphere $\rho = 4$. Set up the triple integral for the volume of $D$ in spherical coordinates.
A) $\int _ { 0 } ^ { 2 \pi } \int _ { 0 } ^ { \pi / 2 } \int _ { 0 } ^ { 4 }$ Q$^ { 2 } \sin \varphi \operatorname { de } d \varphi d \theta$
B) $\int _ { 0 } ^ { 2 \pi } \int _ { 0 } ^ { \pi / 4 } \int _ { 0 } ^ { 4 } \rho ^ { 2 } \sin \varphi \mathrm { de } \mathrm { d } \varphi \mathrm { d } \theta$
C) $\int _ { 0 } ^ { 2 \pi } \int _ { 0 } ^ { 3 \pi / 4 } \int _ { 0 } ^ { 4 } \rho ^ { 2 } \sin \varphi \operatorname { de } d \varphi \mathrm { d } \theta$
D) $\int _ { 0 } ^ { 2 \pi } \int _ { 0 } ^ { \pi } \int _ { 0 } ^ { 4 } \rho ^ { 2 } \sin \varphi \mathrm { de } \mathrm { d } \varphi \mathrm { d } \theta$

Question

Quizplus · Accepted Answer

We need to first visualize the region D. The cone $\varphi=\frac{\pi}{4}$ is a cone with vertex at the origin and an opening angle of $\frac{\pi}{4}$. The sphere $\rho=4$ is a sphere with center at the origin and radius 4. The region D is the part of the sphere that lies above the cone. Thus, we need to integrate the function $\rho^2\sin\varphi$ over the region D. We can see that the limits of integration for $\theta$ are from $0$ to $2\pi$ (covering the entire circle). For $\varphi$, we need to integrate from the bottom of the cone up to the top of the sphere, so the limits are from $0$ to $\frac{\pi}{4}$. Finally, for $\rho$, we need to integrate from the origin up to the surface of the sphere, so the limits are from $0$ to $4$. Thus, the correct integral is $\int_0^{2\pi}\int_0^{\pi/4}\int_0^4\rho^2\sin\varphi d\rho d\varphi d\theta$, which is represented by choice B.

Solve the Problem D\mathrm { D }D Be the Region That Is Bounded Below by the Cone

Solve the Problem $\mathrm { D }$ Be the Region That Is Bounded Below by the Cone