Solve the problem.
-Set up the triple integral for the volume of the sphere $\varrho = 9$ in spherical coordinates.
A) $\int _ { 0 } ^ { 2 \pi } \int _ { 0 } ^ { \pi } \int _ { 0 } ^ { 9 } \mathrm {~d} d \operatorname { d } \varphi \mathrm { d } \theta$
B) $\int _ { 0 } ^ { 2 \pi } \int _ { 0 } ^ { \pi / 2 } \int _ { 0 } ^ { 9 } \varrho ^ { 2 } \sin \varphi \mathrm { de } \mathrm { d } \varphi \mathrm { d } \theta$
C) $\int _ { 0 } ^ { 2 \pi } \int _ { 0 } ^ { \pi } \int _ { 0 } ^ { 9 } \varrho ^ { 2 } \sin \varphi \mathrm { d } \rho \mathrm { d } \varphi \mathrm { d } \theta$
D) $\int _ { 0 } ^ { 2 \pi } \int _ { 0 } ^ { \pi / 2 } \int _ { 0 } ^ { 9 } \mathrm { de } \mathrm { d } \varphi \mathrm { d } \theta$

Question

Quizplus · Accepted Answer

In spherical coordinates, the volume element is given by $\mathrm{d}V = \varrho^2 \sin \varphi \mathrm{d}\varrho \mathrm{d}\varphi \mathrm{d}	heta$. Therefore, the integral for the volume of the sphere of radius 9 is $\int_{0}^{2\pi} \int_{0}^{\pi} \int_{0}^{9} \varrho^2 \sin \varphi \mathrm{d}\varrho \mathrm{d}\varphi \mathrm{d}	heta$. So, the correct choice is C.

Solve the Problem ϱ=9\varrho = 9ϱ=9 In Spherical Coordinates ∫02π∫0π∫09 ddd⁡φdθ\int _ { 0 } ^ { 2 \pi } \int _ { 0 } ^ { \pi } \int _ { 0 } ^ { 9 } \mathrm {~d} d \operatorname { d } \varphi \mathrm { d } \theta∫02π​∫0π​∫09​ dddφdθ

Solve the Problem $\varrho = 9$ In Spherical Coordinates $\int _ { 0 } ^ { 2 \pi } \int _ { 0 } ^ { \pi } \int _ { 0 } ^ { 9 } \mathrm {~d} d \operatorname { d } \varphi \mathrm { d } \theta$