Using spherical coordinates, the volume of the solid inside $x ^ { 2 } + y ^ { 2 } + z ^ { 2 } = 4 z$ and above $x ^ { 2 } + y ^ { 2 } = z ^ { 2 }$ is

A) $8 \pi$
B)7 $\pi$
C) $6 \pi$
D) $5 \pi$
E)4 $\pi$

Question

Using spherical coordinates, the volume of the solid inside $x ^ { 2 } + y ^ { 2 } + z ^ { 2 } = 4 z$ and above $x ^ { 2 } + y ^ { 2 } = z ^ { 2 }$ is

A) $8 \pi$
B)7 $\pi$
C) $6 \pi$
D) $5 \pi$
E)4 $\pi$

Quizplus · Accepted Answer

The given equations can be rewritten in spherical coordinates. The equation $x^2 + y^2 + z^2 = 4z$ becomes $\rho^2 = 4\rho \cos(\phi)$ after substituting $x = \rho \sin(\phi) \cos(\theta)$, $y = \rho \sin(\phi) \sin(\theta)$, and $z = \rho \cos(\phi)$, simplifying to $\rho = 4\cos(\phi)$. The equation $x^2 + y^2 = z^2$ simplifies to $\tan^2(\phi) = 1$, or $\phi = \frac{\pi}{4}$. The volume in spherical coordinates is given by the integral $\int_{0}^{2\pi} \int_{0}^{\pi/4} \int_{0}^{4\cos(\phi)} \rho^2 \sin(\phi) d\rho d\phi d\theta$, which evaluates to $8\pi$.

Using Spherical Coordinates, the Volume of the Solid Inside x2+y2+z2=4zx ^ { 2 } + y ^ { 2 } + z ^ { 2 } = 4 zx2+y2+z2=4z

Using Spherical Coordinates, the Volume of the Solid Inside $x ^ { 2 } + y ^ { 2 } + z ^ { 2 } = 4 z$