Find the volume of the indicated region.
-the region bounded above by the sphere $x ^ { 2 } + y ^ { 2 } + z ^ { 2 } = 4$ and below by the cone $z = \sqrt { x ^ { 2 } + y ^ { 2 } }$
A) $2 \pi ( 2 - \sqrt { 2 } )$
B) $2 \pi ( 2 - \sqrt { 3 } )$
C) $\frac { 8 } { 3 } \pi ( 2 - \sqrt { 2 } )$
D) $\frac { 8 } { 3 } \pi ( 2 - \sqrt { 3 } )$

Question

Quizplus · Accepted Answer

First, we should find the intersection of the sphere and cone. Substituting $z = \sqrt { x ^ { 2 } + y ^ { 2 } }$ into the equation of the sphere gives $x ^ { 2 } + y ^ { 2 } + z ^ { 2 } = 4$ becomes $x ^ { 2 } + y ^ { 2 } + x ^ { 2 } + y ^ { 2 } = 4$ or $2 x ^ { 2 } + 2 y ^ { 2 } = 4$ or $x ^ { 2 } + y ^ { 2 } = 2$, which is the equation of a circle with radius $\sqrt {2}$ centered at the origin. Since the cone is also symmetric about the origin, we only need to consider the region of intersection in the first octant.

Now consider a thin slice of the region parallel to the xy-plane at height z. The area of the circle at height z is $\pi ( 2 ^ { 2 } - z ^ { 2 } )$ and the height of the slice is $z$. Therefore, the volume of the slice is $dV = \pi ( 2 ^ { 2 } - z ^ { 2 } ) dz$. To get the total volume, we integrate from 0 to 2:

$$ V = \int _ { 0 } ^ { 2 } \pi ( 2 ^ { 2 } - z ^ { 2 } ) dz $$

This evaluates to $\frac{8}{3}\pi (2 - \sqrt{2})$, so the answer is choice C.

Find the Volume of the Indicated Region x2+y2+z2=4x ^ { 2 } + y ^ { 2 } + z ^ { 2 } = 4x2+y2+z2=4

Find the Volume of the Indicated Region $x ^ { 2 } + y ^ { 2 } + z ^ { 2 } = 4$