Question

Find the volume of the solid that lies within the sphere \({{\rm{x}}^{\rm{2}}}{\rm{ + }}{{\rm{y}}^{\rm{2}}}{\rm{ + }}{{\rm{z}}^{\rm{2}}}{\rm{ = 4}}\), above the \({\rm{xy}}\)-plane, and below the cone\({\rm{z = }}\sqrt {{{\rm{x}}^{\rm{2}}}{\rm{ + }}{{\rm{y}}^{\rm{2}}}} \).

Short answer

The volume of the solid is \(\frac{{{\rm{8}}\sqrt {\rm{2}} }}{{\rm{3}}}{\rm{\pi }}\).

Step-by-step solution

Sketch the cone

1) Let's start with the equation\({{\rm{x}}^{\rm{2}}}{\rm{ + }}{{\rm{y}}^{\rm{2}}}{\rm{ + }}{{\rm{z}}^{\rm{2}}}{\rm{ = 4}}\). Rewrite

\(\begin{array}{l}{{\rm{\rho }}^{\rm{2}}}{\rm{ = 4}}\\{\rm{\rho = 2}}\end{array}\)

As a result, start with a ball with a radius of\({\rm{2}}\) and is centered at the origin.

2) Consider the second constraint, which states that the solid lies above the\({\rm{x y}}\) plane and below the \({\rm{z = }}\sqrt {{{\rm{x}}^{\rm{2}}}{\rm{ + }}{{\rm{y}}^{\rm{2}}}} \)cone. It's a good idea to recall from the Example \({\rm{4}}\) in the book that the cone\({\rm{z = }}\sqrt {{{\rm{x}}^{\rm{2}}}{\rm{ + }}{{\rm{y}}^{\rm{2}}}} \) is the same as the cone\(\phi {\rm{ = }}\frac{{\rm{\pi }}}{{\rm{4}}}\) gives us.

\(\phi {\rm{ = }}\frac{{\rm{\pi }}}{{\rm{4}}}\)is a sketch of a cone:

Sketch the solid

Let's try adding our ball to this drawing to see how it appears. The grey shade represents the \({\rm{x y}}\) plane.

Step 3:Find the volume of the solid

Now, using spherical coordinates, define our solid as follows:

\({\rm{E = }}\left\{ {{\rm{(\rho ,\theta ,}}\phi {\rm{)}}\mid {\rm{0}} \le {\rm{\rho }} \le {\rm{2,0}} \le {\rm{\theta }} \le {\rm{2\pi ,}}\frac{{\rm{\pi }}}{{\rm{4}}} \le \phi \le \frac{{\rm{\pi }}}{{\rm{2}}}} \right\}\)

Now it's time to figure out how big this solid is:

\(\iiint_{E}{~d}V=\int_{\frac{\pi }{4}}^{\frac{\pi }{2}}{\int_{0}^{2\pi }{\int_{0}^{2}{{{\rho }^{2}}}}}\sin \phi d\rho d\theta d\phi \)

To make the computation easier, separate the integrals.

\(\begin{aligned}\iiint_{E}{~d}V &=\int_{\frac{\pi }{4}}^{\frac{\pi }{2}}{\sin }\phi d\phi \int_{0}^{2\pi }{d}\theta \int_{0}^{2}{{{\rho }^{2}}}~d\rho \\ & =\left. (-\cos \phi ) \right|_{\frac{\pi }{4}}^{\frac{\pi }{2}}\cdot (\theta )_{0}^{2\pi }\cdot \left( \frac{1}{3}{{\rho }^{3}} \right)_{0}^{2} \\ & =\left( \frac{1}{\sqrt{2}} \right)\cdot (2\pi )\cdot \left( \frac{8}{3} \right) \\ & =\left( \frac{\sqrt{2}}{2} \right)\cdot ()2\pi )\cdot \left( \frac{8}{3} \right) \end{aligned}\)

\(\iiint_{E}{~d}V=\frac{8\sqrt{2}}{3}\pi \)

Therefore, the volume of the solid is \(\frac{{{\rm{8}}\sqrt {\rm{2}} }}{{\rm{3}}}{\rm{\pi }}\).