Question

Short answer

The volume of the given solid is \(\frac{{16\pi }}{3}\).

Step-by-step solution

Convert into polar coordinates.

In polar coordinates\(x = r\cos \theta \),\(y = r\sin \theta \).

\( \Rightarrow {x^2} + {y^2} = {r^2}\)

So, the given cone \(z = \sqrt {{x^2} + {y^2}} \)will be,

\(\begin{array}{l}z = \sqrt {{r^2}} \\ \Rightarrow z = r\end{array}\)

And the disk \({x^2} + {y^2} \le 4\)will be,

\(\begin{array}{l}0 \le {r^2} \le 4\\ \Rightarrow 0 \le r \le 2\end{array}\)

The region of integration is a polar rectangle described as \(0 \le \theta \le 2\pi \)and\(0 \le r \le 2\).

Compute the volume.

The volume (V) will be

\begin{aligned}V &= \iint\limits_D {zdA} \\ &= \int\limits_0^{2\pi } {\int\limits_0^2 {r \cdot rdrd\theta } } \\ &= \int\limits_0^{2\pi } {\int\limits_0^2 {{r^2}drd\theta } } \\ \end{aligned}

Evaluate the integral.

\(I = \int\limits_0^{2\pi } {\int\limits_0^2 {{r^2}drd\theta } } \)

Integrating on \(r\) we get,

\(\begin{array}{c}I = \int\limits_0^{2\pi } {\left( {\frac{1}{3}{r^3}} \right)} _0^2d\theta \\ = \int\limits_0^{2\pi } {\left( {\frac{1}{3}\left( {{2^3} - {2^0}} \right)} \right)} d\theta \\ = \int\limits_0^{2\pi } {\frac{8}{3}} d\theta \end{array}\)

Integrating on \(\theta \) we get,

\(\begin{array}{c}I = \frac{8}{3}\left( {\left( \theta \right)} \right)_0^{2\pi }\\ = \frac{8}{3}\left( {2\pi - 0} \right)\\ = \frac{{16\pi }}{3}\end{array}\)

Therefore, volume of the given solid is \(\frac{{16\pi }}{3}\).