Question

Short answer

The volume of the given solid is \(\frac{{4\pi {a^3}}}{3}\).

Step-by-step solution

Convert into polar coordinates.

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

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

The equation for sphere with radius a and centre origin is

\(\begin{aligned}{l}{x^2} + {y^2} + {z^2} = {a^2}\\ \Rightarrow {z^2} = {a^2} - \left( {{x^2} + {y^2}} \right)\\ \Rightarrow z = \sqrt {{a^2} - \left( {{x^2} + {y^2}} \right)} \\ \Rightarrow z = \sqrt {{a^2} - {r^2}} \end{aligned}\)

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

Compute the volume.

The volume (V) will be,

Evaluate the integral.

\(I = - \int\limits_0^{2\pi } {\int\limits_0^a {{{\left( {{a^2} - {r^2}} \right)}^{\frac{1}{2}}} \cdot \left( { - 2r} \right)drd\theta } } \)

Integrating on \(r\) we get,

\(\begin{aligned}{l}I = \int\limits_0^{2\pi } {\left( {\frac{{{a^2} - {r^2}}}{{\frac{1}{2} + 1}}} \right)} _0^ad\theta \\ = \int\limits_0^{2\pi } {\left( {\frac{2}{3}{{\left( {{a^2} - {r^2}} \right)}^{\frac{3}{2}}}} \right)} _0^ad\theta \\ = \int\limits_0^{2\pi } {\left( {\frac{2}{3}{{\left( {{a^2} - {a^2}} \right)}^{\frac{3}{2}}} - \frac{2}{3}{{\left( {{a^2} - 0} \right)}^{\frac{3}{2}}}} \right)} d\theta \\ = \frac{2}{3}\int\limits_0^{2\pi } {{a^3}} d\theta \end{aligned}\)

Integrating on \(\theta \) we get,

\(\begin{aligned}{l}I = \frac{2}{3}{a^3}\left( {\left( \theta \right)} \right)_0^{2\pi }\\ = \frac{2}{3}{a^3}\left( {2\pi - 0} \right)\\ = \frac{{4\pi {a^3}}}{3}\end{aligned}\)

Therefore, volume of the given solid is \(\frac{{4\pi {a^3}}}{3}\).