Q. 37 - Calculus

Question

In Exercises 35–40, find the volume of the solid bounded above by the given function over the specified region$\Omega$. $f(x,y)=\sqrt{4-x^2-y^2}$

Region: 

Step-by-Step Solution

Answer

Volume bounded by given function is$4\mathrm{\pi }$.

Step 1. Given Information

A function, $f(x,y)=\sqrt{4-x^2-y^2}$

Region:

Step 2. Calculating the volume of the solid bounded above by the given function and region

The double integral is given by,

${\int }_{-2}^2{\int }_0^{\sqrt{4-x^2}}\sqrt{4-x^2-y^2}dy dx + {\int }_{-2}^0{\int }_{-\sqrt{4-x^2}}^0\sqrt{4-x^2-y^2}dy dx$

Due to symmetry it can also be written as, 

$3{\int }_0^2{\int }_0^{\sqrt{4-x^2}}\sqrt{4-x^2-y^2}dy dx$

Converting to polar coordinates, the integration becomes,

$3{\int }_0^{\frac{\mathrm{\pi }}{2}}{\int }_0^2\left(\sqrt{4-r^2}\right) r dr d\theta$

Put, $4-r^2=t, -2r dr=dt,$ 

We get, $3{\int }_0^{\frac{\mathrm{\pi }}{2}}{\int }_4^0\left(-\frac{\sqrt{t}}{2}\right) dt d\theta$

$$\begin{gathered} =-\frac{3}{2}{\int }_0^{\frac{\mathrm{\pi }}{2}}\frac{2}{3}{\overline{)\left(t^{\frac{3}{2}}\right)}}_4^0 d\theta \\ =-{\int }_0^{\frac{\mathrm{\pi }}{2}}{\overline{)\left(0-4^{\frac{3}{2}}\right)}}_4^0 d\theta \\ =-3{\int }_0^{\frac{\mathrm{\pi }}{2}}\frac{2}{3}\left(0-4^{\frac{3}{2}}\right)d\theta \\ =8{\int }_0^{\frac{\mathrm{\pi }}{2}}1 d\theta \\ =8 {\overline{)\theta }}_0^{\frac{\mathrm{\pi }}{2}} \\ =8\times \frac{\mathrm{\pi }}{2} \\ =4\mathrm{\pi } \end{gathered}$$