Question

Using polar coordinates to evaluate iterated integrals: Evaluate the given iterated integrals by converting them to polar coordinates. Include a sketch of the region.

${\int }_{-5}^0{\int }_{-\sqrt{25-x^2}}^0\frac{3}{{\left(4+x^2+y^2\right)}^3}dydx$

Short answer

${\int }_{-5}^0{\int }_{-\sqrt{25-x^2}}^0\frac{3}{{\left(4+x^2+y^2\right)}^3}dydx=\frac{2475}{107648}\mathrm{\pi }$

Step-by-step solution

Draw the region

From the limits of integration, the region is shown below,

Convert into polar form

By using the following substitution,

$$\begin{gathered} x=r\cos \theta \\ y=r\sin \theta \\ {x}^2+y^2=r^2 \\ dxdy=rdrd\theta \end{gathered}$$

The equivalent polar integral of the given integral is,

${\int }_{-5}^0{\int }_{-\sqrt{25-x^2}}^0\frac{3}{{\left(4+x^2+y^2\right)}^3}dydx\Rightarrow {\int }_{\mathrm{\pi }}^{\frac{3\mathrm{\pi }}{2}}{\int }_0^{-5}\frac{3}{{(4+r^2)}^3}rdrd\theta$

Calculate the volume

$$">$$\begin{gathered} V={\int }_{\mathrm{\pi }}^{\frac{3\mathrm{\pi }}{2}}{\int }_0^{-5}\frac{3}{{(4+r^2)}^3}rdrd\theta \\ V={\int }_{\mathrm{\pi }}^{\frac{3\mathrm{\pi }}{2}}d\theta {\int }_0^{-5}\frac{3}{{(4+r^2)}^3}rdr \\ V=\left[\frac{3\mathrm{\pi }}{2}-\mathrm{\pi }\right]{\int }_0^{-5}\frac{3}{{(4+r^2)}^3}rdr \\ V=\frac{\mathrm{\pi }}{2}{\int }_0^{-5}\frac{3}{{(4+r^2)}^3}rdr \\ \mathrm{Substitute}, \\ 4+r^2=t \\ 2rdr=dt \\ dr=\frac{1}{2r}dt \\ \mathrm{When}r=0,t=4 \\ \mathrm{When}r=-5,t=4+{(-5)}^2=29V=\frac{\mathrm{\pi }}{2}\frac{3}{2}{\int }_4^{29}\frac{1}{t^3}dt \\ V=\frac{3\mathrm{\pi }}{4}{\left[\frac{t^{-3+1}}{-3+1}\right]}_4^{29} \\ V=-\frac{3\mathrm{\pi }}{8}{\left[\frac{1}{t^2}\right]}_4^{29} \\ V=-\frac{3\mathrm{\pi }}{8}\left[\frac{1}{841}-\frac{1}{16}\right] \\ V=-\frac{3\mathrm{\pi }}{8}\left[\frac{16-841}{13456}\right] \\ V=-\frac{3\mathrm{\pi }}{8}\left[-\frac{825}{13456}\right] \\ V=\frac{2475}{107648}\mathrm{\pi }\mathrm{cubic}\mathrm{units} \end{gathered}$$