Question

Find the mass of the solid in Problem 5 if the density is ${\left(x^2+y^2+z^2\right)}^{\mathbf{-}\mathbf{1}}$. Check your work by doing the problem in both spherical and cylindrical coordinates.

Short answer

The required mass is $\mathrm{\pi }\ln 2$.

Step-by-step solution

Given information

The density of the solid is $\left(x^2+y^2+z^2\right){}^{-1}$.

Concept of the spherical coordinates and cylindrical coordinates

The spherical coordinates$\left(r,\theta ,\varphi \right)$is related to Cartesian coordinates$\left(x,y,z\right)$by:

$$\begin{gathered} \mathit{r}\mathbf{=}\sqrt{{\mathbf{x}}^{\mathbf{2}}\mathbf{+}{\mathbf{y}}^{\mathbf{2}}\mathbf{+}{\mathbf{z}}^{\mathbf{2}}}\mathit{\theta } \\ \mathbf{}\mathbf{}\mathbf{=}\mathit{t}\mathit{a}{\mathit{n}}^{\mathbf{-}\mathbf{1}}\mathbf{\left(}\frac{\mathbf{y}}{\mathbf{x}}\mathbf{\right)}\mathit{\theta } \\ \mathbf{}\mathbf{}\mathbf{=}{\mathbf{cos}}^{\mathbf{-}\mathbf{1}}\left(\frac{z}{r}\right) \end{gathered}$$

The cylindrical coordinates $\left(r,\theta ,z\right)$is related to Cartesian coordinates$\mathbf{(}\mathbf{x}\mathbf{,}\mathbf{y}\mathbf{,}\mathbf{z}\mathbf{)}$by:

$$\begin{gathered} \mathit{r}\mathbf{=}\sqrt{{\mathbf{x}}^{\mathbf{2}}\mathbf{+}{\mathbf{y}}^{\mathbf{2}}}\mathit{\theta } \\ \mathbf{}\mathbf{}\mathbf{=}\mathit{t}\mathit{a}{\mathit{n}}^{\mathbf{-}\mathbf{1}}{\left(\frac{y}{x}\right)}^{\mathbf{z}} \\ \mathbf{}\mathbf{}\mathbf{=}\mathit{z} \end{gathered}$$

Calculation to find the required mass by the use of density in spherical coordinates

In spherical coordinates the density is$\rho =\frac{1}{r^2}$.

Calculate as follows:

$$\begin{gathered} M={\int }_0^{2\tau \tau }d\varphi {\int }_0^{\tau \tau /4}\sin \theta d\theta {\int }_1^{\frac{2}{\cos \theta }}\frac{1}{r^2}{r}^{2}dra \\ =2\tau \tau {\int }_0^{\tau \tau /4}\tan \theta d\theta \\ =2\tau \tau {\int }_0^{\tau \tau /4}\tan \theta d\theta \\ =\tau \tau \ln 2 \end{gathered}$$

Calculation to find the required mass by the use of density in cylindrical coordinates

In the cylindrical coordinates the density is$\rho =\frac{1}{z^2+r^2}$.

Calculate as follows:

$$\begin{gathered} M={\int }_0^{2\tau \tau }d\theta {\int }_1^{2}dz{\int }_0^z\frac{r}{z^2+r^2}{r}^{2}dr \\ =2\tau \tau {\int }_1^{2}dz{\int }_0^z\frac{d\left(z^2+r^2\right)}{z^2+r^2}\left(\frac{1}{2}\right) \\ =2\tau \tau {\int }_1^2\left(\ln 2z^2-\ln z^2\right)dz \\ =\tau \tau \ln 2 \end{gathered}$$

Therefore, the required mass is $\tau \tau$in 2.