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Chapter 5: Q6P (page 268)

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

Short Answer

Expert verified

The required mass is $蟺 \ln 2$ .

Step by step solution

01

Given information

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

02

Concept of the spherical coordinates and cylindrical coordinates

The spherical coordinates $(r, 胃, 蠒)$ is related to Cartesian coordinates $(x, y, z)$ by:

$r = \sqrt{x^{2} + y^{2} + z^{2}} 胃 = t a n^{- 1} (\frac{y}{x}) 胃 = \cos^{- 1} (\frac{z}{r})$

The cylindrical coordinates $(r, 胃, z)$ is related to Cartesian coordinates $(x, y, z)$ by:

$r = \sqrt{x^{2} + y^{2}} 胃 = t a n^{- 1} {(\frac{y}{x})}^{z} = z$

03

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

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

Calculate as follows:

$M = {鈭�}_{0}^{2 蟿蟿} d 蠒 {鈭�}_{0}^{蟿蟿 / 4} \sin 胃 d 胃 {鈭�}_{1}^{\frac{2}{\cos 胃}} \frac{1}{r^{2}} r^{2} d r a = 2 蟿蟿 {鈭�}_{0}^{蟿蟿 / 4} \tan 胃 d 胃 = 2 蟿蟿 {鈭�}_{0}^{蟿蟿 / 4} \tan 胃 d 胃 = 蟿蟿 \ln 2$

04

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

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

Calculate as follows:

$M = {鈭�}_{0}^{2 蟿蟿} d 胃 {鈭�}_{1}^{2} d z {鈭�}_{0}^{z} \frac{r}{z^{2} + r^{2}} r^{2} d r = 2 蟿蟿 {鈭�}_{1}^{2} d z {鈭�}_{0}^{z} \frac{d (z^{2} + r^{2})}{z^{2} + r^{2}} (\frac{1}{2}) = 2 蟿蟿 {鈭�}_{1}^{2} (\ln 2 z^{2} - \ln z^{2}) d z = 蟿蟿 \ln 2$

Therefore, the required mass is $蟿蟿$ in 2.

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Most popular questions from this chapter

Repeat Problem 3for a rod of length / with density varying uniformly from 2to 1.

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