Q. 65 Use a double integral in polar c... [FREE SOLUTION]

Chapter 13: Q. 65 (page 1028)

Use a double integral in polar coordinates to prove that the volume of a sphere with radius $R$ is $\frac{4}{3} 蟺 R^{3}$ .

Short Answer

Expert verified

The volume of a sphere is $V = \frac{4}{3} 蟺 R^{3}$

Step by step solution

Given information

The objective of this problem is to use double integral to prove that the volume of the sphere with radius $R$ is $\frac{4}{3} 蟺 R^{3}$ .

Calculation

In Cartesian system the equation of a sphere with radius $R$ is

$x^{2} + y^{2} + z^{2} = R^{2} z^{2} = R^{2} - x^{2} - y^{2} z = \sqrt{R^{2} - x^{2} - y^{2}}$

Put $x = r \cos 胃$ and $y = r \sin 胃$

$z = \sqrt{R^{2} - r^{2}}$

Volume of solid in double integration

$V = 鈭� z d x d y$

In polar form

$V = {鈭�}_{0}^{2 蟺} {鈭�}_{- R}^{2} \sqrt{R^{2} - r^{2}} r d r d 胃$

Volume of a sphere is symmetrical about any axis.

Therefore,

$V = 2 {鈭�}_{0}^{2 蟺} {鈭�}_{0}^{z} \sqrt{R^{2} - r^{2}} r d r d 胃$

Here, $胃_{1} = 0, 胃_{2} = 2 蟺$ and $r_{1} = - R, r_{2} = R$

Put $R^{2} - r^{2} = t^{2}$

- 2 r d r = 2 t d t r d r = - t d t

For $r = 0, t = R$ and for $r = R, t = 0$

Then

V = 2 {鈭�}_{0}^{2 z} {鈭�}_{R^{0}}^{0} t^{2} (- d t) d 胃 V = 2 {鈭�}_{0}^{2 蟺} [{鈭�}_{0}^{蟺} t^{2} d t d 胃 V = 2 {鈭�}_{0}^{2 蟺} {[\frac{t^{3}}{3}]}_{0}^{R} d 胃 V = 2 {鈭�}_{0}^{2 蟺} [\frac{R^{3}}{3}] 胃 V = 2 \frac{R^{3}}{3} {鈭�}_{0}^{2 r} d 胃 V = \frac{2 R^{3}}{3} [胃]_{0}^{2 蟺} V = \frac{4}{3} 蟺 R^{3}

Thus, the volume of a sphere is

$V = \frac{4}{3} 蟺 R^{3}$

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Use a double integral in polar coordinates to prove that the volume of a sphere with radius $R$ is $\frac{4}{3} 蟺 R^{3}$ .

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Use a double integral in polar coordinates to prove that the volume of a sphere with radius Ris43蟺R3.

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Use a double integral in polar coordinates to prove that the volume of a sphere with radius $R$ is $\frac{4}{3} 蟺 R^{3}$ .