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 $\frac{4}{3} π R^{3}$ .

Short Answer

Expert verified

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

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 $\frac{4}{3} π R^{3}$ .

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 $x = r \cos θ$ and $y = r \sin θ$

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

Volume of solid in double integration

$V = \iint z d x d y$

In polar form

$V = \int_{0}^{2 π} \int_{- R}^{2} \sqrt{R^{2} - r^{2}} r d r d θ$

Volume of a sphere is symmetrical about any axis.

Therefore,

$V = 2 \int_{0}^{2 π} \int_{0}^{z} \sqrt{R^{2} - r^{2}} r d r d θ$

Here, $θ_{1} = 0, θ_{2} = 2 π$ 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 \int_{0}^{2 z} \int_{R^{0}}^{0} t^{2} (- d t) d θ V = 2 \int_{0}^{2 π} [\int_{0}^{π} t^{2} d t d θ V = 2 \int_{0}^{2 π} {[\frac{t^{3}}{3}]}_{0}^{R} d θ V = 2 \int_{0}^{2 π} [\frac{R^{3}}{3}] θ V = 2 \frac{R^{3}}{3} \int_{0}^{2 r} d θ V = \frac{2 R^{3}}{3} [θ]_{0}^{2 π} V = \frac{4}{3} π R^{3}

Thus, the volume of a sphere is

$V = \frac{4}{3} π R^{3}$

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

Short Answer

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Use a double integral in polar coordinates to prove that the volume of a sphere with radius Ris43πR3.

Short Answer

Step by step solution

Given information

Calculation

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