Chapter 13: Q 18. (page 1066)

Show that the first moment of $\in$ is $M_{Y Z} = \frac{1}{16} πk$

Short Answer

Expert verified

It is done by using the formula $M_{Y Z} = \int \int \int_{\in} x ρ (x, y, z) d x d y d z$

Step by step solution

Given Information

$\in$ is the portion of unit sphere centered at the origin of first octant.

Simplification and determination of limits

To determine the first moment of solid w.r.t $y z$ plane, formula to be used:

$M_{Y Z} = \int \int \int_{\in} x ρ (x, y, z) d x d y d z$

Here, $k$ is uniform density of $\in$

$\Rightarrow ρ (x, y, z) = k$

We know the relation:

$x = ρ \sin ϕ \cos θ, y = ρ \sin ϕ \sin θ, z = ρ \cos ϕ$

Here, $ρ = \sqrt{x^{2} + y^{2} + z^{2}}, \tan θ = \frac{y}{z}, \cos ϕ = \frac{z}{ρ}$

Also $d x d y d z = ρ^{2} \sin ϕ d ρ d ϕ d θ$

$0 < θ < \frac{π}{2}, 0 < ϕ < \frac{π}{2}, 0 < ρ < 1$ are limits of spherical coordinates $ρ, ϕ, θ$

Calculating the First Moment

First moment of $\in = \int \int \int_{V} x ρ (x, y, z) d x d y d z$

Converting into spherical coordinates

$= \int_{ϕ = 0}^{\frac{π}{2}} \int_{θ = 0}^{\frac{π}{2}} \int_{ρ = 0}^{1} ρ^{2} \sin ϕ (ρ \sin ϕ \cos θ) k d ρ d ϕ d θ$

$= k [\int_{ϕ = 0}^{\frac{π}{2}} \sin^{2} ϕ d ϕ] \times [\int_{θ = 0}^{\frac{π}{2}} \cos θ d θ] \times [\int_{ρ = 0}^{1} ρ^{3} d ρ]$

Solving the integral

$= k (\frac{1}{2} \times \frac{π}{2}) {(\frac{ρ^{4}}{4})}_{0}^{1} {(\sin θ)}_{0}^{\frac{π}{2}}$

By application of limits, we get

$M_{Y Z} = k (\frac{π}{4}) (\frac{1}{4}) (1)$

Hence, first moment is $M_{Y Z} = \frac{πk}{16}$

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Show that the first moment of $\in$ is $M_{Y Z} = \frac{1}{16} πk$

Short Answer

Step by step solution

Given Information

Simplification and determination of limits

Calculating the First Moment

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Most popular questions from this chapter

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Show that the first moment of $\in$ is $M_{Y Z} = \frac{1}{16} πk$