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Elements Of Electromagnetics

7th Edition

ISBN:9780190698614

Author:Sadiku, Matthew N. O.

Publisher:Sadiku, Matthew N. O.

ChapterMA: Math Assessment

Section: Chapter Questions

Problem 1.1MA

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Question

Determine by direct integration the location of the centroid of one-half of a thin, uniform hemispherical shell of radius R.

Expert Solution

Step 1

Let's assume a small element at angle ' $θ$ ' with the $y$ -axis and the radius of the small width element is $r$ .

Let's assume the mass of the one half of the hemisphere is $M$ .

Therefore the mass of the small element is:

$\begin{array}{rcl} d m & = & \frac{M}{{πR}^{2}} \cdot πr \cdot Rdθ \\ = & \frac{M}{{πR}^{2}} \cdot πRsinθ \cdot Rdθ \\ = & M \cdot sinθ \cdot dθ \end{array}$

Step 2

Now the $y$ coordinate of the location of the centroid is:

$\begin{array}{rcl} y_{c} & = & \frac{1}{M} \int d m \cdot y \\ = & \frac{1}{M} \int_{0}^{\frac{π}{2}} M \cdot \sin θ \cdot d θ \cdot R \cos θ \\ = & \int_{0}^{\frac{π}{2}} R \cdot \sin θ \cdot \cos θ \cdot d θ \\ = & \frac{R}{2} \int_{0}^{\frac{π}{2}} \sin 2 θ \cdot d θ \\ = & \frac{R}{2} {[- \frac{\cos 2 θ}{2}]}_{0}^{\frac{π}{2}} \\ = & \frac{R}{2} [\frac{1}{2} + \frac{1}{2}] \\ = & \frac{R}{2} \end{array}$

Therefore, y coordinate of the centroid is a distance of $\frac{R}{2}$ below the origin.

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