Volume of an ellipsoid Suppose that the ellipse x2/a2 +

Question

Problem 79E

Volume of an ellipsoid Suppose that the ellipse $x^2/a^2 + y^2/b^2 = 1$ is revolved about the x-axis. What is the volume of the ellipsoid that is generated? Is the volume different if the same ellipse is revolved about the y-axis?

Answer 1

Step-by-Step Solution:

Solution 79E

Step 1 :

Given : $\frac{{x}^{2}}{{a}^{2}}+\frac{{y}^{2}}{{b}^{2}}=1$ .

a. To calculate the volume of the ellipsoid, we know that the volume of the solid generated by revolving the curve y=f(x) about the x-axis is given as follows

$V=\int\limits_{a}^{b}\Pi{}{y}^{2}dx$

Here the volume generated by $\frac{{x}^{2}}{{a}^{2}}+\frac{{y}^{2}}{{b}^{2}}=1$ is calculated by putting ${y}^{2}=\frac{{b}^{2}}{{a}^{2}}({a}^{2}-{x}^{2})$ and using the limit from (-a,a) as follows

$V=\int\limits_{-a}^{a}\Pi{}\frac{{b}^{2}}{{a}^{2}}({a}^{2}-{x}^{2})dx$
$=\frac{\Pi{}{b}^{2}}{{a}^{2}}\int\limits_{-a}^{a}({a}^{2}-{x}^{2})dx$
$=\frac{\Pi{}{b}^{2}}{{a}^{2}}(\int\limits_{-a}^{a}{a}^{2}dx-\int\limits_{a}^{-a}{x}^{2}dx)$
$=\frac{\Pi{}{b}^{2}}{{a}^{2}}({a}^{2}x-\frac{{x}^{3}}{3}){{|}^{a}}_{-a}$
$=\frac{\Pi{}{b}^{2}}{{a}^{2}}({a}^{3}-\frac{{a}^{3}}{3})-({-{a}^{3}-(\frac{{-a}^{3}}{3})}^{\ })$
$=\frac{2\Pi{}{b}^{2}}{{a}^{2}}(\frac{2{a}^{3}}{3})=\frac{4\Pi{}{ab}^{2}}{3}$
Hence the volume generated by revolving the ellipse $\frac{{x}^{2}}{{a}^{2}}+\frac{{y}^{2}}{{b}^{2}}=1$ about the x-axis is $\frac{4\Pi{}{ab}^{2}}{3}$ .

b. To calculate the volume of the solid generated by revolving the curve $\frac{{x}^{2}}{{a}^{2}}+\frac{{y}^{2}}{{b}^{2}}=1$ about the y-axis is given as follows

$V=\int\limits_{a}^{b}\Pi{}{x}^{2}dy$
Here the volume generated by $\frac{{x}^{2}}{{a}^{2}}+\frac{{y}^{2}}{{b}^{2}}=1$ is calculated by putting ${x}^{2}=\frac{{a}^{2}}{{b}^{2}}({b}^{2}-{y}^{2})$ and using the limit from (-b,b) as follows

$V=\int\limits_{-b}^{b}\Pi{}\frac{{a}^{2}}{{b}^{2}}({b}^{2}-{y}^{2})dy$
$=\frac{\Pi{}{a}^{2}}{{b}^{2}}\int\limits_{-b}^{b}({b}^{2}-{y}^{2})dy$
$=\frac{\Pi{}{a}^{2}}{{b}^{2}}(\int\limits_{-b}^{b}{b}^{2}dy-\int\limits_{-b}^{b}{y}^{2}dy)$
$=\frac{\Pi{}{a}^{2}}{{b}^{2}}({b}^{2}y-\frac{{y}^{3}}{3}){{|}^{b}}_{-b}$
$=\frac{\Pi{}{a}^{2}}{{b}^{2}}({b}^{3}-\frac{{b}^{3}}{3})-({-{b}^{3}-(\frac{{-b}^{3}}{3})}^{\ })$
$=\frac{\Pi{}{a}^{2}}{{b}^{2}}(\frac{2{b}^{3}}{3})=\frac{4\Pi{}{{a}^{2}b}^{\ }}{3}$
Hence the volume generated by the ellipse revolved around the y-axis is $\frac{4\Pi{}{{a}^{2}b}^{\ }}{3}$ .

Yes, the volume is different if ellipse is revolved about y-axis.

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