Volume inside an ellipsoid

  • #1
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Homework Statement



Find the volume between the planes ##y=0## and ##y=x## and inside the ellipsoid ##\frac{x^2}{a^2} + \frac{y^2}{b^2} + \frac{z^2}{c^2} = 1##


The Attempt at a Solution



I understand we can approach this problem under the change of variables:

$$x=au; y= bv; z=cw$$

Thus we get:

$$V= \iiint_R \,dxdydz = abc\iiint_S \,dudvdw$$

At this point the ellipsoid has become a sphere. Thus we could use spherical coordinates to compute the volume.

My issue is with the extremes of the integral; concretely with the $\theta$ angle. I would set up the integral like this:

$$\int_{0}^{\pi / 4} d\theta \int_{0}^{\pi / 2} d\phi \int_{0}^{1} dr$$

But the stated solution is:

$$\int_{0}^{\tan^{-1} (a/b)} d\theta \int_{0}^{\pi / 2} d\phi \int_{0}^{1} dr$$

My extremes make sense to me; it is just about visualizing a sphere and two intersecting planes. But ##\tan^{-1} (a/b)## confuses me.

What's wrong and why?
 

Answers and Replies

  • #2
BvU
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Hi,

It would appear that if ##\ \ y = x \quad \Rightarrow\ \ \arctan(1) ## is a bound for the coordinates before the change. Guess what the corresponding bound is afterwards ?
 
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  • #3
Ray Vickson
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Homework Statement



Find the volume between the planes ##y=0## and ##y=x## and inside the ellipsoid ##\frac{x^2}{a^2} + \frac{y^2}{b^2} + \frac{z^2}{c^2} = 1##


The Attempt at a Solution



I understand we can approach this problem under the change of variables:

$$x=au; y= bv; z=cw$$

Thus we get:

$$V= \iiint_R \,dxdydz = abc\iiint_S \,dudvdw$$

At this point the ellipsoid has become a sphere. Thus we could use spherical coordinates to compute the volume.

My issue is with the extremes of the integral; concretely with the $\theta$ angle. I would set up the integral like this:

$$\int_{0}^{\pi / 4} d\theta \int_{0}^{\pi / 2} d\phi \int_{0}^{1} dr$$

But the stated solution is:

$$\int_{0}^{\tan^{-1} (a/b)} d\theta \int_{0}^{\pi / 2} d\phi \int_{0}^{1} dr$$

My extremes make sense to me; it is just about visualizing a sphere and two intersecting planes. But ##\tan^{-1} (a/b)## confuses me.

What's wrong and why?
Assuming that ##\theta## is the polar angle, the volume element in spherical coordinates is ##r^2 \sin \theta \, dr \, d\theta \, d\phi,## so you should have ##\int_0^1 r^2 \, dr,## not ##\int_0^1 dr.##
 
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  • #4
Bosko
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In polar ( spherical ) coordinates .... is not simple $$ d\theta\space d\phi \space dr $$
but $$ r d\theta\space r\sin \theta d\phi \space dr $$

Let me find the picture ... aa here it is ..
main-qimg-577afc605d298f12a16d483d0c10af7f.gif
 

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  • #5
Bosko
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But also you can do it without calculus ( to check the result ) by careful thinking
if you compress ellipsoid
in x direction by a times
in y direction by b times
in z direction by c times you will get nice round ball

you want to calculate volume of the slice of ball between
y=0 ... x, z plane and
x/a=y/b plane => b/a=y/x

##\theta## goes from 0 to ## \arctan{ \frac{b}{a}} ##

Volume of the ball with radius r=1 is ##V= \frac{4\pi}{3} ##
Volume of the slice is ##V= \frac{4\pi}{3} \space \frac{\arctan{\frac{b}{a}} }{2\pi} ##
##V= \frac{2\pi}{3} \arctan{\frac{b}{a}}##
and when you stench back the ball to the ellipsoid
##V=abc \frac{2\pi}{3} \arctan{\frac{b}{a}}##
 
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