What is the volume inside an ellipsoid between two intersecting planes?

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Homework Help Overview

The problem involves finding the volume between the planes \(y=0\) and \(y=x\) within the ellipsoid defined by \(\frac{x^2}{a^2} + \frac{y^2}{b^2} + \frac{z^2}{c^2} = 1\). The original poster attempts to use a change of variables to transform the ellipsoid into a sphere for easier volume calculation.

Discussion Character

  • Exploratory, Mathematical reasoning, Assumption checking

Approaches and Questions Raised

  • Participants discuss the setup of integrals for volume calculation in spherical coordinates, questioning the limits of integration, particularly the angle \(\theta\). Some suggest that the bounds should reflect the geometry of the intersecting planes.

Discussion Status

The discussion is ongoing, with participants providing insights into the setup of the volume integral and the implications of the change of variables. There is a focus on visualizing the geometry involved and clarifying the limits of integration without reaching a consensus on the correct approach.

Contextual Notes

Participants note the potential confusion surrounding the transformation of the ellipsoid into a sphere and the corresponding adjustments needed for the limits of integration. The discussion also touches on the volume element in spherical coordinates and the implications of the geometry defined by the planes.

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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?
 
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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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JD_PM said:

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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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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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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