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

In summary, the problem involves finding the volume between two intersecting planes and within an ellipsoid using a change of variables. The resulting volume can be calculated using spherical coordinates, but the bounds for the theta angle may be confusing. It is important to note that the volume element in spherical coordinates is not simply dθdφdr, but rather rdrdθsinθdφ. Additionally, the volume can also be calculated without calculus by visualizing the ellipsoid as a compressed ball and finding the volume of the corresponding slice.
  • #1
JD_PM
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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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  • #2
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
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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  • #4
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
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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What is an ellipsoid?

An ellipsoid is a three-dimensional geometric shape that resembles a flattened sphere. It is defined as the surface formed by rotating an ellipse about one of its axes.

How is the volume inside an ellipsoid calculated?

The volume inside an ellipsoid can be calculated using the formula V = (4/3)πabc, where a, b, and c are the semi-major, semi-minor, and semi-intermediate axes of the ellipsoid, respectively.

What are some real-world applications of ellipsoids?

Ellipsoids are commonly used in geodesy, cartography, and astronomy to model the shape of the Earth, planets, and other celestial bodies. They are also used in engineering and architecture to design structures with curved surfaces.

How does the volume inside an ellipsoid differ from that of a sphere?

The volume inside an ellipsoid is generally larger than that of a sphere with the same radius. This is because an ellipsoid has three different axes of varying lengths, while a sphere has only one radius.

Can the volume inside an ellipsoid be negative?

No, the volume inside an ellipsoid cannot be negative. It is always a positive value, as it represents the amount of space contained within the surface of the ellipsoid.

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