Set up but do not evaluate the integral representing the volume of the region $ax^2 + by^2 = r^2$

Question

In my class right now we are doing volumes using cross sections.

The question is: Set up but do not evaluate the integral representing the volume of the region $$ax^2 + by^2 = r^2$$

It's rotating around an unspecified axis.

I've solved for y:

$ax^2+by^2=r^2$ $\ $$\Rightarrow$ $y=\sqrt{\frac{r^2 -ax^2}{b}}$

Some class mates are saying to set $y=r$ but I don't really understand why I would do that or what it would accomplish.

I carried the problem out

$$A(x)= \int_{c}^{d}\left |\sqrt{\frac{r^2-ax^2}{b}} \right |dx$$ $$V(x)= \pi \int_{c}^{d}\left |(\sqrt{\frac{r^2-ax^2}{b}})^2 \right |dx$$ but i feel like this isn't the correct answer. What should i do?

Answer 1

The equation you mention is just an ellipse without 3D information. I assume that you need the volume for the solid on rotating this along some axis of symmetry, which I represent as
\begin{equation} ax^{2}+by^{2}+cz^{2}=r^{2} \end{equation} Then, the integral for one octant of the solid is \begin{eqnarray} V_{1/8}&=&\int_{0}^{r/\sqrt{c}} dV(z)=\int_{0}^{r/\sqrt{c}}A(z)dz\\&=&\int_{0}^{r/\sqrt{c}}\int_{0}^{\sqrt{(r^{2}-cz^{2})/b}}\int_{0}^{\sqrt{(r^{2}-cz^{2})/a}}\sqrt{\frac{r^{2}-ax^{2}-cz^{2}}{b}}dx\ dy\ dz \end{eqnarray}

Then $V=8V_{1/8}$.

Answer 2

It wasn't clear to me whether this question had been resolved; now that it has been revised to indicate that we are working with an ellipsoid (actually either an oblate or prolate spheroid):

The region in the plane is an ellipse with the "standard-form" equation,

$$ \frac{x^2}{r^2 / a} \ + \ \frac{y^2}{r^2 / b} \ = \ 1 \ \ , $$

so the ellipse is centered on the origin and has "semi-horizontal" and "semi-vertical" axes of $ \ \frac{r}{\sqrt{a}} \ \text{and} \ \frac{r}{\sqrt{b}} \ , $ respectively. It's a bit unclear what the intent of the problem is, since rotating an ellipse (unequal lengths of axis) about the $ \ x-$ axis will produce an ellipsoid with a different volume from one produced by rotation about the $ \ y-$ axis. So we'll address different possibilities.

If we are evaluating the volume by the use of "cross-sections", that presumably means that we are taking "slices", along the $ \ z-$ direction, which all have the same proportions between their axes as has the ellipse in the $ \ xy-$ plane. If that ellipse is being rotated about the $ \ x- $ axis, the ellipsoid will extend along the $ \ z- $ axis from $ \ -\frac{r}{\sqrt{b}} \ \ \text{to} \ \frac{r}{\sqrt{b}} \ . $ The cross-section of the ellipsoid in the $ \ yz-$ plane is then $ \ \frac{x^2}{r^2 / a} \ + \ \frac{z^2}{r^2 / b} \ = \ 1 \ $ . Upon solving this equation for $ \ x \ , $ we find that the elliptical cross-section at a height $ \ |z| \ $ has "semi-horizontal axis" $ \ x \ = \ \sqrt{\frac{r^2 - bz^2}{a}} \ $ ; the "semi-vertical axis" is then $ \ y \ = \ \sqrt{\frac{r^2 - bz^2}{b}} \ $ (as may be found by using the cross-section proportions, or solving the cross-section in the $ \ xz-$ plane for $ \ y \ . $

The area of the elliptical cross-sections in the $ \ z-$ direction are thus given by

$$ \ A(z) \ = \ \pi \cdot \frac{r^2 - bz^2}{\sqrt{ab}} \ . $$

Since the ellipsoid is symmetrical about the $ \ xy-$ plane, we can integrate the "upper half-ellipsoid" and multiply the result by 2 to find the volume,

$$ V \ = \ 2 \ \int_0^{r/\sqrt{b}} \ \ \pi \cdot \frac{r^2 - bz^2}{\sqrt{ab}} \ \ dz \ \ = \ \ \frac{4 \pi r^3}{3 \sqrt{ab^2}} \ \ . $$

For the ellipsoid produced by rotating our ellipse about the $ \ y-$ axis, the cross-section in the $ \ yz-$ plane becomes $ \ \frac{x^2}{r^2 / a} \ + \ \frac{z^2}{r^2 / a} \ = \ 1 \ $ ; an argument analogous to the above leads us to

$$ \ A(z) \ = \ \pi \cdot \frac{r^2 - az^2}{\sqrt{ab}} $$

and

$$ V \ = \ 2 \ \int_0^{r/\sqrt{a}} \ \ \pi \cdot \frac{r^2 - az^2}{\sqrt{ab}} \ \ dz \ \ = \ \ \frac{4 \pi r^3}{3 \sqrt{a^2b}} \ \ . $$

These look a bit different from the standard ellipsoid volume formulas because of the peculiar form of the given ellipse equation. Were we to use $ \ a^2x^2 \ + \ b^2y^2 \ = \ a^2b^2 \ , $ our results would become $ \ \frac{4 \pi}{3} \ a^2b \ \ \text{and} \ \ \frac{4 \pi}{3} \ ab^2 \ , $ one being the volume of a oblate, the other, a prolate spheroid, depending upon which of $ \ a \ $ or $ \ b \ $ is larger. (These of course go over to the formula for the volume of a sphere for $ a \ = \ b \ . $ )