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

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