Find the volume of $D\{(x,y,z)\in \mathbb{R}^3:\frac{x^2}{a^2} +\frac{y^2}{b^2}\leq z\leq 1 \}$
It looks like
(1) I believe this could be solve with a double integral an considering the function $f(x,y) = \displaystyle\frac{x^2}{a^2} + \displaystyle\frac{y^2}{b^2}$ as the height and then substracting this from the volume of the rectangle that contains the volume.
Then I should integrate between $0\leq \displaystyle\frac{x^2}{a^2}+\displaystyle\frac{y^2}{b^2}\leq 1$
(2) Another way may be defining g(x,y) = 1- f(x,y), I believe this should give the same answer as (1).
I'm not sure wheter I should use polar coordinates or spherical coordinates instead -or a similar change, since the symmetry the the volume isn't exactly spheric-. But after the changes $x=r\cos\theta$ and $y=r\sin\theta$ doesn't seem wasy to determine the limits of $r$ and $\theta$ because would lead to the inequality $0\leq b^2r^2\cos^2\theta+a^2r^2\sin^2\theta\leq a^2b^2$ and I would conclude -probably wrong- $0\leq r \leq \left(\displaystyle\frac{cos^2\theta}{a^2} +\displaystyle\frac{\sin^2\theta}{b^2} \right)^{-1}$ which seems odd and also would give me some troubles if I try to compute the integral with that limit -because I'd be integrating $\displaystyle\int\int \displaystyle\frac{r^3\cos^2\theta}{a^2}+\displaystyle\frac{r^3\sin^2\theta}{b^2}\;drd\theta$-.
Probably my approach is wrong and I don't get how to do this, but there is another way or it must be done as I said?.
Hint: For any particular $z$, what is the area of the ellipse that is the cross-section?
Integrate that result over $z\in[0,1]$.