Volume of the intersection two elliptical cylinders How can I find the volume of the intersection of the following elliptical cylinders?
$$\dfrac{x^2}{9} + \dfrac{y^2}{4} \leq 1$$
$$\dfrac{y^2}{4}+\dfrac{z^2}{9} \leq 1$$
In the first octant.

This is what I have done:
Let $A$ be the projection of the intersection in the XY plane. Then:
$$A = \{ (x, y, 0) \in \mathbb{R}^2 : x, y \geq 0, \dfrac{x^2}{9} + \dfrac{y^2}{4} \leq 1\}$$
And from the second cylinder I get:
$$z = 3\sqrt{1-\dfrac{y^2}{4}}$$
So the volume $V$ will be:
$$\displaystyle \int_A 3\sqrt{1-\dfrac{y^2}{4}} dxdy $$
Is that right?

That is right, but most of the work is actually to split $A$ into the proper bounds for each variable, and then do the actual integration. So write $A$ as \begin{equation} A = \{ (x, y) \in \mathbb{R}^2 : 0\leq x \leq 3\sqrt{1-y^2/4}\;,\; 0\leq y \leq 2\}, \end{equation} and then we have \begin{eqnarray} V&=&\int_0^2 \int_0^{3\sqrt{1-y^2/4}} 3\sqrt{1-\dfrac{y^2}{4}} dxdy \\ &=& \int_0^2 \left (3\sqrt{1-\dfrac{y^2}{4}}\right)^2 dy \\ &=& \int_0^2 9-\dfrac{9y^2}{4} dy \\ &=& (9y-\dfrac{3y^3}{4}) |_0^2 \\ &=& 12 \end{eqnarray}
This problem happens to work out nicely because the ellipse equations for $x$ in terms of $y$ matches with the equation of $z$ in terms of $y$ - so you can get rid of the square root. In general when ellipses or circle are involved I would go for a coordinate transformation to cylindrical coordinates.
