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?

  • $\begingroup$ While this question is unsolved for general elliptical axes lengths, I'd like to make a link to the related post. Also see the meta post. $\endgroup$ – Lee David Chung Lin Jan 22 '19 at 11:49

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.

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