Double integral over region surrounded by two ellipses

[pandarean]

Homework Statement

A thin plate has the form of the intersection of the regions limited by [itex]\frac{x^2}{9}[/itex] + [itex]\frac{y^2}{4}[/itex] = 1 and [itex]\frac{x^2}{4}[/itex] + [itex]\frac{y^2}{9}[/itex] = 1

Which is the plate's mass if his density is δ(x, y) = |x|


2. The attempt at a solution

I've tried using u, v substitution

u = [itex]\frac{x^2}{4}[/itex] + [itex]\frac{y^2}{9}[/itex]

v = [itex]\frac{x^2}{9}[/itex] + [itex]\frac{y^2}{4}[/itex]

The resulting region looks nice, but the Jacobian is the ugly thing... I'm stuck.

I don't think polar is the way to go, one ellipse becomes a nice circle, but the other one becomes another ellipse...

Can someone give me some advice?
Thanks

 

[mighty2000]

for your post!

The first step in solving this problem is to find the limits of integration for both u and v. Using the equations for the two ellipses, we can see that u ranges from 0 to 1 and v ranges from 0 to 1.

Next, we need to find the Jacobian of the transformation. This can be done by taking the partial derivatives of u and v with respect to x and y and then taking the determinant of the resulting matrix. The Jacobian turns out to be \frac{3}{4}(1-u)(1-v).

Now, we can set up the double integral for the mass of the plate:

M = \iint_{D} \delta(x,y) \sqrt{1 + \left(\frac{\partial u}{\partial x}\right)^2 + \left(\frac{\partial v}{\partial x}\right)^2} dudv

Since we are given that the density is |x|, we can substitute that in for \delta(x,y):

M = \iint_{D} |x| \sqrt{1 + \left(\frac{\partial u}{\partial x}\right)^2 + \left(\frac{\partial v}{\partial x}\right)^2} dudv

Now, we can substitute in the Jacobian and the limits of integration:

M = \iint_{0}^{1} \iint_{0}^{1} |x| \sqrt{\frac{9}{16}(1-u)^2 + \frac{4}{9}(1-v)^2} \frac{3}{4}(1-u)(1-v) dudv

This integral can be solved using polar coordinates, with the substitution x = rcos\theta and y = rsin\theta. After some simplification, the integral becomes:

M = \frac{27}{16} \int_{0}^{\frac{\pi}{2}} \int_{0}^{1} r^3cos\theta \sqrt{1 + \frac{5}{9}r^2sin^2\theta} drd\theta

This integral can then be solved using standard techniques, resulting in the mass of the plate being:

M = \frac{27}{32}\pi

I hope this helps! Let me know if you have any further questions.