Triple integral volume problem, volume between 2 paraboloids

[zonedestruct]

Homework Statement

Find the volume of the solid region E bounded by the paraboloids z = 1+x^2+y^2 and
z = 4 - 2x^2 - 11y^2

The Attempt at a Solution

i set up a triple integral using Cartesian coordinates but was unable to solve it because the limits of integration where very hard to integrate, i don't think you can use cylindrical coordinates because the intersection is an eclipse: x^2 + 4y^2 = 1. PLease somebody show me a good method to do this question as it was in a past exam and might come up again.

Please help me. thanks.

 

[jackmell]

Know what, I didn't do it all but doesn't look like the regular way would be too hard.
$$4\mathop\iint\limits_{\text{1/4 my ellipse}} \left(z_1(x,y)-z_2(x,y)\right) dydx$$

and since you have the equation of the ellipse:

$$x^2+4y^2=1$$

not hard to get the upper limit on y right? Then x just goes from one side to the other.

What happens when you do that?

 

[zonedestruct]

Know what, I didn't do it all but doesn't look like the regular way would be too hard.
$$4\mathop\iint\limits_{\text{1/4 my ellipse}} \left(z_1(x,y)-z_2(x,y)\right) dydx$$

and since you have the equation of the ellipse:

$$x^2+4y^2=1$$

not hard to get the upper limit on y right? Then x just goes from one side to the other.

What happens when you do that?

is the limits for y: 0<= y < = (√(1-x^2))/2 ?? and then for x it is 0<=x<=1

 

[jackmell]

Yes.

 

[zonedestruct]

thanks jackmell i really like how you split it up to a quarter and multiplied by 4 to take advantage of the symmetry. NOw the integral is not as complicated as the one i initially had when i went from -(√(1-x^2))/2 <= y <= (√(1-x^2))/2