Integrating Over an Elliptical Region in Space

[hyper]

Hello, I need to find the volume between the plane z=1 and z=10-x*x-4*y*y.

I have tried using polar coordinates:
first integrate from z=1 to z=10-x*x-4*y*y
The I integrate in the plane z=1, but here I need to integrate over an ellipse, how do I do this? When I wrote the expression to polar coordinates I got an expression for r that I couldn't integrate.

I have also tried integrating over the ellipse using cartesian coordinates, but I got stuck aswell, can someone pleas help?

 

[Cyosis]

Could you show the actual function you have to integrate over? I suspect it has the form \sqrt{9-4y^2}? If so use the substitution u=\frac{3}{2} \sin \theta. Still integrating like this isn't very pretty. A more elegant method and by far the easiest to integrate is to notice that this is a paraboloid with elliptical cross section. You can then slice the paraboloid parallel to the x-y plane in a lot of slices and then integrate from z=1 to 10. The only thing you have to do is find an expression for the semi major axis and the semi minor axis as a function of z and use that the area of an ellipse is given by pi a b, with a the semi major axis and b the semi minor axis.

 

[HallsofIvy]

Could you show the actual function you have to integrate over? I suspect it has the form \sqrt{9-4y^2}? If so use the substitution u=\frac{3}{2} \sin \theta. Still integrating like this isn't very pretty. A more elegant method and by far the easiest to integrate is to notice that this is a paraboloid with elliptical cross section. You can then slice the paraboloid parallel to the x-y plane in a lot of slices and then integrate from z=1 to 10. The only thing you have to do is find an expression for the semi major axis and the semi minor axis as a function of z and use that the area of an ellipse is given by pi a b, with a the semi major axis and b the semi minor axis.

No, finding the volume bounded by z= 1 and z= 10- x^2- 4y^2 involves integrating the difference in z values, (10- x^2- 4y^2)- 1= 9- x^2- y^2[/tex] over the circle x^2+ 4y^2= 9, where the two surfaces cross.<br /> <br /> Now the limits of integration, in xy-coordinates will involve something like y= \frac{1}{2}\sqrt{9- x^2}.<br /> <br /> Hyper, you might consider using parameters r and \theta such that x= r cos(\theta) and y= 4r sin(\theta), <b>not</b> standard polar coordinates, with r going from 0 to 1 and \theta from 0 to 2\pi. Do you know how to find the &quot;differential of area&quot; in r and \theta using the Jacobian?

 

[Cyosis]

Why wouldn't slices be applicable, it yields the same answer?