Please help if you know how to solve it. Thank you very much.
The transformation x=au, y=bv,z=cw (a>0 , b> 0) can be rewritten as x/a =u, y/b =v, z/c=w and hence it maps the circular region u^2+v^2+w^2=1
into the elliptical region x^2/a^2 +y^2/b^2 +z^2/c^2 < or = to 1.
In this excercise, perform the integration by transforming the ellipsoidal region of integration into a spherical region of integration and then evaluating the transformed integral in spherical coordinates.
triple integral of G of x^2 dv
Where G is the region in the first quadrant enclosed by the ellipsoid 9x^2 + 4y^2 +z^2 =36.
Below is my partial solution to this question
since 9x^2 + 4y^2 +z^2 =36.
dividing both sides by 36 : x^2/4 + y^2/9 +z^2/36 =1
Let u^2 =x^2/4 so x=2u
v^2 =y^2/9 so y=3v
w^2 =z^2/36 so z=6w
In spherical coordinates , rho is from 0 to 1 , phi is from 0 to pi, theta is from 0 to pi/2
also the Jacobian = 36
I got stuck in theta is from 0 to pi/2 .
I think theta should be from 0 to pi/2 because Where G is the region in the first quadrant enclosed by the ellipsoid 9x^2 + 4y^2 +z^2 =36.
* but I have the clue from my teacher that theta is from 0 to 2pi.
I don't understand why theta is from 0 to 2pi not from 0 to pi/2. Please tell me if you know why. Thank you very much.
If you can please show me how to set up the integral. Thank you very much.