Hi,

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.

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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

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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.

2. Originally Posted by kittycat

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.
i don't see why your professor said that, it seems to me $0 \le \theta \le \frac {\pi}2$ as well.

what exactly did he say? did you try the problem with your values to see if you get the correct answer? (you have the answer in the back of the book, right?)