Volume of Solid Region Between Sphere and Paraboloid
Taking the equation $z=4x^2y^2$, we can rewrite it as $x^2+y^2=4z$. We can now substitute this value for $(x^2+y^2)$ into the equation $x^2+y^2+z^2=6$, we obtain:
$(4z)+z^2=6$,
or
$z^2z2=0$
This leads to the solutions $z=1$ and $z=2$. Intersecting these with our sphere, we obtain two circles: $x^2+y^2=5$ (when we plug in $z=1$), and $x^2+y^2=2$ (when we plug in $z=2$). The region is described as being "above the sphere", so I'd use the top one. When you want to integrate, your bounds on $r$ will be $0$ to $\sqrt{2}$.
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Alex over 1 year
"Find the volume of the solid region above the sphere $x^2+y^2+z^2 = 6$ and below by the paraboloid $z = 4x^2y^2$"
I am, of course, going to be solving this double integral by converting to polar coordinates. My problem lies mainly in determining the bounds I will use. If I take the equations as: $$z_1 = \sqrt{6x^2y^2} (sphere) $$$$z_2 = 4x^2y^2 (paraboloid)$$ Then my answer should be in the form: $$ V = \int _D\int(z_1z_2)dA$$
Of course, my leftmost integral will be $\int_0^{2\pi}$, since I am setting part of my domain as $0<=\theta<=2\pi$
I took the domain for my radius as the intersection between $z_1 and z_2$: $$4x^2y^2 = \sqrt{6x^2y^2}$$ Which, in the end, gives: $$(4x^2y^2)(3x^2y^2)=0$$ And if I take $r^2 = x^2+y^2$: $$(4r^2)(3r^2)=0$$
So, $$D = [(r, \theta)  0 <= \theta <= 2\pi, 2 <= r <= \sqrt{3}]$$
Something about this seems off, but I'm not entirely sure what I'm overlooking.Am I on the right track, or have I missed some critical step or piece of information?

colormegone almost 9 yearsShouldn't you have the equation $ (4r^2)^2 \  \ (6r^2) \ = \ 0 \ $ ? Since you have gone over to polar coordinates (since the volume has axial symmetry), the "shell method" would be appropriate for calculating the volume.

G Tony Jacobs almost 9 yearsThe paraboloid intersects the sphere in two distinct circles. You should figure out which of your solutions for $r$ corresponds to a positive $z$value, and use that one.

G Tony Jacobs almost 9 yearsLooking at a graph will help, certainly.
