"Find the volume of the solid region above the sphere $x^2+y^2+z^2 = 6$ and below by the paraboloid $z = 4-x^2-y^2$"
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: $$z_1 = \sqrt{6-x^2-y^2} (sphere) $$$$z_2 = 4-x^2-y^2 (paraboloid)$$
Then my answer should be in the form: $$ V = \int _D\int(z_1-z_2)dA$$
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 $z_1 and z_2$: $$4-x^2-y^2 = \sqrt{6-x^2-y^2}$$
Which, in the end, gives: $$(4-x^2-y^2)(3-x^2-y^2)=0$$
And if I take $r^2 = x^2+y^2$: $$(4-r^2)(3-r^2)=0$$
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?
Best Answer
Taking the equation $z=4-x^2-y^2$, we can rewrite it as $x^2+y^2=4-z$. We can now substitute this value for $(x^2+y^2)$ into the equation $x^2+y^2+z^2=6$, we obtain:
$(4-z)+z^2=6$,
or
$z^2-z-2=0$
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 $\sqrt{2}$.