Volume between paraboloid and sphere

Question

I need to find the volume between a parabola and a sphere given by $2-2x^2-2y^2=z$ for the paraboloid and $z \ge 2-2x^2-2y^2$, $x^2+y^2+z^2=1$ for the sphere.

The intersection point between the sphere and parabola finds place at $z=\frac{1}{2}$ therefore $\frac{1}{2} \le z \le 2$. This is where I think I made my mistake, at finding my interval for $r$

I got $r^2=1-z^2$ for the equation of the paraboloid and $r^2=\frac{2-z}{2}$ for the equation of the sphere

therefore $\sqrt{1-z^2}\le r\le \sqrt{\frac{2-z}{2}}$

for theta I got $0\le \theta\le 2\pi$

Now the integral, which gave me the wrong answer

$$\int _{\frac{1}{2}}^2\int _0^{2\pi }\int _{\sqrt{1-z^2}}^{\sqrt{\frac{2-z}{2}}}rdrd\theta dz = \frac{27}{16}\pi$$

Answer 1

Approach using usual multiple integrals not using the disk method for anyone interested:

We have $x^2 + y^2 + z^2 = 1$ and $z \leq 2 - 2(x^2+y^2)$ so $z \leq 2 - 2(1-z^2) \implies 2z^2+z \leq 0$ which means $0 \leq z \leq \frac{1}{2}$.

Leveraging symmetry, we transition to cylindrical coordinates $(x,y,z) \mapsto (r\cos\theta, r\sin\theta, z)$. $\theta$ limits are standard $0 \leq \theta \leq 2\pi$. $r$ limits are a bit tricky. You notice that between $0 \leq z \leq \frac{1}{2}$, the parabola is inside the sphere, so the $r$ is bounded below by the parabola and bounded above by the sphere. So $\sqrt{1 - \frac{z}{2}}\leq r \leq \sqrt{1 - z^{2}}$ (since that's the region between $r^2+z^2\leq 1$ and $z+2r^2 \geq 2$ i.e between the parabola and the sphere). So the integral becomes:

$$\int\limits_{0}^{\frac{1}{2}}\int\limits_{0}^{2 \pi}\int\limits_{\sqrt{1 - \frac{z}{2}}}^{\sqrt{1 - z^{2}}} r\, dr\, d\theta\, dz = \frac{\pi}{48}$$

Answer 2

Given $z \ge 2-2x^2-2y^2$, the integration range for $z$ is $0\le z\le \frac 12$. With the disk method, the enclosed volume is \begin{align} V= \int_{0}^{\frac12} \pi( r_2^2-r_1^2)dz = \pi \int_{0}^{\frac12} \left[(1-z^2)-(1-\frac12z)\right]dz=\frac\pi{48} \end{align}