# Finding the bounds to solve for the volume using cylindrical coordinates

My goal is to set up the triple integral that will solve for the volume inside the sphere $$x^2 + y^2 + z^2 = 2z$$ and above the paraboloid $$x^2 + y^2 = z$$ using cylindrical coordinates.

Upon solving, I know that the intersection of the sphere and the paraboloid is r = 1 and the bounds for $$\theta$$ is $$0 \leq \theta \leq 2\pi$$.

However, I am uncertain for the bounds of the z variable. I solve for the value of z in converted equation of the sphere $$r^2 + z^2 = 2z$$. By quadratic formula, I got the value of

$$z = \frac{2 \pm \sqrt{4 - 4r^2}}{2}$$

Should I only include $$z = \frac{1}{2} (2 + \sqrt{4-4r^2})$$ as one of the bounds in the z variable? Or there will be a split integral? My yielding triple integral in this case if I only include $$z = \frac{1}{2} (2 + \sqrt{4-4r^2})$$ would be

$$\int_0^{2\pi} \int_0^1 \int_{r^2}^{\frac{1}{2}(2+\sqrt{4-4r^2})} r dzdrd\theta$$

Yes you are correct. Paraboloid $$z = x^2 + y^2 = r^2$$ and Sphere $$r^2 + z^2 = 2z$$

To find $$z$$ at intersection, plug in $$r^2 = z$$ from equation of paraboloid into the equation of sphere. We get $$z^2 - z = 0 \implies z = 0, 1$$. So at intersection above origin, $$z = 1$$. We also find that at intersection $$r = 1$$ which is the radius of the sphere too.

Now as you mentioned, from the equation of the sphere, $$r^2 + (z-1)^2 = 1 \implies z = 1 \pm \sqrt{1-r^2}$$

As the sphere is centered at $$(0, 0, 1)$$, for any $$0 \leq r \lt 1$$, there are two $$z$$ values on sphere - one below $$z = 1 \ , (z = 1 - \sqrt{1-r^2})$$ and one above $$z = 1 \ , (z = 1 + \sqrt{1-r^2})$$. The intersection of paraboloid and sphere is also at $$z = 1$$, $$z$$ is bound below by paraboloid and above by sphere for $$0 \leq r \leq 1$$. So the point that you should take on sphere as upper bound is $$z = 1 + \sqrt{1-r^2}$$. Hence it should be,

$$\displaystyle \int_0^{2\pi} \int_0^1 \int_{r^2}^{1 + \sqrt{1-r^2}} r \ dz \ dr \ d\theta$$

Edit: Now knowing that intersection is at $$r = 1, z = 1$$, the combined volume is volume of paraboloid between $$0 \leq z \leq 1$$ and half of the unit sphere above $$z = 1$$. So if you are allowed to use basic formula for volume, you can use the fact that volume of half unit sphere is $$\frac{2 \pi }{3}$$. In that case, the volume can be obtained as,

$$\displaystyle \int_0^{2\pi} \int_0^1 \int_{r^2}^1 r \ dz \ dr \ d\theta + \frac{2 \pi}{3}$$