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

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


1 Answer 1


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,

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


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