# Volume of paraboloid $z = x^2+y^2$ with heigth $h$

I am asked to find the Volume of paraboloid $z = x^2+y^2$ with heigth $h$. How would be the best way to approach that problem (cartesian/cylindrical)?

My reasoning using cylindrical coordinates doesn't seem to work, is there a particular reason?

$$\int_{0}^{2\pi} \int_{0}^{\pi/2} \int_{0}^{h} r^2 \ r \ dz dr d\theta$$

Textbook answer: $\frac{\pi h^2}{2}$

First, note that we want to find a volume. Volume is always $V=\iiint \; dV$. We just need to set this up. You had the right idea of using cylindrical coordinates. So thus far we have $\iiint r \; dzdrd\theta$. Notice that for our region, $z$ always 'starts' at the paraboloid and continues up until we hit the plane (the picture should help you see this). So $z$ runs from $x^2+y^2=r^2$ up to $z=h$. Then you have already correctly noted in the comments the radius ranges from $0$ to $\sqrt{h}$. This gives $$V= \int_0^{2\pi} \int_0^\sqrt{h} \int_{r^2}^h r \; dz\;dr\;d\theta$$ which gives the correct answer.
Note: My initial unit analysis was a careless cursory look at the problem. While $h^2$ has 'units' meters$^2$, the constant $\frac{\pi}{2}$ (really the $\frac{1}{2}$ portion) comes from something that had units - namely meters. So the answer has units meters$^3$, which does represent a volume.
• but according to wolfram the volume of a paraboloid with radius $a$ and height $h$ is $\frac{\pi a^2 h}{2}$, but in this case when the height is $h$, the radius is $\sqrt{h}$, so shouldn't the volume be $\frac{\pi h^2}{2}$ ? Commented May 20, 2016 at 4:07
• @NeisySofíaVadori Because the bottom of the volume is given by the paraboloid. So the 'minimum' $z$'s are the ones along the paraboloid and the 'maximum' $z$'s are given by the sheet $z=h$. Commented Jul 10, 2018 at 21:13
This problem only requires single variable calculus. Note that the paraboloid exhibits radial symmetry. Consider the shapped formed by rotating a parabola in 2d space around the y axis, we clearly attain a paraboloid, as it is a solid of revolution. Using disk integration with $y = x^2$ \begin{align} V &= \int_0^h \pi x^2 dy\\ &= \int^h_0 \pi x^2 \frac{dy}{dx} dx\\ &= 2\pi\int^\sqrt{h}_0 x^3dx\\ &= \frac{2\pi x^4}{4} \bigg|_0^\sqrt{h} \\ &= \frac{\pi h^2}{2} \end{align}