coordinate systemsmultiple integral
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: $\frac{\pi h^2}{2}$
Using cylinder coordinate is the best way to solve this problem : $$\left\{(r\cos\theta ,r\sin\theta +a ,z)\mid \theta \in [0,2\pi], r\in[0,a], z\in \left[0,\frac{2r^2+2ar\sin \theta+a^2}{a}\right]\right\}.$$
Since the graph of $r = 2\sin \phi$ for $\phi \in [\pi,2\pi]$ overlaps with that for $\phi \in [0,\pi]$,
graph from spark notes
the upper limit for $\phi$ should be $\pi$ instead of $2 \pi$.
\begin{align} V &= \int_{0}^{\pi}\int_{0}^{2\sin\phi}\int_{0}^{2r^2}\,dz\,rdr\,d\phi \\ &= \int_{0}^{\pi}\int_{0}^{2\sin\phi}2r^3 dr\,d\phi \\ &= \int_{0}^{\pi} \frac{(2 \sin \phi)^4}{2} \,d\phi \\ &= 8\int_{0}^{\pi} \sin^4 \phi \,d\phi \\ &= 16 \int_{0}^{\pi/2} \sin^4 \phi \,d\phi \quad (\text{symmetry of sine}: \sin\left(\frac\pi2+\phi\right) = \sin\left(\frac\pi2-\phi\right)) \\ &= 16 \cdot \frac{3 \cdot 1}{4 \cdot 2} \cdot \frac{\pi}{2} \quad \text{(Wallis' formula)} \\ &= 3\pi \end{align}
Best Answer
This is a bit 'tricky' as it is the 'opposite' of what a student often thinks to do. You want the volume of the paraboloid piece but what you are calculating is really more the stuff 'underneath' the paraboloid.
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.