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

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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.

enter image description here

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  • $\begingroup$ 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 $\frac{\pi h^2}{2}$ ? $\endgroup$ Commented May 20, 2016 at 4:07
  • $\begingroup$ @JeevanDevaranjan I read the problem far too quickly, took a look at the answer you posted, did a unit analysis, and ran with it. The answer has been corrected and uses the method you were asking about. Good luck with your studies! $\endgroup$ Commented May 20, 2016 at 4:41
  • $\begingroup$ Can you explain me please why z moves from r^2 @mathematics2x2life? $\endgroup$ Commented Jul 10, 2018 at 20:31
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    $\begingroup$ @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 $z=h$. $\endgroup$ Commented Jul 10, 2018 at 21:13
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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}

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