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I need to calculate the volume enclosed by: $$x^2 + y^2 = z, \space y = x^2, \space z=0, \space y = 1$$

The shape of the volume I get gets me confused. It is a paraboloid ($x^2 + y^2 = z$) intersected with cylinder ($y = x^2$) and limited by specific $z$ and $y$ plains. When I tried drawing this I saw that the volume is not limited by the "upper" $z$ plain, therefore it seems to be infinite. Did the lecturer provide us with "wrong" conditions, so the volume is infinite?

Am I right? If yes, how can I calculate the volume if I change my previous condition ($z = 0, \space y = 1$) to $0\le z \le 1$? I tried approaching this "updated" problem, but also didn't have any luck.

Any help would be appreciated.

EDIT: The answer including the integral solution was posted - see below. The whole problem was caused by me thinking about the volume "inside" the paraboloid, while the task was to calculate it "outside", enclosed by the cylinder.

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  • $\begingroup$ Image for illustration. $\endgroup$ – user202729 Jun 25 '18 at 16:19
  • $\begingroup$ Sorry for the inconvenience, I did it the correct way now; how do you create plots like this? Is this Wolfram? This can be really useful for me in the future. $\endgroup$ – Sqoshu Jun 26 '18 at 8:09
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    $\begingroup$ Yes, Wolfram Mathematica. They has a free online version at sandbox.open.wolframcloud.com , or tio.run/#mathematica (doesn't support image output unfortunately) $\endgroup$ – user202729 Jun 26 '18 at 8:12
  • $\begingroup$ Thanks for your input, that is really useful. $\endgroup$ – Sqoshu Jun 26 '18 at 8:16
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Look first at the $xy$-plane (the bottom). The condition limits the area $D$ between $y=x^2$ and $y=1$. It is bounded in $(x,y)$. Now look at what happens along the vertical $z$ axis. It says: take those points $(x,y,z)$ that are between $z=0$ and $z=x^2+y^2$. The set (and the volume) is finite, it is between two surfaces ($xy$ plane and the paraboloid).

enter image description here

Try to split integration as $$ \iint_D\int_{z=0}^{z=x^2+y^2}\,dz\,dxdy. $$

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  • $\begingroup$ My first try is something like: $\int_{-1}^{1} dx \int_{0}^{x^2} dy \int_{0}^{x^2 + y^2} dz$. $\endgroup$ – Sqoshu Jun 25 '18 at 10:58
  • $\begingroup$ @Sqoshu The second part is wrong. It is not really between $y=x^2$ and $y=1$ (i.e. above the parabola), you got it below the parabola between $y=x^2$ and $y=0$. $\endgroup$ – A.Γ. Jun 25 '18 at 11:01
  • $\begingroup$ If I think correctly the $y$ value changes from $x^2$ to $1$, so shall the integral be "reversed" like: $\int_{x^2}^{1}dy$? $\endgroup$ – Sqoshu Jun 25 '18 at 11:07
  • $\begingroup$ Does this "step by step" approach always work with integrals like this (going thru $(x,y)$, then focusing on $(x,y,z)$)? It seems much more logical now. $\endgroup$ – Sqoshu Jun 25 '18 at 11:10
  • $\begingroup$ @Sqoshu Typically, there are two principles to split a triple integration to iterated: (1) double-single, and (2) single-double. Here we applied the first one (it looks like summing up a spaghetti looking small parts to the whole), but the second one is often used as well, it reminds slicing a bread. The choice between them depends most on how the set description looks like. $\endgroup$ – A.Γ. Jun 25 '18 at 11:20
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First of all the volume is not infinite because it is bounded at $z=2$.
Draw the figure on a $xy$-plane at $z=t$ where $0\leq t\leq2 $ you will see that the figure gets closed.
$\int_{0}^{2}S(t)dt$ where $S(t)$ is the area enclosed by $y=1,y=x^2$ and $x^2+y^2=t$.
Keep in mind that you will have to consider two cases separately when $0\leq t\leq1 $ and $1\leq t\leq2 $.

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  • $\begingroup$ Why should I consider the two cases separately? What changes at $t = 1$? $\endgroup$ – Sqoshu Jun 25 '18 at 10:50
  • $\begingroup$ The area is divided in two pieces. $\endgroup$ – Chris2018 Jun 25 '18 at 10:52
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Thanks to help from A.Γ. and Chris2006:

It turns out that the volume is enclosed and can be calculated. The answer was accepted. Using the hints given there: $$\int_{-1}^{1} dx \int_{x^2}^{1} dy \int_{0}^{x^2 + y^2} dz = \int_{-1}^{1} dx \int_{x^2}^{1} (x^2 + y^2) dy = \int_{-1}^{1} (x^2 y + \frac{y^3}{3})_{y =(x^2, 1)} dx= \int_{-1}^{1} (x^2 + \frac{1}{3} - x^4 - \frac{x^6}{3})dx = (\frac{x^3}{3} + \frac{1}{3} x - \frac{x^5}{5} - \frac{x^7}{21})_{x =(-1, 1)} = \frac{88}{105}$$

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