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Problem:

Find the volume of the solid which is below the paraboloid $z=x^{2}+y^{2}$ and above of the region bounded by $y=x^{2}$ and $x=y^{2}$. Draw the the solid.

Solution:

First, I have tried to draw the given surfaces:

Surfaces

But I don't how to define the integral to find such volume of the described segment.

Ideas, suggestions,...?

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    $\begingroup$ Nice drawing :) (+1) $\endgroup$ – iamvegan Sep 30 '16 at 23:57
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The graph of the region of integration in the $xy$ plane is shown below. I will let you draw the 3D graph of the solid lying above the region and below the graph of $z=x^2+y^2$

The volume is found by integrating over the region

$$ \int_0^1\int_{x^2}^{\sqrt{x}}x^2+y^2\,dy\,dx$$

which I assume you can handle.

Region bounded by y=x^2 and x=y^2

I edited this answer to replace my poor hand drawn image of the solid with a wire frame image constructed using the free Geogebra software. Solid under a curve

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  • $\begingroup$ Do I need to revolutionize that marked area to create the requested solid? $\endgroup$ – InfZero Oct 1 '16 at 17:53
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    $\begingroup$ No, it is not a solid of revolution. The shaded region is the base of the solid and lies in the $xy$ plane. The top of the solid is that portion of the paraboloid $z=x^2+y^2$ lying above the shaded region. The solid is a type of 'cylinder' but the top, instead of being congruent to the base is the portion of $z=x^2+y^2$ lying above the base. The volume must be computed using the double integral shown in the answer and you should get a resulting volume of $\frac{6}{35}$. I can edit my answer to include those steps if you like. $\endgroup$ – John Wayland Bales Oct 1 '16 at 17:58
  • $\begingroup$ Thanks. I can solve the double integral, but now I have some difficulties to image/draw the solid. $\endgroup$ – InfZero Oct 1 '16 at 18:00
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    $\begingroup$ Yes, I can imagine the shape but I have no artistic ability. The height is $0$ at the origin and $2$ at $(1,1)$ and the top is shaped somewhat like the base but is curved instead of flat because it follows the contours of the paraboloid. $\endgroup$ – John Wayland Bales Oct 1 '16 at 18:05
  • $\begingroup$ This is what imagine with a piece a paper: dropbox.com/s/czd2ltpje3a531e/… $\endgroup$ – InfZero Oct 1 '16 at 18:31

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