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Use double integrals to calculate the volume of the following region.

The solid beneath the cylinder $z=y^2$ and above the region $R = \{ (x,y) : 0 \leq y \leq 1, y \leq x \leq 1 \}$.

I just need help setting the integral up, I can calculate the volume.

Thank you.

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  • $\begingroup$ Imagine the square between the $ \ x-$ axis and $ \ y = 1 \ $ and between the $ \ y-$ axis and $ \ x = 1 \ . $ Add the diagonal line $ \ y = x \ . $ Since we have the inequality $ \ y \ \le \ x \ \le \ 1 , $ the region in the $ \ xy-$ plane is the "lower right" triangle in the square. The "roof" of the solid is a parabolic cylinder, with the bottom of the parabolic cross-sections sitting on the $ \ x-$ axis; the parabolas "open upward". $\endgroup$ Commented Apr 1, 2014 at 22:31

1 Answer 1

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It is $$\int_0^1\int_y^1y^2dxdy$$Depending on a comment: In the upper triangle values of $y$ is greater than values of $x$ enter image description here

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  • $\begingroup$ I'm still not understanding why it would be the lower right triangle. Let's say i want to use the other triangle. Can i use it and integrate with respect to dy first? 0 to 1 for first integral and x to 1 for second? $\endgroup$ Commented Apr 1, 2014 at 23:09
  • $\begingroup$ Nvm, i think i figured it out. I took the integral of both and got the same volume. $\endgroup$ Commented Apr 1, 2014 at 23:17
  • $\begingroup$ Sorry for inactivity, a added a new part $\endgroup$
    – Semsem
    Commented Apr 2, 2014 at 6:58
  • $\begingroup$ in general they are different you have to use the lower one $\endgroup$
    – Semsem
    Commented Apr 2, 2014 at 7:01

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