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Integrate enter image description here over the region in the first octant enter image description here above the parabolic cylinder enter image description here and below the paraboloid enter image description here

I could not get the limits right even that I tried many one but I still could not get it

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  • $\begingroup$ The region in the $xy$-plane is given by the equation $ x^2 + y^2 = 4 $. $\endgroup$ – Mhenni Benghorbal Mar 27 '14 at 3:16
  • $\begingroup$ alright but I still have not got how we could find the limits $\endgroup$ – user131040 Mar 27 '14 at 3:29
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We can set up the integral as follows...

$$ \iiint F(x,y,z)dzdydx$$

For the z integration or bounds it is simply from the lower surface ($z=y^2$) to the upper surface $(z=8-2x^2-y^2)$ So now we have...

$$ \iint \!\!\int_{y^2}^{8-2x^2-y^2}(8xz) dzdydx $$

Now, we can think of the integral as being resolved onto the xy-plane with z=0. Setting the two functions of x and y equal to each other and simplifying, we get $x^2+y^2=4$. Now, trying to find the y bounds we solve to $y$ in terms of $x$. Thus, $y=\sqrt{4-x^2}$. Next, we need to integrate along the x axis where $y=0$. This is from $0$ to $2$. And our integral ends up being:

$$\int_0^2\!\!\!\int_0^{\sqrt{4-x^2}}\!\!\int_{y^2}^{8-2x^2-y^2}(8xz)dzdydx$$

It has been a while since I have done these so please correct me if there is an error or a better way to do it.

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  • $\begingroup$ thank you i it the right answer $\endgroup$ – user131040 Mar 27 '14 at 4:42

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