Integrate over the region in the first octant above the parabolic cylinder and below the paraboloid I could not get the limits right even that I tried many one but I still could not get it

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

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.

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