# Triple Integral of $2y+x^2$

Find the volume of the solid region that is bounded above by the surface $f(x,y)=2y+x^2$, bounded below by the $xy$-plane and lies over the region in the $xy$-plane given by $[2,4]\times[-3,2]$ BE CAREFUL!!

This seems really basic, is it really as easy as:

$$\int_2^4\int_{-3}^2\int_0^{2y+x^2} 2y+x^2 dz dy dx$$

This seems to easy and i was told "Be careful!!!" so I think I may be missing something.

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Hint: You only need to integrate in two dimensions. The naive approach would be $$\int_2^4\int_{-3}^2 2y+x^2 dy dx$$ The "be careful" comes because some of the volume is below the $xy$ plane (for example,the point $(2,-3)$ and an area around it) so you need to work harder on the limits of integration.
@user73064: no, the bounds are not a nice rectangle. Note that $(2,-2)$ is in the area of integration. If you are going to integrate in $y$ first (as this is written, and is a good idea) you consider $x$ fixed for that integral. The limits in $y$ can depend on $x$. So you should solve $2y+x^2\ge 0 \cap y \in [-3,2]$ for $y$ in terms of $x$ and use that for your limits. –  Ross Millikan Apr 17 '13 at 17:44
@user73064: you will have to split the range in $x$. For some of it, the lower limit on $y$ needs to be modified. For some of it, not, because $2y+x^2 \ge 0$ for all the $y \in [-3,2]$ –  Ross Millikan Apr 17 '13 at 17:50
@user73064: that is the right idea, but when $x \ge 2\sqrt 3$ you don't change the lower limit. So you want $\left(\int_2^{2\sqrt 3}dx\int_{-x^2/2}^2dy + \int_{2\sqrt 3}^4 dx \int_{-3}^2 dy\right)2y+x^2$ –  Ross Millikan Apr 18 '13 at 4:20
The volume of an element is given by $\iiint dxdydz$. On $z$-axis, it is bound below by $0$ and on top by $2y + x^2$ so your volume is given by $$\int_2^4 \int_{-3}^2\int_0^{2y + x^2}dzdydx$$