# What is wrong with the limits of this triple integral?

I need to integrate using cilindrical coordinates some function on the region $E$ under the paraboloid $z = 4 - x^2 - y^2$ and the first octant, so I say $$E = \{ (r, \theta, z): 0 \leq r \leq 4 - z, 0 \leq \theta \leq \frac{\pi}{2}, 0 \leq z \leq 4 \}.$$ But I integrate and my answer is not the one provided in the book. I checked the integral with Wolfram Alpha and the my result is correct, so I have to have something wrong in the way I determined $E,$ but have no Idea what could be. Can someone help me?

• We can't actually help you until you tell us what the answer in your book is and what is the integral you set up. – fixedp Apr 9 '14 at 21:48
• How did you come up with the bounds on $r$? The upper one is not correct. In fact, it might be simpler if you integrate with respect to $z$ on the inside instead of $r$. – Santiago Canez Apr 9 '14 at 21:50
• Guys, the OP did state his region of integration: "under the paraboloid $z=4-x^2-y^2$ in the first octant". – Bill Cook Apr 9 '14 at 21:53

Your relation: $r=4-z$ is incorrect. If $z=4-x^2-y^2$, then $z=4-r^2$ so that $r^2=4-z$ and thus $r=\sqrt{4-z}$.
Anyway, it is usually better (and easier) to set up the bounds beginning with $z$ instead of $r$. So you know that the "bottom" of your region is determined by the $xy$-plane (i.e. $z=0$) and the "top" is determined by $z=4-x^2-y^2=4-r^2$. Thus your $z$-bounds are $z=0$ and $z=4-r^2$.
Next, intersect the top and bottom to find bounds for $r$: $0=z=4-r^2$ implies $r=2$. So $0 \leq r \leq 2$ and then you've got the angles right (since you're operating in the first octant - containing to the first quadrant of the $xy$-plane.
So your bounds are: $0 \leq z \leq 4-r^2$, $0 \leq r \leq 2$, and $0 \leq \theta \leq \pi/2$. You were close. :)