# Triple integral for Volume of a region

$Q$ is the solid bounded by the plane $x+2y+2z=2$ and above paraboloid $x=z^2+y^2$. Setup the triple integral for the volume of $Q$.

I tried to find vertices and go from there bu t got confused. I need help with this problem.

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Have you tried drawing a picture? –  wj32 Nov 5 '12 at 1:48

When I approach a problem like this I consider a point $(x,y,z)$ and work towards discovering a set of 3 inequalities which describe an arbitrary point in the given region. The paraboloid $x=z^2+y^2$ opens into the positive $x$-half volume. On the other hand $x = 2-2y-2z$ intersects the paraboloid when $z^2+y^2=2-2y-2z$ which is also written as $(z+1)^2+(y+1)^2=4$. Observe $z^2+y^2 \leq x \leq 2-2y-2z$ where $y,z$ are bounded by $0 \leq (z+1)^2+(y+1)^2 \leq 4$. We can unwrap the $y,z$ inequality as $(z+1)^2 \leq 4-(y+1)^2$ or $-\sqrt{4-(y+1)^2} -1 \leq z \leq \sqrt{4-(y+1)^2} -1$ where $-3 \leq y \leq 1$. To summarize:
1. $-3 \leq y \leq 1$ puts the point $(x,y,z)$ between the $y=-3$ and $y=1$ plane.
2. $-\sqrt{4-(y+1)^2} -1 \leq z \leq \sqrt{4-(y+1)^2} -1$ places $(x,y,z)$ inside the cylinder $(z+1)^2+(y+1)^2=4$.
3. $z^2+y^2 \leq x \leq 2-2y-2z$
Also, pragmatically speaking, if you were to calculate this then I would urge you to use polar coordinates $y+1=r\cos \theta$ and $z+1 = r\sin \theta$ so the $y,z$ bounds are replaced with $0 \leq r \leq 2$ and $0 \leq \theta \leq 2\pi$. Then only the $x$ bounds would involve variables.