Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

$Q$ 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.

Thanks in advance.

share|cite|improve this question
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. $z^2+y^2 \leq x \leq 2-2y-2z$

When setting up the volume integral you want to put the numerical bounds on the outside and the double-variable inequality on the inside. Remember the integral ,if it exists, is a number so the bounds must be ordered so that the result is a number.

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.

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.