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

Use polar coordinates to find the volume of the given solid bounded by the paraboloid $z=1+2x^2+2y^2$ and the plane $z=7$ in the first octant.

I did it. Is that right ?

$$\int_0^{\pi \over 2} \int_0^{\sqrt{3}}(7-(1+2r^2))r dr d\theta = \frac{9\pi}{4} $$


share|cite|improve this question
but can't we do like this? $2*x^2+2*y^2+1=7$ so we have $x^2+y^2=6$ or $x^2+y^2=3$ from which we get $R=\sqrt{3}$? – dato datuashvili Dec 7 '13 at 13:51
up vote 2 down vote accepted


You are almost right except that the integrand is ($7-(1+2r^2)$) = $(6-2r^2)$

share|cite|improve this answer
Why is it the reverse order in integrand ? The paraboloid isn't under of the plane ? – Ewin Dec 7 '13 at 13:56
Oops, I changed it – satish ramanathan Dec 7 '13 at 13:58
@satishramanathan can we express it using circle area? – dato datuashvili Dec 7 '13 at 13:58
Ok. Anyway Thank you for helping : ) – Ewin Dec 7 '13 at 13:58
@Ewin i tried to use circle method,i think it is wrong right? – dato datuashvili Dec 7 '13 at 13:59

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.