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 ?

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


  • $\begingroup$ 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 $R=\sqrt{3}$? $\endgroup$ – dato datuashvili Dec 7 '13 at 13:51


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

  • $\begingroup$ Why is it the reverse order in integrand ? The paraboloid isn't under of the plane ? $\endgroup$ – Ewin Dec 7 '13 at 13:56
  • $\begingroup$ Oops, I changed it $\endgroup$ – Satish Ramanathan Dec 7 '13 at 13:58
  • $\begingroup$ @satishramanathan can we express it using circle area? $\endgroup$ – dato datuashvili Dec 7 '13 at 13:58
  • $\begingroup$ Ok. Anyway Thank you for helping : ) $\endgroup$ – Ewin Dec 7 '13 at 13:58
  • $\begingroup$ @Ewin i tried to use circle method,i think it is wrong right? $\endgroup$ – dato datuashvili Dec 7 '13 at 13:59

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