4
$\begingroup$

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 ?

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

Thanks

$\endgroup$
1
  • $\begingroup$ 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
3
$\begingroup$

Answer:

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

$\endgroup$
9
  • $\begingroup$ 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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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