Find the volume of the region in the first octant underneath the paraboloid $$z = 1 - \frac{x^2}{81} - \frac{y^2}{36}$$

I've been given the hint to use the change of variables $x = 9rcos(\theta)$ and $y = 6rsin(\theta)$

I know that $x > 0, y > 0, z > 0$ since we're in the first octant. Still not quite sure how to go about finding the limits, and how to use the substitution.

  • $\begingroup$ Well done you have made by yourself a good work only from a few hints (+1)! $\endgroup$ – user Mar 6 '18 at 11:03


  • draw the region by plot on $x-y$, $z-x$, $z-y$ plane and find the limit of integration

  • set up the integral in cartesian coordinates $\iiint_V 1 dxdydz$

  • set up the integral in polar coordinates $\int_0^{\frac{\pi}2}\int_0^{1}\int_0^{r(z)} |J| dr\,dz\,d\theta$

  • $\begingroup$ How do you find $r(z)$? I guess the jacobian is $r$ which means the integrand is also $r$? $\endgroup$ – Pame Mar 5 '18 at 23:35
  • $\begingroup$ @Pame in polar coordinates the region becomes $z=1-r^2$ thus $r=\sqrt{1-z}$. Jacobian is r for ordinary change in polar coordinates here we have some constant term more, evaluate |J| by definition. $\endgroup$ – user Mar 5 '18 at 23:48
  • $\begingroup$ Okay so i assume the jacobian must be $54r$ based on the substitutions in the hint. So the integrand becomes $54r^2$ which after computing the triple integral gives me the answer $6\pi$ which seems to be incorrect. $\endgroup$ – Pame Mar 6 '18 at 10:36
  • $\begingroup$ @Pame why $54r^2$? it should be simply $54 r$ if I'm not wrong $\endgroup$ – user Mar 6 '18 at 10:45
  • $\begingroup$ Yup, $54r$ was correct. Btw, why was the integrand 1 to begin with? How did we actually apply the substitutions? $\endgroup$ – Pame Mar 6 '18 at 11:00

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