1
$\begingroup$

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.

$\endgroup$
  • $\begingroup$ Well done you have made by yourself a good work only from a few hints (+1)! $\endgroup$ – user Mar 6 '18 at 11:03
0
$\begingroup$

HINT

  • 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 $\int_0^{\frac{\pi}2}\int_0^{1}\int_0^{r(z)} |J| dr\,dz\,d\theta$

$\endgroup$
  • $\begingroup$ 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

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.