0
$\begingroup$

This is probably a very easy/silly question, but still I'm not sure about it.

I want to calculate the volume of a body bound between the graph of $x^2+y^2-z^4=1$ (what does it look like?) and the planes $z=-5, z=5$.

I thought I would use the following limits:

$-\sqrt{1-x^2+z^4}\leq y \leq \sqrt{1-x^2+z^4}$

$-\sqrt{1+z^4}\leq x \leq \sqrt{1+z^4}$

$-5\leq z\leq 5$

But when integrating I get an undefined value.

What is wrong?

$\endgroup$

1 Answer 1

0
$\begingroup$

The $x^2+y^2$ in the equation and the simple $z$-limits hint at using cylindrical polar coordinates. Letting $x=r\cos \theta$ and $y=r\sin\theta$ yields the equations $z=-5$, $z=5$ and $r^2-z^4=1$ as describing the boundaries of the region of integration.

To find the limits, note that $z$ varies from $-5$ to $5$; for fixed $z$, $r$ varies from $0$ to $\sqrt{1+z^4}$, and $\theta$ varies from $0$ to $2\pi$. So your integral is given by $$\int_{-5}^5 \int_0^{\sqrt{1+z^4}} \int_0^{2\pi} r\, d\theta\, dr\, dz$$ You can take it from here.

$\endgroup$
2
  • $\begingroup$ Oh yes, how silly of me. Thank you very much! $\endgroup$
    – Whyka
    Commented Dec 21, 2015 at 16:52
  • $\begingroup$ And my first solution was also right, I just messed up the calculation of the integral. Better use cylindrical coordinates! $\endgroup$
    – Whyka
    Commented Dec 21, 2015 at 17:11

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .