0
$\begingroup$

I need to find the volume of solid enclosed by the cone $z =\sqrt{x^2+ y^2}$ between the planes $z =1$ and $z =2$

Now using Spherical Coordinates I can set up the integral as:

$\displaystyle \int_{0}^{2\pi}\int_{0}^{\pi/4}\int_{sec\phi}^{2sec\phi} {\rho}^2sin\phi\text{ } d{\rho}\text{ } d{\phi}\text{ } d\theta$

Just for my practice I also want to find this via cylindrical Coordinates. But, I don't understand how should I express the region in terms of $dz$ $dr$

Can anyone please explain to me step by step how should I express this in cylindrical coordinates ?

Thank You.

$\endgroup$
  • $\begingroup$ If you do the order $dz\,dr$, you will need to break it up into two separate integrals ($0\le r\le 1$ and $1\le r\le 2$). But if you do the order $dr\,dz$ then $z$ goes from $1$ to $2$ and $r$ goes from $0$ to $z$ ... $\endgroup$ – Ted Shifrin Oct 8 '19 at 16:25
  • $\begingroup$ @TedShifrin: so, if I do it by the order $dzdr$ the integral setup will be : $\displaystyle\int_{0}^{2\pi}\int_{0}^{1}\int_{0}^{1} rdrdzd\theta + \int_{0}^{2\pi}\int_{1}^{2}\int_{0}^{2} rdrdzd\theta$ Is this correct ? $\endgroup$ – sat091 Oct 8 '19 at 17:00
  • $\begingroup$ That isn't the order $dz\,dr$, is it?!! You want to edit and proofread to make sure? $\endgroup$ – Ted Shifrin Oct 8 '19 at 17:01
  • $\begingroup$ @TedShifrin:: Ah sorry , I actually meant to do it by the order $dr dz$ and not $dzdr$ $\endgroup$ – sat091 Oct 8 '19 at 17:15
  • $\begingroup$ No, as I already said, we are told that $z$ goes from $1$ to $2$, so you're just going to have a single integral. $\endgroup$ – Ted Shifrin Oct 8 '19 at 17:18
1
$\begingroup$

If you insist in cylindrical coordinates you need two integrals $$ \int _0^{2\pi} \int _0^1\int_1^2dzrdrd\theta +\int _0^{2\pi} \int _1^2\int_r^2dzrdrd\theta=$$

$$\pi + 4\pi/3 = 7\pi/3$$

where the first integral evaluates the middle cylinder and the second one evaluates the rest of the volume.

However the easier way is to use the solid of revolution formula.

The volume is found by the integral $$\int _1^2 \pi r^2 dz $$ where $r^2 = z^2$

Therefore the answer is $$\int _1^2 \pi r^2 dz =\int _1^2 \pi z^2 dz =\frac {7\pi }{3} $$

$\endgroup$
  • $\begingroup$ Although the OP accepted the answer, from my perspective it did not begin to address the question as asked. $\endgroup$ – Ted Shifrin Oct 8 '19 at 17:04
  • $\begingroup$ @TedShifrin Thanks for the informative comment. Please check my editted version. $\endgroup$ – Mohammad Riazi-Kermani Oct 8 '19 at 19:15
  • $\begingroup$ Yes, thanks. As I commented to the OP, it is actually easier to do the integral in the (less standard) order $dr\,dz$; then only one integral is required. $\endgroup$ – Ted Shifrin Oct 8 '19 at 19:22

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.