1
$\begingroup$

I'm stuck on what the boundaries are for the volume bounded by the cone $z=-\sqrt{(x^2+y^2)}$ and the surface $z=-\sqrt{(9-x^2-y^2)}$ $\,\,$-essentially an upside down ice cream cone

Remember that $r^2=x^2+y^2$

I assumed for cylindrical coordinates the triple integral boundaries would be

$-\sqrt{(9-r^2)}\le z \le -r$

$0\le r \le 3$

$0\le \theta \le 2\pi$

And for spherical coordinates the triple integral boundaries would be

$0\le r \le 3$

$\pi/2\le \phi \le \pi$

$0\le \theta \le 2\pi$

However upon entering these values into MATLAB, the cylindrical coordinates integral equals to zero, whilst the spherical coordinates integral equals to $18\pi$.

So somethings wrong with my boundaries, as both integrals should equal the same volume.

This is my working out for cylindrical coordinate integral so far:

$\int_0^{2\pi}\int_0^3\int_{-\sqrt{9-r^2}}^{-r} r\,\, dzdrd\theta$

$\int_0^{2\pi}\int_0^3\ [rz]_{-\sqrt{9-r^2}}^{-r} \,\, dzdrd\theta$

$\int_0^{2\pi}\int_0^3\ -r^2+{r\sqrt{9-r^2}} \,\, dzdrd\theta$

$\int_0^{2\pi}[(\int_0^3\ -r^2 \,dr)+(\int_0^3\ {r\sqrt{9-r^2} \,dr)}]d\theta$

Let $u=9-r^2$

$du=-2r\,dr$

$\int_0^{2\pi}[[\frac{-r^3}{3}]_0^3\,-\frac{1}{2}\int_0^3\ u^{\frac{1}{2}} \,du)]\,d\theta$

$\int_0^{2\pi}[-\frac{27}{3}\,-\frac{1}{2}[\frac{2}{3}u^{\frac{3}{2}}]_{r=0}^{r=3} ]\,d\theta$

$\int_0^{2\pi}[-9\,-\frac{1}{2}[{\frac{2}{3}}(9-r^2)^{\frac{3}{2}}]_{0}^{3} ]\,d\theta$

$\int_0^{2\pi}[-9\,-\frac{1}{2}[{\frac{2}{3}}(9-3^2)^{\frac{3}{2}}\,-{\frac{2}{3}}(9-0^2)^{\frac{3}{2}}] ]\,d\theta$

$\int_0^{2\pi}[-9\,-\frac{1}{2}{\frac{2}{3}}[(9-9)^{\frac{3}{2}}\,-(9)^{\frac{3}{2}}] ]\,d\theta$

$\int_0^{2\pi}[-9\,-\frac{1}{3}[(0)^{\frac{3}{2}}\,-(9)^{\frac{3}{2}}] ]\,d\theta$

$\int_0^{2\pi}[-9\,-\frac{1}{3}[-(9)^{\frac{3}{2}}] ]\,d\theta$

$\int_0^{2\pi}[-9\,-\frac{1}{3}[-27] ]\,d\theta$

$\int_0^{2\pi}[-9\,+9]\,d\theta$

$\int_0^{2\pi}0\,d\theta$

$=0$

So as you can see I can't proceed to the third integral since the second integral equals zero

Any help would be greatly appreciated :)

$\endgroup$
1
  • $\begingroup$ Hi again. I don't think $18\pi$ is correct either. Consider a sphere with radius of $3$. The volume would be $\frac43\pi\times3^3=36\pi$. Your answer is half of this while your shape is clearly less than half a sphere. The reason for your spherical answer giving this is incorrect limits on $\phi$. You should use $\frac{3\pi}{4}\leq\phi\leq\pi$ as your cone has an angle of $\frac{\pi}{4}$. (I'll try to give you an actual answer once I get back to my computer.) $\endgroup$
    – Ian Miller
    Commented Apr 7, 2016 at 8:23

2 Answers 2

1
$\begingroup$

By cylindrical coordinates the set up of the integral should be

$$\int_0^{2\pi}\int_{-3}^{-\frac3{\sqrt2}}\int_{0}^{\sqrt{9-z^2}} r\,\, dzdrd\theta+\int_0^{2\pi}\int_{-\frac3{\sqrt2}}^0\int_{0}^{-z} r\,\, dzdrd\theta$$

$\endgroup$
2
  • $\begingroup$ Wait, the polar coordinates are not for double integrals? I think it should be cylindrical coordinates, altough the first two coordinates are the same for the third integral we need a third variable. $\endgroup$
    – manooooh
    Commented Nov 26, 2018 at 3:08
  • 1
    $\begingroup$ @manooooh Yes of course, I’ve used the uncorrect term here! I fix that, thanks. $\endgroup$
    – user
    Commented Nov 26, 2018 at 6:06
0
$\begingroup$

The surfaces intersect when $$-r=-\sqrt{9-r^2}$$ which is $r^2=\frac92$, not $r^2=9$ as you have in your cylindrical attempt.

For spherical coordinates $\phi$ should be from $\frac34\pi$ to $\pi$. I believe the rest is correct, as long as you are writing $r^2=x^2+y^2+z^2$ in the spherical case. (It is common to call this $\rho^2$ in order to distinguish it from $r^2=x^2+y^2$.)

$\endgroup$

You must log in to answer this question.

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