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 🙂
Best Answer
By cylindrical coordinates the set up of the integral should be
$$\int_0^{2\pi}\int_{-3}^{-\frac3{\sqrt2}}\int_{0}^{\sqrt{9-z^2}} r\,\, dzdrd\theta+\int_0^{2\pi}\int_{-\frac3{\sqrt2}}^0\int_{0}^{-z} r\,\, dzdrd\theta$$