Vtyjkyuk

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

$-sqrt(9-r^2)le z le -r$

$0le r le 3$

$0le theta le 2pi$

And for spherical coordinates the triple integral boundaries would be

$0le r le 3$

$pi/2le phi le pi$

$0le theta le 2pi$

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

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:

$int_0^2piint_0^3int_-sqrt9-r^2^-r r,, dzdrdtheta$

$int_0^2piint_0^3 [rz]_-sqrt9-r^2^-r ,, dzdrdtheta$

$int_0^2piint_0^3 -r^2+rsqrt9-r^2 ,, dzdrdtheta$

$int_0^2pi[(int_0^3 -r^2 ,dr)+(int_0^3 rsqrt9-r^2 ,dr)]dtheta$

Let $u=9-r^2$

$du=-2r,dr$

$int_0^2pi[[frac-r^33]_0^3,-frac12int_0^3 u^frac12 ,du)],dtheta$

$int_0^2pi[-frac273,-frac12[frac23u^frac32]_r=0^r=3 ],dtheta$

$int_0^2pi[-9,-frac12[frac23(9-r^2)^frac32]_0^3 ],dtheta$

$int_0^2pi[-9,-frac12[frac23(9-3^2)^frac32,-frac23(9-0^2)^frac32] ],dtheta$

$int_0^2pi[-9,-frac12frac23[(9-9)^frac32,-(9)^frac32] ],dtheta$

$int_0^2pi[-9,-frac13[(0)^frac32,-(9)^frac32] ],dtheta$

$int_0^2pi[-9,-frac13[-(9)^frac32] ],dtheta$

$int_0^2pi[-9,-frac13[-27] ],dtheta$

$int_0^2pi[-9,+9],dtheta$

$int_0^2pi0,dtheta$

$=0$

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 :)

The surfaces intersect when
$$-r=-sqrt9-r^2$$
which is $r^2=frac92$, not $r^2=9$ as you have in your cylindrical attempt.

For spherical coordinates $phi$ should be from $frac34pi$ 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 $r^2=x^2+y^2$.)

In polar coordinates the set up of the integral should be

$$int_0^2piint_-3^-frac3sqrt2int_0^sqrt9-z^2 r,, dzdrdtheta+int_0^2piint_-frac3sqrt2^0int_0^-z r,, dzdrdtheta$$