I'm having a hard time with this problem, i have some ideas, but i don't know how to continue. Here is the exercise:

"Consider the solid $S$ bounded by the walls of the superior cone whose equation is $z = \sqrt{3}\sqrt{x^2+y^2} $ and inside the sphere of equation $x^2+y^2+(z-2)^2=4$ whose density at a point $P$ is given by $\delta (x,y,z) =z$. Write (without calculating!) the mass of $S$ as triple iterated integral in both cylindrical and spherical coordinates. (Hint: in spheric coordinates, $dV = \rho^2 sin(\phi) d\rho d\theta d\phi$)."

First of all, i noticed that that $(z-2)^2$ would give me a headache to write in spherical coordinates, so i did the linear variable substitution: $$u = x$$ $$ v = y$$ $$ w = z -2$$

So, the cone equation became $w+2 = \sqrt{3}\sqrt{u^2+v^2}$ and the sphere $u^2+v^2+w^2 = 4$. Now, there are 2 problems:

1) I can't visualize the limits of integration. (this is more general, it happens in a LOT of problems that i have to change from rectangular to polar/cylindrical/spherical coordinates, so any tips here are welcome). Also, this is the biggest problem of this exercise, for me.

2) How can i make just one integral for the mass? I mean, in the 1st octant that will be okay, because all of the variables are positive. But, for example, in the 5º octant, wouldn't the volume (then the mass) be negative? When i did some problems calculating mass in spherical/cylindrical coordinates, i always "broke" the sphere into 8 parts (when the density was symmetric with respect to the origin) but in this case, i must write it all in just 1 integral.

Any help would be great. Thanks!


This is not a complete answer, but at least a hint to get you started:

The trick here is not to change to $w=z-2$, but to directly go to ordinary spherical coordinates centered at the origin. The equation for the sphere can be written as $x^2+y^2+z^2=4z$, which in spherical coordinates becomes $r^2 = 4 r \cos\phi$, where $\phi$ is the angle from the $z$ axis (which seems to be the convention that you're using).

In other words, the sphere is described by the equation $r=4\cos \phi$. (We may cancel $r$, since $r>0$ except at the origin, which is included in the equation $r=4\cos\phi$ anyway.)

Regarding the density, the sphere lies wholly in the upper half space $z \ge 0$ (since it's centered at $(0,0,2)$ and has radius $2$), so you are only integrating over points where the density $\delta(x,y,z)=z$ is nonnegative.

  • $\begingroup$ Thanks a lot! But now it comes the bigger problem: o dont know which parts of the sphere are inside or outside the cone, so i can't see the limits of integration. $\endgroup$ – Vitor C Goergen Oct 21 '16 at 18:31
  • $\begingroup$ Actually, the formulation is ambiguous! The cone divides the inside of the sphere into two parts, one above the cone ($z > \sqrt{\dots}$) and one below it ($z < \sqrt{\dots}$), and either one is “bounded by the sphere and the cone”. I would guess that whoever set the problem had the upper part in mind, though. $\endgroup$ – Hans Lundmark Oct 21 '16 at 19:15
  • $\begingroup$ Now it comes another problem: i know that the density will always be positive, by my question is: when i'm at a point that one of the coordinates (not $z$) is negative, say (3,-7,1). Isn't the volume of a set in that region negative? $\endgroup$ – Vitor C Goergen Oct 22 '16 at 1:30
  • $\begingroup$ No, volumes are by definition never negative. $\endgroup$ – Hans Lundmark Oct 22 '16 at 6:30
  • $\begingroup$ So, for example, while calculating the volume of a sphere, i will never have to divide into various pieces and calculate the integrals of all the pieces? I can always make the range of $\phi$ going from $-\pi$ to $pi$, and $\theta$ from $0$ to $2 \pi$ ? $\endgroup$ – Vitor C Goergen Oct 22 '16 at 14:49

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