Volumes using triple integration

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 SS bounded by the walls of the superior cone whose equation is z=√3√x2+y2z = \sqrt{3}\sqrt{x^2+y^2} and inside the sphere of equation x2+y2+(z−2)2=4x^2+y^2+(z-2)^2=4 whose density at a point PP is given by δ(x,y,z)=z\delta (x,y,z) =z. Write (without calculating!) the mass of SS as triple iterated integral in both cylindrical and spherical coordinates. (Hint: in spheric coordinates, dV = \rho^2 sin(\phi) d\rho d\theta d\phidV = \rho^2 sin(\phi) d\rho d\theta d\phi).”

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

So, the cone equation became w+2 = \sqrt{3}\sqrt{u^2+v^2}w+2 = \sqrt{3}\sqrt{u^2+v^2} and the sphere u^2+v^2+w^2 = 4u^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 5آ؛ 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!

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1 Answer
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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-2w=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=4zx^2+y^2+z^2=4z, which in spherical coordinates becomes r^2 = 4 r \cos\phir^2 = 4 r \cos\phi, where \phi\phi is the angle from the zz axis (which seems to be the convention that you’re using).

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

Regarding the density, the sphere lies wholly in the upper half space z \ge 0z \ge 0 (since it’s centered at (0,0,2)(0,0,2) and has radius 22), so you are only integrating over points where the density \delta(x,y,z)=z\delta(x,y,z)=z is nonnegative.

  

 

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.
– Dovah-king
2 days ago

  

 

Actually, the formulation is ambiguous! The cone divides the inside of the sphere into two parts, one above the cone (z > \sqrt{\dots}z > \sqrt{\dots}) and one below it (z < \sqrt{\dots}z < \sqrt{\dots}), and either one is “bounded by the sphere and the coneâ€‌. I would guess that whoever set the problem had the upper part in mind, though. – Hans Lundmark 2 days ago      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 zz) is negative, say (3,-7,1). Isn't the volume of a set in that region negative? – Dovah-king 2 days ago      No, volumes are by definition never negative. – Hans Lundmark yesterday      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\phi going from -\pi-\pi to pipi, and \theta\theta from 00 to 2 \pi2 \pi ? – Dovah-king yesterday