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Finding the volume inside a cone bounded by the edge of a sphere

  1. Aug 15, 2012 #1
    1. The problem statement, all variables and given/known data
    Find the volume of the region D in R^3 which is inside the sphere x^2 + y^2 + z^2 = 4 and also inside the cone z = sqrt (x^2 + y^2)



    2. Relevant equations



    3. The attempt at a solution
    So I decided that the best approach might be finding the area under the sphere and then subtracting the area under the cone. The first step then seemed to be to find the corresponding projection onto the x-y plane that both of these have. Since the cone defines z as the sqrt (x^2 + y^2) plugging this into the equation for the sphere yields x^2 + y^2 = 2. I took this to mean that the projection is the circle of radius two centered on the origin.

    Now for the area under the sphere. In polar coords, r varies from 0 to 2, and ∅ from 0 to 2∏. Translating the equation for the sphere into polar coords I got z = sqrt (4 - r^2). So then altogether I have [itex]\int[/itex][itex]^{2\pi}_{0}[/itex][itex]\int[/itex][itex]^{2}_{0}[/itex]sqrt(4 - r^2)r dr d∅.

    For the cone it seemed like cylindrical coords would be the way to go. I thought z varies from 0 to r, r from 0 to 2, and ∅ from 0 to 2∏. So I had the integrand [itex]\int[/itex][itex]^{2\pi}_{0}[/itex][itex]\int[/itex][itex]^{2}_{0}[/itex] [itex]\int[/itex][itex]^{r}_{0}[/itex] r dz dr d∅.

    Then I would just subtract the two. I'm really rusty on this stuff so I'm unsure in particular about my limits in the integration and the projection space. Does this seem correct? Is there an easier approach? This is a question from an old placement exam. Thanks!
     
  2. jcsd
  3. Aug 15, 2012 #2
    I'm not entirely sure what you're doing. But it might be easier if you first figured out where the sphere and the cone intersect. Then calculating the volume should be straightforward, as the volume is bounded by the cone below this value, and by the sphere above it.
     
  4. Aug 15, 2012 #3
    That is what I was trying. So the cone being defined as z = sqrt (x^2 + y^2) -> z^2 = x^2 + y^2. Plugging this into the equation for the sphere yields x^2 + y^2 + x^2 + y^2 = 4 → x^2 + y^2 = 2. Hence that is where they intersect (x^2 + y^2 = 2 and z = sqrt of 2). This also defines the domain on the xy plane that it is being integrated over: the circle of radius 2.
     
  5. Aug 15, 2012 #4
    Oh, I think I see what you mean now. I will try playing with it. Thanks, sorry for my misunderstanding.
     
  6. Aug 15, 2012 #5
    OK, and now you want to calculate the volume of the sphere which is above the intersection. It's probably easiest to do in polar coordinates (although surely spherical coords would work as well). So in polar coordinates you'd integrate something like
    V = [tex]\int_0^{2\pi} d \theta \int_{\sqrt{2}}^2 dz \int dr r [/tex]
    where the limits for r you get from the requirement that the volume is a sphere.
     
  7. Aug 15, 2012 #6
    Is that Cylindrical Coords then? That is something I keep confusing, I thought Polar would be a function f(∅,r) where the z is the output rather than an input.

    Okay, so I thought that ∅ varies from 0 to 2\pi, z varies from r to sqrt (4 - r^2), and r from 0 to sqrt(2). So then [itex]\int[/itex][itex]^{\sqrt{2}}_{0}[/itex][itex]\int[/itex][itex]^{2\pi}_{0}[/itex][itex]\int[/itex][itex]^{\sqrt{4-r^2}}_{r}[/itex] r dz d∅ dr. Does that seem correct? I also have a hard time deciding the order in which to do the integration. Thanks so much for your help!
     
  8. Aug 15, 2012 #7

    LCKurtz

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    @Fractal20 This problem is best done in spherical coordinates. You can calculate the required volume directly with one triple integral with constant limits on all three integrals. The angle at the base of the cone tells you the ##\phi## limit and the radius of the sphere gives you the ##\rho## limit.
     
  9. Aug 15, 2012 #8
    Okay, in spherical coords I thought it would be [itex]\int[/itex][itex]^{\pi/4}_{0}[/itex][itex]\int[/itex][itex]^{2 \pi}_{0}[/itex][itex]\int[/itex][itex]^{2}_{0}[/itex] p^2 sin[itex]\varphi[/itex] dp d[itex]\vartheta[/itex] d[itex]\varphi[/itex]. Upon evaluating this I got [16 [itex]\pi[/itex]([itex]\sqrt{2}[/itex] -1)]/(3[itex]\sqrt{2}[/itex]).

    I wanted to check this with my attempt in cylindrical. First off I realized that I neglected to actually put the function into my earlier attempt. So I thought x^2 + y^2 + z^2 translates to r^2 + z^2 in cylindrical. Then I would have: [itex]\int[/itex][itex]^{r}_{0}[/itex][itex]\int[/itex][itex]^{2\pi}_{0}[/itex][itex]\int[/itex][itex]^{\sqrt{2}}_{0}[/itex] (r^2 + z^2)r dr d[itex]\vartheta[/itex] dz. But evaluating this yielded 8 [itex]\pi[/itex]/3. So I am at least doing one of these wrong but probably both.

    It is probably a lot to ask to check my work on this level. But if anybody feels up for it, it is greatly appreciated. Thanks for everything y'all!
     
  10. Aug 15, 2012 #9

    LCKurtz

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    That is correct. See how easy the appropriate setup makes it?

    Your first mistake is in the integrand. When calculating a volume with a triple integral you always use ##1\, dV## for the integrand. There should be no ##r^2+z^2## there. Next, you don't want to integrate in the ##r## direction first because if you do, you will have to break the problem into two parts for ##r## on the cone and ##r## on the sphere, an unnecessary complication. Try setting it up in the order ##dzdrd\theta##. Your ##z## limits won't be constant.

    [Edit] Didn't you already do that in post #6?
     
    Last edited: Aug 15, 2012
  11. Aug 15, 2012 #10
    Hmm, now I am confused about the 1 dV thing. For the spherical coord expression I had meant to include an additional p^2 since p^2 = x^2 + y^2 + z^2 but I guess my oversight was actually correct.

    So in my book Calculus (Boyce and Diprima) Cylindrical is defined as in general by triple integral f(r,theta,z) dV and Spherical as triple integral f(p, theta, phi) dV. That being said, most of the examples were then about center of masses and rather than f in the integral they just have a density expression. But they are really expicit about representing f in terms of the respective terms and including them in the integral?
     
  12. Aug 15, 2012 #11
  13. Aug 15, 2012 #12

    LCKurtz

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    First, if you work out the cylindrical version you have in post #6 you will get the same answer as the one in spherical coordinates.

    Now, as to your confusion about the integrand:$$
    \iiint_V 1 dV$$always gives volume. If ##\delta## is a mass density function, then$$
    \iiint_V \delta\, dV$$gives the mass. If ##f = x^2+y^2 = r^2## then$$
    \iiint_V f\, dV$$gives the second moment about the ##z## axis, and so on. What is in the integrand determines what the integral is calculating. And, of course, you always express the integrand and ##dV## in terms of the variables you are using for the integration.
     
  14. Aug 15, 2012 #13

    LCKurtz

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    But he isn't claiming that is a volume. It is just a practice integral he made up. You could think of that ##\rho^2## as a density function, which would mean the volume is heavier away from the origin. With that interpretation, he would be calculating the mass.
     
  15. Aug 16, 2012 #14
    Thanks that really clears it up. I forgot the whole point is that you are adding up those little volumes that are determined by all those d's (dx dy dz boxes in normal cartesian, dr etc... in others). It is funny how plug and chug problems remove you from the fundamental idea.
     
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