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Sphere using mulitiple integration

  1. May 15, 2006 #1
    I've proved the volume of a sphere using multiple integration but now i need to use that for the following:

    find a so that the volume inside the hemisphere z=sqrt(16-x^2-y^2) and outside the cylinder [tex]x^2 + y^2= a^2[/tex] is one quarter of the hemisphere.

    I'm having trouble even formulating an integral for this. Once it is set up as a multiple integral i'll be okay with it. Any help would be appreciated. Thanks
     
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  3. May 15, 2006 #2

    HallsofIvy

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    My first reaction when I read this was "find a what??":rofl: you are to find a number a so that the volume inside the hemisphere and outside the cylinder is 1/4 the volume of the hemisphere. That is the same as saying the volume inside the hemisphere and inside the cylider is 3/4 the volume of the hemisphere. Of course, a sphere of radius 4 has volume [itex]\frac{4}{3}\pi(4)^3= \frac{216}{3}\pi[/itex] and the hemisphere has volume [itex]\frac{128}{3}\pi[/itex]so you are looking for a that will make the volume 3/4 of that: [itex]32\pi[/itex].

    I would recommend using cylindrical coordinates (obviously, a is less than 4). Then the integral with respect to r is from 0 to a (unknown), the integral with respect to [itex]\theta[/itex] is from 0 to [itex]2\pi[/itex], and the integral with respect to z is from 0 to the hemisphere: 0 to [itex]\sqrt{16- r^2}[/itex].

    [tex]\int_0^{2\pi}\int_0^a\int_0^{\sqrt{16-r^2}}rdzdrd\theta= 2\pi\int_0^a\sqrt{16-r^2}rdr[/tex]

    That, of course, is a very simple integral which will give you the volume in terms of a. Set it equal to [itex]32\pi[/itex] and solve for a.
     
    Last edited: May 16, 2006
  4. May 15, 2006 #3
    shouldn't that be setting it equal to 32/pi not 32pi?

    also where does the equation [itex]\sqrt{16- r^2}[/itex] come from?

    is this possible to do in cartesian coords? or is this way just easier.
     
  5. May 15, 2006 #4

    HallsofIvy

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    Why in the world would it be 32/pi? The volume of a hemisphere of radius 4 is pi times(2/3)(43) and 3/4 of that is [itex]32\pi[/itex].
    The equation you gave for the hemisphere was [itex]z= \sqrt{16- x^2- y^}[/itex]. In polar coordinates (cylindrical coordinates without z) [itex]x^2+ y^2= r^2[/itex] so that becomes [itex]z= \sqrt{16- r^2}[/itex].

    Yes, you can do it in Cartesian coordinates: the integral would be
    [tex]\int_{x=-a}^a\int_{y=-\sqrt{a^2- x^2}}^{\sqrt{a^2-x^2}}\int_{z=0}^{\sqrt{16-x^2-y^2}}dzdydx= \int_{x=-a}^a\int_{y= -\sqrt{a^2- x^2}}^{\sqrt{16-x^2}}\sqrt{16- x^2- y^2}dydx[/tex]
    and you will have to do a rather complicated trig substitution to integrate that. I think you will find it is far simpler in polar coordinates!
     
  6. May 15, 2006 #5
    i thought it should be 32pi, but you'd got 32/pi further up in the question, so i was just checking.

    and asking where the 16-r^2 part came from was a stupid question that i realised the answer to as soon as i clicked on submit reply lol

    thanks for the help
     
  7. May 16, 2006 #6
    the final integral looks simple enough, but believe it or not...i just can't seem to do it. the more i look at it the more i can't do it.

    [tex]2\pi\int_0^a\sqrt{16-r^2}rdr[/tex]

    could i get a hint as to how to do this? i'm guessing integration by parts, but i can't figure out how to integrate [tex]\int_0^a\sqrt{16-r^2}dr[/tex]
     
  8. May 16, 2006 #7

    HallsofIvy

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    Try the substitution u= 16- r2!

    ([itex]\int_0^a\sqrt{16-r^2}dr[/itex] is a standard trig substitution: let [itex]x= 4 sin(\theta)[/itex] but having the r outside the integral makes [itex]2\pi\int_0^a\sqrt{16-r^2}rdr[/itex] much easier.)
     
  9. May 16, 2006 #8
    i was never particularly good with integration by substitution, but i'll give this a go. thanks
     
  10. May 16, 2006 #9
    okay, i've had a go at this and i end up with
    [itex]2pi\frac{(16-r^2)^2}{4}=32pi[/itex] between the limits of a and 0, where a is the number i'm trying to find. then i'd go on to substitue a in and solve.

    could someone tell me if this is right, or have i gone drastically wrong somewhere.
     
  11. May 16, 2006 #10

    HallsofIvy

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    No, the limits of integration, after the substitution, are no longer 0 and a. Also, the anti-derivative of u1/2 will involve a factor of 2/3 and I don't see that anywhere. If u= 16- r2 then du= -2r dr so -(1/2)du= rdr. Of course, [itex]\sqrt{16-r^2}= u^{\frac{1}{2}}[/itex]. When r= 0, u= 16 and when r= a, u= 16-a2 so the integral becomes
    [tex]\pi\int_{16}^{16-a^2}u^{\frac{1}{2}}du= \frac{2\pi}{3}(16-a^2)^{\frac{3}{2}}= 32\pi[/tex]

    That should be easy to solve for a.
     
  12. Mar 10, 2007 #11
    But solving this

    [tex]\int_0^a\sqrt{16-r^2}dr[/tex]

    is not the answer to this

    [tex]2\pi\int_0^a\sqrt{16-r^2}rdr[/tex]

    because the r times root 16 - r squared, why have you ignored the r.

    How do you incorporate that?

    Cheers Ash
     
  13. Mar 10, 2007 #12

    HallsofIvy

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    Which post are you referring to? Both Almost Famous and I, in our last posts, were referring to
    [tex]2\pi\int_0^a\sqrt{16-r^2}rdr[/tex]
     
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