1. Limited time only! Sign up for a free 30min personal tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Homework Help: 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
  2. jcsd
  3. May 15, 2006 #2


    User Avatar
    Science Advisor

    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 by a moderator: 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


    User Avatar
    Science Advisor

    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.


    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


    User Avatar
    Science Advisor

    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


    User Avatar
    Science Advisor

    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


    is not the answer to this


    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


    User Avatar
    Science Advisor

    Which post are you referring to? Both Almost Famous and I, in our last posts, were referring to
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook