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Homework Help: Volumes of cross section

  1. Jun 12, 2010 #1
    Hey guys im new here i was wondering if anyone can help me understand this problem better
    1. The problem statement, all variables and given/known data
    The base of a certain solid is the circle x^2 + y^2 = 9. Cross sections of the solid with planes perpendicular to the y-axis are semicircles with their diameter in the base of the solid. find the volume of the solid.

    2. Relevant equations


    3. The attempt at a solution

    I cannot figure out where to start this problem. I know this will be an integral from -3 to 3 since its a circle with radius 3, but im not sure how to proceed after that. Can anyone point me in the right direction?
  2. jcsd
  3. Jun 12, 2010 #2
    It sounds like you need to know the area of half a sphere of radius 3 centred at the origin. It is a solid with a base which is a circle of this radius, further it says that cross sections of the solid are semicircles with diameter in the base, which I take to mean, having also a radius 3?
  4. Jun 12, 2010 #3
    im not exactly sure..i cant figure out what the problem exactly wants i asked my teacher for a hint and he said "plot it on a xyz axis and figure out what the picture looks like and sum all the cross sections" that didnt help much though so im still pretty confused
  5. Jun 13, 2010 #4


    Staff: Mentor

    Your teacher gave you a place to start - by graphing the solid to get an idea of what it looks like, and then adding the volume elements. Per the rules of the forum you have to make an attempt to solve the problem before we can help.
  6. Jun 13, 2010 #5
    Since the cross section is a semi circle the area would be (pie r^2)/2. And that will go inside the integral from -3 to 3. The part where i get confused is the problem says the diameter of the cross section is in the base of the solid. How do i go figure that out? I mean by lookin at it it seems like it'll be x+x or 2x but what do i do after that?
  7. Jun 13, 2010 #6
    i worked the problem a little further. Since the D= 2x i would get (pie (D/2)^2)/2. So it would be (pie * D^2)/8 and diameter is 2x so that would give me pie * x^2/2. Then i get (pie * X^2)/2 and since X^2 + Y^2 = 9 is the circles formula X^2 would be (9-y^2). i pulled out pie/2 outside of the integral as a constant and integrated (9-y^2) from -3 to 3. I ended up with 18pie im not sure if this is correct though. Any help?
  8. Jun 13, 2010 #7


    Staff: Mentor

    The name of this Greek letter -- [itex]\pi[/itex] -- is pi, not pie.

    Yes, 18 [itex]\pi[/itex] is correct. The typical volume element has a cross-sectional area of [itex](1/2) \pi x^2[/itex] and a thickness of [itex]\Delta y[/itex], so the volume of a typical volume element is [itex](1/2) \pi x^2~\Delta y = (1/2) \pi (9 - y^2) \Delta y[/itex].

    The definite integral is
    [tex]\int_{-3}^3 (1/2) \pi (9 - y^2)~dy[/tex]

    Notice that when you get an expression for the typical volume element, you have almost the exact same thing as the integrand.
  9. Jun 13, 2010 #8
    Thanks for correcting me i always say pie for some reason. This problem seemed pretty difficult but drawing it does make a lot of difference..Thanks for your help
  10. Jun 13, 2010 #9


    Staff: Mentor

    Pie is a kind of pastry.

    . Well, yes. The first thing I do in these kinds of problems is draw a sketch of the region. It's possible to work problems like this out without a reasonable visual representation, but it's much harder to do.
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