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Find the volume of the solid (strange one )

  1. Oct 11, 2007 #1
    ...whose base is enclosed by the circle [tex]x^2+y^2=1[/tex] and whose cross sections taken perpendicular to the x-axis are semi-circles.

    So I understand that the volume of each "sample" semi-circle will be [tex]\frac{\pi r^2}{2}*\Delta x[/tex]

    The radius is y. Now this is where I lose the solution manual.....Do I need to take into account that [tex]y=^+_-\sqrt{1-x^2}[/tex] ? I am having trouble setting up this integral.

    I am working on the bounds now...but^^^^ that part is effing me up.

  2. jcsd
  3. Oct 11, 2007 #2
    [tex] V = \pi \int y^2 dx [/tex]

    Just let [itex] y^2 = r^2 - x^2 [/itex] where r is a constant, and then its just a case of picking the limits of the integral to suit the problem. The integral itself shouldn't present any problems.
  4. Oct 11, 2007 #3
    Well, the radius IS y..so I don't follow you.

  5. Oct 11, 2007 #4


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    I'm trying to picture this. It just seems like its half a sphere, or a dome shape. Maybe I'm visualizing it wrong. If I'm right, they could have just said a dome.

    If I'm right and you want to do it without calculus, its just the formula for a sphere divided by 2: 2*pi*r^3/3. Assuming I'm picturing the right shape, you could at least use this to check your answer if you use integration. What does the solution manual say is the answer? Does it match 2*pi*r^3/3, which for r=1 would be 2*pi/3?

    In your example, becareful because there are 2 r's you're dealing with, one is a fixed value which is the radius of the base which = 1 from the question. The r in your integral symbol is a variable representing the radius of a slice.

    So substitute your r in the integral with the sqrt(1-x^2).
  6. Oct 12, 2007 #5


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    Yes, and since the "thickness" is [itex]\Delta x[/itex] you will need to write y2 as a function of x. That's easy!
  7. Oct 12, 2007 #6
    I got it. It's just twice the volume of the "half-dome' whose base is the half-circle y=(1-x^2)^.5 from 0 to 1....like tony said! Sometimes I just need to move on an look at the problem again later (whenI have gotten some sleep:/ )
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