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Volume of solid with given base and cross sections

  1. Sep 17, 2011 #1
    1. The problem statement, all variables and given/known data
    The base is the semicircle y = [itex]\sqrt{9-x2}[/itex](Square root of 9-x2.. i dont know why the formatting isnt showing up) where -3 <= x <= 3. The cross section perpendicular to the x axis are squares.



    2. Relevant equations

    -3[itex]\int[/itex]3 = A(y)dy
    A(y) = area of cross section

    3. The attempt at a solution
    Okay so i know the cross section is a square. And i see that y = [itex]\sqrt{9-x2}[/itex] creates a semi-circle. I am not sure how to find the cross sectional area. I'm assuming it needs to be in terms of y. And the area of a square is a (side)2. But how do i know what the side is equal to?
     
  2. jcsd
  3. Sep 17, 2011 #2

    phinds

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    Looks to me to be a fairly simply solid (half a cylinder) so I think you're going about it the hard way.
     
  4. Sep 17, 2011 #3
    hmm well the reason I am going about it this way is because the problem is in the section of the book where we are finding volumes by integrating the area of the cross section. I am not aware of another way to do this problem.
     
  5. Sep 17, 2011 #4

    phinds

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    I'm not a big fan of simple problems requiring the use of more advanced math than is needed but I do understand that the math you're studying needs practice problems so I guess you'll need to do it the hard way.
     
  6. Sep 17, 2011 #5

    HallsofIvy

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    You are told that the base of the square is in the semi-circle from y= 0 up to [itex]y= \sqrt{9- x^2}[/itex]. That is the length of the side. The area of the square is [itex]9- x^2[/itex].
     
  7. Sep 17, 2011 #6

    phinds

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    but of course when you do it the easy way, the area of the square is irrelevant and the answer pretty much just pops out. When you do it the hard way you should get that the volumn is 27*pi
     
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