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Cross-sectional area

  1. Dec 2, 2014 #1
    1. The problem statement, all variables and given/known data
    Find the volume of the solid whose base is a 4 by 4 square. Cross sections perpendicular to one diagonal of the square base are semi-circles with diameter on the base.

    2. Relevant equations
    V=pi r^2
    A=S^2

    3. The attempt at a solution
    The cross sections are perpendicular to the x axis, so I need to integrate with respect to x. I centered the square base on a cartesian coordinate system and found via the pythogorean theorem that half the diagonal, which I believe is the radius of the semicircle, has a value of 2*2^(1/2). V=pi r^2, so [tex] \int_{0}^{4}(pi*(2\sqrt{2})^2) dx[/tex] =100.53
    Am I on the right track with this solution?
     
  2. jcsd
  3. Dec 3, 2014 #2

    Mark44

    Staff: Mentor

    I don't think so.
    According to your problem description, the cross sections are perpendicular to the diagonal of the square. As you move away from one corner of the square, the semicircular slices get larger in diameter until you get to the other diagonal of the square, then they get smaller again.

    I set up the problem by placing the diagonal of the square on the x-axis so that one corner is at (0, 0) and two other corners are at ##(\sqrt{2}, \sqrt{2})## and ##(\sqrt{2}, -\sqrt{2})##. I also used the symmetry of the situation by doubling the volume I get over half of the square.

    Edit: The value I get is about 24.
     
    Last edited: Dec 3, 2014
  4. Dec 3, 2014 #3
    Now th
    Now that I've had some coffee, I can see that I should have integrated along the diagonal like you did. Also, I forgot to half the volume formula because I'm dealing with semi-circles. In this situation, how do I find the equation that produces the radius of the semi-circles?

    Edit: The best I could do was the following: I centered one of the corners on the origin and used the y-axis to divide the square into two symmetric halves. From here, I called the upper boundary of the square f(x) and the bottom boundary g(x) and then I integrated [tex]2 \int_{-2 \sqrt{2}}^{0} \frac {( \pi (.5[f(x)-g(x)])^2)}{2}[/tex] Is there a better way to do that?

    Here's a picture of what I did if the text isn't clear
    http://postimg.org/image/w0w7pwuy9/6e31b24f/
     
    Last edited by a moderator: Dec 3, 2014
  5. Dec 3, 2014 #4

    Mark44

    Staff: Mentor

    That will work, but it isn't the most convenient way to set things up. I put the left corner of the square at the origin, which makes it easier to find the equations of the lines that make up the edges of the square.

    BTW, I had an error in my result. I now get a value between 23 and 24. I have edited my earlier post to correct it.
     
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