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Volume by parallel cross-section.

  1. Oct 30, 2009 #1
    The questions: the base of a solid is the region bounded y=x^2 and y=4. Find the volume of the solid given that the cross sections perpendicular to the x-axis are squares.

    The answer is 512/15.

    I set A(x)= (2y)^2 = 4(x^2)^2=4x^4, and the answer is wrong when I integrate this from x = -2 to x = 2.

    How can I solve this?

    Please help me.
     
    Last edited: Oct 30, 2009
  2. jcsd
  3. Oct 30, 2009 #2

    Mark44

    Staff: Mentor

    The typical volume element, [itex]\Delta V[/itex], has a volume of (4 - x2)(4 - x2)[itex]\Delta x[/itex]. The first factor of (4 - x2) is the distance between the line y = 4 and the curve y = x2. The second factor of (4 - x2) is the vertical height of the volume element (perpendicular to the x-y plane). The [itex]\Delta x[/itex] factor is the thickness of the volume element.
     
  4. Oct 30, 2009 #3
    That's more confusing.... can you explain it with integrals?
     
  5. Oct 30, 2009 #4

    LCKurtz

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    Maybe this picture will help you:

    forumslab.jpg
     
  6. Oct 30, 2009 #5
    Thank you for the picture, but I do understand how it looks...

    just don't know how it goes with the integrals...
     
  7. Oct 30, 2009 #6
    so A(x) = (4-x2)2 ?
    2[tex]\int[/tex](4-x2)2dx (x is 0~2)
     
  8. Oct 30, 2009 #7

    LCKurtz

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    Yes, since you have symmetry about x = 0.
     
  9. Oct 31, 2009 #8
    But, I am not getting 512/15.
     
  10. Oct 31, 2009 #9

    Mark44

    Staff: Mentor

    Then you have done something wrong.
    [tex]\int_{x = -2}^2 (4 - x^2)^2 dx~=~2\int_{x = 0}^2 (4 - x^2)^2 dx~=~\frac{512}{15}[/tex]
     
  11. Oct 31, 2009 #10
    Yes. made some calculation error
     
  12. Oct 31, 2009 #11

    Mark44

    Staff: Mentor

    Did you find out where you went wrong?
     
  13. Oct 31, 2009 #12
    Yes. Thank you. I have posted another question this time regarding shell method
     
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