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Volumes by Slicing and Rotation About an Axis

  1. Oct 10, 2009 #1
    1. The problem statement, all variables and given/known data
    I'm having issues with this section in calc. I'm not at all sure what I'm doing!

    Here is the problem I'm having trouble with:

    Directions: Find the volume of the solids.

    Problem: The base of the solid is the disk X^2 + Y^2 <= 1. The cross-sections by planes perpendicular to the Y-axis between Y= -1 and Y= 1 are isosceles right triangles with one leg in the disk.

    So essentially it shows a right isosceles triangle where right angle touches the left edge of the circle and the base of the right triangle extends to the other side of the circle.

    2. Relevant equations

    about all I really understand what to do is find the area of the triangle.

    A(x) = 1/2 BH
    1/2(2x)(2x) = 2x^2

    3. The attempt at a solution

    I honestly don't have a clue what I'm doing in this section. But if I were to guess I'd take the integral of the area from 1to-1 since that's my interpretation of what the book says.

    So V = int from 1to-1(2x^2) = 4/3

    But that doesn't sound right at all.

    It also talks about the disk and washer methods in this section but I'm not sure if those apply. Any help would be appreciated.
  2. jcsd
  3. Oct 10, 2009 #2


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    You have the idea, but the plane cross sections are perpendicular to the y axis, not parallel to it. So the base of the triangle is 2y, which you can get in terms of x from the equation of the circle. Your cross section slab has thickness dx. So your element of volume is

    dV = (1/2)(2y)2 dx

    Get it all in terms of x and go.
  4. Oct 10, 2009 #3
    ok so
    dV = 1/2(2(sqrt(1-x^2))

    so V = int from -1 to 1(1/2(2(sqrt(1-x^2))) = pi/2 ?
  5. Oct 10, 2009 #4


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    Did you miss a squared term?
  6. Oct 10, 2009 #5
    Ah.. right...


    V = int from -1 to 1(1/2(2(sqrt(1-x^2))^2) = 8/3
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