Calculus

The base of a solid in the xy-plane is the first-quadrant region bounded y = x and y = x^2. Cross sections of the solid perpendicular to the x-axis are equilateral triangles. What is the volume, in cubic units, of the solid?
So I got 1/30
because (integral from 0 to 1) (x-x^2)^2=1/30

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  1. Not quite. You have figured the volume consisting of squares. For the triangle, the base is x-x^2, so the altitude is (x-x^2)√3/2. The volume is then

    ∫[0,1] (1/2)(x-x^2)(x-x^2)√3/2 dx
    = ∫[0,1] √3/4 (x-x^2)^2 dx = √3/120

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  2. for the second integral i got (sqrt 3)/60 how did you get (sqrt 3)/120?

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  3. √3/4 (x^5/5 - x^4/2 + x^3/3) [0,1]
    = √3/4 (1/5 - 1/2 + 1/3)
    = √3/4 * 1/30
    = √3/120

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  4. Oh thank you!!!

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