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24 October, 04:05

Determine the volume of the solid that lies between planes perpendicular to the x-axis at x=0 and x=4. The cross sections perpendicular to the x-axis on the interval 0≤x≤4 are squares whose diagonals run from the curve y=x√ to the curve y=-x√.

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  1. 24 October, 04:20
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    Volume = 16 unit^3

    Step-by-step explanation:

    Given:

    - Solid lies between planes x = 0 and x = 4.

    - The diagonals rum from curves y = sqrt (x) to y = - sqrt (x)

    Find:

    Determine the Volume bounded.

    Solution:

    - First we will find the projected area of the solid on the x = 0 plane.

    A (x) = 0.5 * (diagonal) ^2

    - Since the diagonal run from y = sqrt (x) to y = - sqrt (x). We have,

    A (x) = 0.5 * (sqrt (x) + sqrt (x)) ^2

    A (x) = 0.5 * (4x) = 2x

    - Using the Area we will integrate int the direction of x from 0 to 4 too get the volume of the solid:

    V = integral (A (x)). dx

    V = integral (2*x). dx

    V = x^2

    - Evaluate limits 0 < x < 4:

    V = 16 - 0 = 16 unit^3
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