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√.

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