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

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step-by-step explanation:

the height of the surface increases from the center out to the sides of the road.

step-by-step explanation:

as we move out we move away from the x-axis.

and as mentioned in the problem statement that height of a road increases when we move away from the center x.

hence proved that 3rd option

the height of the surface increases from the center out to the sides of the road.

is the correct option.