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

example 2 – simplify: step 1: completely factor both the numerators and denominators of all fractions. step 2: change the division sign to a multiplication sign and flip (or reciprocate) the fraction after the division sign; essential you need to multiply by the reciprocal. step 3 : cancel or reduce the fractions.