Answer:
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
Sixty-three, 60 + 3, 9 x 7, 63
We can use the FOIL method to simplify.
= (3x – 1)(2x + 9)
= (3x * 2x) + (3x * 9) + (-1 * 2x) + (-1 * 9)
= 6x + 27x + (-2x) + (-9)
= 33x - 11
Hope This Helped! Good Luck!
For m it’s 1,1/3and -5/3 you will get different values when plug the different values of n