Question

The solid S has a base region B defined by the curves y = 5x − x 2 and y = x. (A) Find the volume of S if the cross-sections through S perpendicular to the x-axis are squares with an edge along the base. (B) Find the volume of S if the cross-sections through S perpendicular to the x-axis are equilateral triangles with an edge along the base. 3 (C) Find the volume of S if the cross-sections through S parallel to the x-axis are semicircles with their diameter along the base. (D) Find the volume of S if the cross-sections through S parallel to the x-axis are isosceles right triangles with a leg along the base.

Answer 1

a) The volume of S is 34.13

b) The volume of S is 14.8

c) The volume of S is 5.17

d) The volume of S is 11.33

Explanation:

a) The cross section area is equal to:

`A=a^(2) =((5x-x^(2))-x)^(2) =(4x-x^(2) )^(2)`

The volume of S is equal to:

`Vol_(S) =\int\limits^4_0 {A(x)} \, dx =\int\limits^4_0 {(4x-x^(2))^(2) } \, dx =34.13`

b) The cross section area is equal to:

`A=(a^(2)√(3) )/(4) =(√(3) )/(4) ((5x-x^(2) )-x)^(2) =(√(3) )/(4) (4x-x^(2) )^(2)`

The volume of S is equal to:

`Vol_(S) =\int\limits^4_0 {A(x)} \, dx =(√(3) )/(4) \int\limits^4_0 {(4x-x^(2))^(2) } \, dx =14.8`

c)

`y=5x-x^(2) \\(dy)/(dx) =0\\5x-x^(2) =0\\x=5/2\\y(5/2)=25/4\\y=5x-x^(2) \\x^(2) -5x+y=0\\x=(5+-√(25-4y) )/(2)`

The cross section area is equal to:

`A_(1) =(1)/(2) \pi r_(1)^(2) =(1)/(2) \pi ((1)/(2) ((5+√(25-4y) )/(2) -(5-√(25-4y) )/(2) ))^(2) =(1)/(8) \pi (25-4y)\\A_(2) =(1)/(2) \pi r_(2)^(2)=(1)/(2)\pi ((1)/(2) (y-(5-√(25-4y) )/(2) ))^(2) =(1)/(32) \pi (2y-5+√(25-4y) )^(2)`

The volume of S is equal to:

`Vol_(S) =\int\limits^a_b {A_(1)(y) } \, dy+\int\limits^4_0 {A_(2)(y) } \, dy ,where-a=25/4,b=4\\Vol_(S) =\int\limits^a_b {(1)/(8)\pi (25-4y)} \, dy +\int\limits^a_b {(1)/(32)\pi (2y-5+√(25-4y) )^(2) } \, dy =5.17`

d) The cross section area is:

`A_(1) =(1)/(2)ab=(1)/(2) a^(2) =(1)/(2) ((5+√(25-4y) )/(2)-(5-√(25-4y))/(2) )^(2) =(1)/(2) (25-4y)\\A_(1)=(1)/(2)ab=(1)/(2) a^(2) =(1)/(2)(y-(5-√(25-4y))/(2)} )^(2) =(1)/(8) (2y-5+√(25-4y)})^(2)`

The volume of S is equal to:

`Vol_(S) =\int\limits^a_b {A_(1)(y) } \, dy +\int\limits^4_0 {A_(2)(y) } \, dy ,where-a=25/4,b=4\\Vol_(S)=\int\limits^a_b {(1)/(2)(25-4y) } \, dy +\int\limits^4_0 {(1)/(8)(2y-5+√(25-4y))^(2) } \, dy =11.33`