Use the method of cylindrical shells to find the volume generated by rotating the region bounded by the given curves about the y-axis.

Question

1. y =square root cude x, y = 0, x= 1
2. y = e^-x^2 , y=0,x=0,x=1
3. y= x^2 ,y = 6x-2x^2

Answer 1

Recall the formula for cylindrical/shell method is A = 2π∫_a^b r(x)h(x)dx

Problem 1

Height = h(x) = ³√x

Radius = r(x) = x

Bounds = [a,b] = [0,1]

A = 2π∫_a^b r(x)h(x)dx

A = 2π∫₀¹ (x·³√x)dx

A = 2π∫₀¹ (x·x^1/3)dx

A = 2π∫₀¹ x^4/3dx

A = 2π(x^7/3/(7/3)) [0,1]

A = 2π(3/7)x^7/3 [0,1]

A = (6π/7)x^7/3 [0,1]

A = (6π/7)(1)^7/3 - (6π/7)(0)^7/3

A = 6π/7 cubic units

Problem 2

Height = h(x) = e^{-x^2}

Radius = r(x) = x

Bounds = [a,b] = [0,1]

A = 2π∫_a^b r(x)h(x)dx

A = 2π∫₀¹ xe^{-x^2}dx

A = 2π(-1/2)e^{-x^2} [0,1]

A = -πe^{-x^2} [0,1]

A = [-πe^{-1^2}] - [-πe^{-0^2}]

A = -π/e + π

A = π - π/e

A = π(1-1/e) cubic units

Problem 3

Here, since we are given two curves, our height will be the difference between the upper and lower functions:

Height = h(x) = f(x)-g(x) = (6x-2x²)-(x²) = 6x-3x²

Radius = r(x) = x

Bounds = [a,b] = [0,2]

A = 2π∫_a^b r(x)h(x)dx

A = 2π∫₀² x(6x-3x²)dx

A = 2π∫₀² 6x²-3x³)dx

A = 2π[2x³-(3/4)x⁴] [0,2]

A = 2π[2(2)³-(3/4)(2)⁴] - 2π[2(0)³-(3/4)(0)⁴]

A = 2π[2(8)-(3/4)(16)]

A = 2π(16-48/4)

A = 2π(16-12)

A = 2π(4)

A = 8π cubic units



Hope these answers and explanations helped!