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

Question

1. y = x^3 , y = 8, x = 0; about x = 3
2. y=4x-x^2 , y=3, about x =1
3. x = 2y^2 ,y > 0, x = 2; about y = 2

Answer 1

Recall the formula for cylindrical/shell method is A = 2π∫_a^b r(x)h(x)dx when rotating about a vertical axis and A = 2π∫_c^d r(y)h(y)dy when rotating about a horizontal axis

Problem 1

Height = h(x) = 8-x³ (distance between y=8 and y=x³)

Radius = r(x) = 3-x (distance between x=3 and the y-axis)

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

A = 2π∫_a^b r(x)h(x)dx

A = 2π∫₀² (3-x)(8-x³)dx

A = 2π∫₀² (24-3x³-8x+x⁴)dx

A = 2π∫₀² (x⁴-3x³-8x+24)dx

A = 2π[x⁵/5-(3/4)x⁴-4x²+24x] [0,2]

A = 2π[2⁵/5-(3/4)(2)⁴-4(2)²+24(2)] - 2π[0⁵/5-(3/4)(0)⁴-4(0)²+24(0)]

A = 2π[32/5-(3/4)(16)-4(4)+48]

A = 2π(32/5-48/4-16+48)

A = 2π(32/5-12+32)

A = 2π(32/5+20)

A = 2π(32/5+100/5)

A = 2π(132/5)

A = 264π/5 cubic units

Problem 2

Height = h(x) = 3-(4x-x²) = 3-4x+x² = x²-4x+3 (distance between y=4x-x² and y=3)

Radius = r(x) = 1-x (distance between x=1 and the y-axis)

Bounds = [a,b] = [1,3]

A = 2π∫_a^b r(x)h(x)dx

A = 2π∫₁³ (1-x)(x²-4x+3)dx

A = 2π∫₁³ (x²-4x+3-x³+4x²-3x)dx

A = 2π∫₁³ (-x³+5x²-7x+3)dx

A = 2π[-x⁴/4+(5/3)x³-(7/2)x²+3x] [1,3]

A = 2π[-3⁴/4+(5/3)(3)³-(7/2)(3)²+3(3)] - 2π[-1⁴/4+(5/3)(1)³-(7/2)(1)²+3(1)]

A = 2π[-81/4+(5/3)(27)-(7/2)(9)+9] - 2π[-1/4+5/3-7/2+3]

A = 2π[-81/4+135/3-63/2+9] - 2π[-1/4+5/3-7/2+3]

A = 2π[-243/12+540/12-378/2+108/12] - 2π[-3/12+20/12-42/12+36/12]

A = 2π(27/12) - 2π(11/12)

A = 54π/12 - 22π/12

A = 32π/12

A = 8π/3 cubic units

Problem 3

Height = h(y) = 2-2y² (distance between x=2 and x=2y²)

Radius = r(y) = 2-y (distance between y=2 and the x-axis)

Bounds = [c,d] = [0,1]

A = 2π∫_c^d r(y)h(y)dy

A = 2π∫₀¹ (2-y)(2-2y²)dy

A = 2π∫₀¹ (4-4y²-2y+2y³)dy

A = 2π∫₀¹ (2y³-4y²-2y+4)dy

A = 2π[(2/4)y⁴-(4/3)y³-y²+4y] [0,1]

A = 2π[(1/2)y⁴-(4/3)y³-y²+4y] [0,1]

A = 2π[(1/2)(1)⁴-(4/3)(1)³-(1)²+4(1)] - 2π[(1/2)(0)⁴-(4/3)(0)³-(0)²+4(0)]

A = 2π[1/2-4/3-1+4]

A = 2π[6/12-16/12-12/12+48/12]

A = 2π(26/12)

A = 52π/12

A = 13π/3 cubic units

Hope the answers and explanations helped!

Answer 2

1. v= 2π∫₀²x³(3 - x)dx = 2π∫₀²(3x³ - x⁴)dx = 2π(3x⁴/4 - x⁵/5)₀² = 2π(12 - 32/5) = 56π/5.

2.3 = 4x - x², x = 1 or x = 3; v = 2π∫₁³(x - 1)(4x - x²)dx = 2π∫₁³(- x³ + 5x² - 4x)dx = 2π(- x⁴/4 +5x³/3 - 2x²)₁³ =

2π[(- 81/4 + 45 - 18) - ( - 1/4 + 5/3 - 2)] = 44π/3.

3.v = π∫₂⁸(4 - x/2)dx = (4x - x²/4)₂⁸ = π[(32 - 16) - (8 -1)] = 9π (must be given y = 0)