compute the volume of the solid obtained by rotating the region bounded by y=6x^2, 2x+y=8, and the y-axis about the line x=-1.

Answer 1

Recall Shell/Cylindrical Method (Vertical)

A = 2π∫_a^b r(x)h(x)dx

Determine height and radius functions

Height = h(x) = (8-2x)-6x² = 8-2x-6x² (distance between y=8-2x and y=6x²)

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

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

Evaluate integral

A = 2π∫_a^b r(x)h(x)dx

A = 2π∫₀¹ (x+1)(8-2x-6x²)dx

A = 2π∫₀¹ (8x-2x²-6x³+8-2x-6x²)dx

A = 2π∫₀¹ (-6x³-8x²+6x+8)dx

A = 2π[-(6/4)x⁴-(8/3)x³+3x²+8x] [0,1]

A = 2π[-(3/2)x⁴-(8/3)x³+3x²+8x] [0,1]

A = 2π[-(3/2)(1)⁴-(8/3)(1)³+3(1)²+8(1)] [0,1]

A = 2π(-3/2-8/3+3+8)

A = 2π(-9/6-16/6+18/6+48/6)

A = 2π(41/6)

A = 82π/6

A = 41π/3 cubic units

Hope this helped!