Set up the integral that would give the volume V generated by rotating the region bounded by the given curves about the y-axis.

Question

y = x⁵,    y = 0,    x = 2

Disk/Washer Method

Answer 1

This is an interesting revolution about the y-axis. We are not interested in what is “inside” y = x⁵ and x = 0, but in rotating what is “outside” y = x⁵ up to x = 2.

y = x⁵ ==> x = y¹/⁵

Integration on y, from y = 0 to y-max = 2⁵

Each “disk" (or “washer”) looks like: Vᵢ = [ π(2)² - π(xᵢ)² ]Δy = [ π(2)² - π(yᵢ¹/⁵)² ]Δy= [ π(2)² - π(yᵢ²/⁵) ]Δy

V = π ∫₀ʸ⁻ᵐᵃˣ (2² - y²/⁵)dy , y-max = 2⁵

===== We stop here, because we are asked only to “setup” the Integral.

But, this is like inviting a Mathematician over to dinner, and eating their dessert in front of them.

So, for posterity, written in invisible ink, I get...

V = π ∫₀ʸ⁻ᵐᵃˣ (2² - y²/⁵)dy = π [2²y - (y⁷/⁵ / (7/5)) ] | ₀ʸ⁻ᵐᵃˣ = π [2²(2⁵) - ((2⁵)⁷/⁵ / (7/5)) ]

= π [2⁷ - (5/7)2⁷ ] = π(2/7)2⁷ = 2⁸π/7 ==> V = 256π/7

Answer 2

The graphs of y = x⁵ and x = 2 intersect at the point (2, 32)

If y = x⁵, then x = y^1/5.

Use the disk method:

Take a thin horizontal cross section of the region of height Δy and length x.

Rotate the cross section to obtain a disk of radius x and thickness Δy

Volume of disk = πx²Δy = πy^2/5Δy

V = π∫_{(0 to 32)} y^2/5dy