Find the volume of the solid obtained by rotating the region in the first quadrant bounded by 𝑦=𝑥^5, 𝑦=1, and the 𝑦-axis about the line 𝑦=−5.

Answer 1

By disk Method V = π ∫₀¹[6² - (x⁵ + 5)²]dx = π∫₀¹(11 - x¹⁰ - 10x⁵)dx = π(11x - x¹¹/11 - 5x⁶/3)₀¹ =

π(11 - 1/11 - 5/3) = 305π/33

Answer 2

You can do a volume by "washers" which will use the formula V = π ∫ r₁² - r₂² dx.

In this case, our washers will be vertical, of infinitesimal width, dx. The bigger radius is constant (6), which will mean our solid of revolution will end up being a cylinder with radius 6 and height 1 (since the bounds of integration are x = 0 and x = 1). This cylinder, of volume 36π, will have a "tornado-shaped" volume carved out of its center:

V = π ∫₀¹ (36 - (5 + x⁵)² dx

V = π ∫₀¹ (36 - (25 + 10x⁵ + x¹⁰))dx

V = π ∫₀¹ (11 - 10x⁵ - x¹⁰)dx

V = π [ 11x - 5/3x⁶ - 1/11x¹¹ ]₀¹ = π (11 - 5/3 - 1/11) = 305/33 π