Question

Find the volume of the solid whose base is the region in the first quadrant bounded by y=x^4, y=1 and the y-axis and whose cross-sections perpendicular to the x axis are semicircles.

Answer 1

The base of the solid - call it B - is the set of points

B = {(x, y) : 0 ≤ x ≤ 1 and x ⁴ ≤ y ≤ 1}

Recall the area of a circle with radius r is πr ²; in terms of the diameter d = 2r, the area is π (d/2)² = π/4 d ². Then the area of a semicircle with the same diamater is half of this, π/8 d ².

Cross sections of the solid in question are semicircles arranged perpendicular to the x-axis, which means the diameters of each cross section corresponds to the vertical distance between y = x ⁴ and y = 1 for any given values of x between 0 and 1. So d = 1 - x ⁴, which makes the area of each cross section come out to π/8 (1 - x ⁴)².

Split up the solid into very thin cross sections with "base" area π/8 (1 - x ⁴)² and thickness ∆x. Take the sum of these half-cylinders' volumes, then let ∆x converge to 0. In short, we get the total volume by integrating,

$\displaystyle \int_0^1\frac\pi8(1-x^4)^2\,\mathrm dx = \frac\pi8\int_0^1(1-2x^4+x^8)\,\mathrm dx = \boxed{\frac{4\pi}{45}}$