Consider the region R bounded by y = x^2 and y = 2 − x^2.

Question

Consider the region R bounded by y = x² and y = 2 − x².

(a) Sketch the graphs of the two curves and shade the region R.

(b) Find the volume of the solid obtained by revolving the region R about y-axis.

(c) Find the volume of the solid obstained by revolving the region R about x-axis.

Answer 1

b) When revolved around the y-axis, the shapes of each cross section are circles where the radius of each circle is the x-value of each function (so for y = x², that mean x = √y or r = √y). The only thing is that the functions switch over when you get to y = 1. So one way to do this is to break it into 2 pieces:

₀∫¹πr²dy + ₁∫²πr²dy = π[(₀∫¹(√y)²dy + ₁∫²(√(2-y)²dy] = π(₀∫¹ydy + ₁∫²(2-y)dy) = [π(y²/2) evaluated between 0 and 1] + [π(2y - y²/2) evaluated between 1 and 2] = π/2 + π/2 = π

An easier way is to use symmetry.

c) When revolved around the x-axis, you get a donut-shape where the cross sections taken in the x-direction are "washers" where the outside radius is "2-x²" and the inside radius is "x²" so the integral becomes:

_-1∫¹π(r_o² - r_i²) dx = π[_-1∫¹((2-x²)² - (x²)² dx] = π[_-1∫¹(4-4x²+x⁴ - x⁴) dx] = π[_-1∫¹(4 - 4x²) dx] =16π/3