Question

Let 𝑅 be the region bounded by the four straight lines 𝑦=𝑥, 𝑥+𝑦=4, 𝑦=𝑥−2 and 𝑥+
𝑦 = 2. Find the surface area of the surface obtained by rotating the region 𝑅 about the 𝑥-axis for 1 complete revolution.

Answer 1

The region is just a square of side √2, with its center at (2,1).

Using the Theorem of Pappus, the surface area is thus the perimeter of the square times the distance traveled by its centroid:

a = 4√2 * 2π = 8π√2

Doing it using the usual definition,

area swept out by the line y=x for x in [1,2],

a = ∫[1,2] 2πy√(1+(y')^2) dx
= ∫[1,2] 2πx√2 dx
= 3π√2

the area swept out by the line y=2-x for x in [1,2] is

∫[1,2] 2π(2-x)√2 dx = π√2

That takes care of the left half of the area. By symmetry, the right half is the same, so the total area is as first calculated: 8π√2