Calculus

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.

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  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

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    posted by Steve

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