Question P.175: Let f be a continuous function on [0, 1], which is bounded b......
Let f be a continuous function on [0, 1], which is bounded below by 1, but is not identically 1. Let R be the region in the plane given by 0 ≤ x ≤ 1, 1 ≤ y ≤ f(x). Let
R_1 = \{(x, y) ∈ R|y ≤ \bar{y}\}~~~and~~~R_2 = \{(x, y) ∈ R|y ≥\bar{ y}\},
where \bar{y} is the y-coordinate of the centroid of R. Can the volume obtained by rotating R_1 about the x-axis equal that obtained by rotating R_2 about the x-axis?
Solution 1. To find an example, we let R be the triangle with vertices (0, 1), (1, 1), and (0, b), where the parameter b > 1 will be chosen later. Then
f(x) = b − (b − 1)x.
There are several expressions for \bar{y} in terms of integrals. For instance,
The horizontal line y = (b + 2)/3 intersects y = f(x) at (2/3,(b + 2)/3). The volume obtained by rotating R_2 about the x-axis is then
\int_{0}^{2/3}\left(\pi\left(f(x)\right)^{2}-\pi\left({\frac{b+2}{3}}\right)^{2}\right)\,d x={\frac{20b^{2}-4b-16}{81}}\,\pi.To find the volume obtained by rotating R_1 about the x–axis, we compute the volume obtained by rotating R, and subtract the volume obtained by rotating R_2. The calculation yields
{\frac{7b^{2}+31b-38}{81}}\,\pi.The two volumes are equal when b = 22/13 and when b = 1, but the solution b = 1 is degenerate.
Solution 2. As in the first solution, we let R be the triangle with vertices (0, 1), (1, 1), and (0, b), but this time we avoid computing integrals by using geometry. Recall that by Pappus’ theorem, the volume swept out by rotating a region about an exterior axis is the product of the area of the region and the distance traveled by its centroid. Therefore, if \bar{y}_1~and~\bar{y}_2 are the y-coordinates of the centroids of R_1~and~R_2 respectively, the two volumes will be equal if and only if
(Area~of~R_1) · 2π\bar{y}_1 = (Area~of~R_2) · 2π\bar{y}_2.
Now R is a triangle, so its centroid, being the intersection of its medians, is two-thirds of the way down from (0, b) to the line y = 1; that is, \bar{y} = (b+2)/3. Similarly, for the triangle R_2 we have
{\overline{{y}}}_{2}={\frac{b+2{\overline{{y}}}}{3}}={\frac{5b+4}{9}}\,.Also, \bar{y} is the weighted average of \bar{y}_1~and~\bar{y}_2, where the respective weights are the areas of R_1~and~R_2. By similar triangles, the area of R_2 is 4/9 times the area of R, so
\overline{{{y}}}=\frac{5}{9}\overline{{{y}}}_{1}+\frac{4}{9}\overline{{{y}}}_{2}\,;this yields \bar{y}_1 = (7b+ 38)/45. The condition for the volumes to be equal now becomes
{\frac{5}{9}}\cdot{\frac{7b+38}{45}}={\frac{4}{9}}\cdot{\frac{5b+4}{9}},and we again find b = 22/13.