Площа поверхні поверхонь обертання

Consider a straight segment rotated about an axis. The segment will sweep out the shape of a frustum. Our first step in computing general surface areas of surfaces of revolution is to find the lateral surface area of this frustum (call this $S$).

Suppose that the large radius is $r_2$, the small radius is $r_1$, and the length of the segment is $L$.

In the trivial case in which $r_2 = r_1 = r$, the frustum degenerates into a cylinder, so the area is simply $S = 2\pi rL$.

In the nontrivial case in which $r_2 > r_1$, we can calculate the area of the frustum as the area of a large cone minus the area of a small cone. The large cone has radius $r_1$ and slant height $s_1 = \frac{r_2}{r_2 - r_1}L$. The small cone has radius $r_2$ and slant height $s_2 = \frac{r_1}{r_2 - r_1}L$.

The lateral surface area of the frustum is: $$S = \pi r_2s_2 - \pi r_1s_1$$ $$S = \frac{\pi L}{r_2 - r_1}(r_2^2 - r_1^2)$$ $$S = (r_2 + r_1)\pi L$$ $$S = 2\pi rL$$ where $r = \frac{r_1+r_2}{2}$ is the average radius.

Now, suppose that we have a curve, described by the graph of a function $y = f(x)$ in the Cartesian coordinate plane, for $a\le x\le b$. We break the curve into tiny segments, and we rotate these segments around the $x$-axis along with the curve itself, creating lots of tiny frusta. These frusta approximate the area of the resulting surface, and the approximation becomes exact as all of the segments approach a length of $0$ and as we increase the number of segments without bound.

Each segment has an average radius $r = f(x)$ The width of the segment is $dx$, and the change in the $y$-coordinate is $dy = f'(x)dx$. Thus, by the Pythagorean Theorem, the length of the segment is: $$dL = \sqrt{dx^2 + dy^2} = \sqrt{1+[f'(x)]^2}dx.$$ Thus, the area of the tiny frustum is: $$dS = 2\pi r dL = 2\pi f(x)\sqrt{1+[f'(x)]^2}dx.$$ It follows that the area of the surface is: $$S = \int_{a}^{b} 2\pi f(x)\sqrt{1+[f'(x)]^2}dx.$$