Surfaces of revolution

Given a region in a plane, you could consider its area directly or revolve it around a line and consider the volume of the resulting solid of revolution. (Of course, you could do other things, but these are the only things that we're considering in this course. Come back in Calculus 3 if you want more.) For simplicity, suppose that the region has two parallel line segments as opposite sides and every line between those passes through the region in only one segment. Then if you set up a coordinate system so that one axis is parallel to these two parallel sides, then there are numbers a and b (with a ≤ b) and functions f and g (defined on [a, b], with f ≥ g), so that the region is given by the inequalities a ≤ x ≤ b and f(x) ≥ y ≥ g(x) in the variables x and y. If f and g are continuous, then we can also say that the region is bounded by x = a, x = b, y = f(x), and y = g(x), and the area of the region is
∫ab (f(x) − g(x)) dx.
If a region doesn't have the appropriate shape, it may still be possible to divide it into regions with such a shape. You can also swap x and y if that makes the region easier to describe or the integral easier to compute.

If the region is revolved around a line in the plane, then this is simple to describe only when this line of revolution is parallel or perpendicular to the region's parallel boundary lines, in which case you can set up the coordinate system so that the line of revolution is one of the coordinate axes. If you revolve the region described above around the x-axis, then the volume of the resulting solid of revolution is

∫ab π(f(x)2 − g(x)2) dx,
assuming that f and g are continuous and g ≥ 0. If instead you revolve this region around the y-axis, then the volume of the resulting solid of revolution is
∫ab 2πx(f(x) − g(x)) dx,
assuming that f and g are continuous and a ≥ 0. If the line of revolution is not parallel or perpendicular to parallel boundary lines, then you need to describe the region in a more complicated way by dividing it into regions with boundary lines that are parallel or perpendicular to the line of revolution. However, you will learn how to find the volumes of much more general solids if you take Calculus 3.

If instead of a region in the plane, you start with a curve in the plane, then we can only handle this for now if the curve is a graph of a differentiable function f. Specifically, if a ≤ b and f is defined on [a, b], then the length of the graph of y = f(x) is

∫ab √(f′(x)2 + 1) dx,
assuming that f′ is continuous on [a, b]. You'll learn how to handle more general curves if you take Calculus 2 (at the very end) and even more in Calculus 3 (towards the beginning).

If instead you revolve this graph around the x-axis, then the area of the resulting surface of revolution is

∫ab 2πf(x)√(f′(x)2 + 1) dx,
assuming that f′ is continuous on [a, b] and f ≥ 0. Finally, if you revolve this same curve around the y-axis, then the area of the resulting surface is
∫ab 2πx√(f′(x)2 + 1) dx,
assuming that f′ is continuous on [a, b] and a ≥ 0. (For some reason, this one's not in the textbook.) You will learn how to find the areas of much more general surfaces if you take Calculus 3.
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This web page was written in 2018 by Toby Bartels, last edited on 2018 December 10. Toby reserves no legal rights to it.

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