Surface of a solid of revolution

Consider the graph of a function f on an interval [a,b].

If we rotate this graph around a horizontal axis of rotation given by y = A, where A < min(f),

the surface area of the resulting surface is given by

If we rotate this graph around a vertical axis of rotation given by x = A, where A < a,

the surface area of the resulting surface is given by

More complicated surfaces have to be decomposed into surfaces of the above type.

Example: Consider the graph of f (x) = cosh(x) on the interval [0,1]. Find the surface area of the surface obtained by rotating this graph about the x-axis.

Solution: We start with a picture:

The axis of rotation is given by y = 0, so we calculate:

Surface given by a revolving parametric curve

Consider a parametric curve x = x(t), y = y(t) for t from [α,β]. Assume that y(t) ≥ 0 for all t and x(t) is monotone:

The surface area of the surface obtained by revolving this curve about a horizontal axis of rotation given by y = A, where A < min(y(t)), is

The surface area of the surface obtained by revolving this curve about a vertical axis of rotation given by x = A, where A < min(x(t)), is

Mass and centre of gravity
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