# Area of a Surface Revolution

Hello

can someone please give me some advice on applying the area of a surface revolution. I never seem to plugin the right formula.

In the formula S(from a to b) 2pi f(x)**sqrt(1 + f'(x)^2) I never know what to put for f(x)** some times the answers have just x and sometime it has the function of f(x).[/quote]
I don't know what you mean by "some times the answers have just x and sometime it has the function of f(x)." What does the problem itself have? In order to find the surface area of a "surface of revolution", you have to know what is being revolved! If you are given a function y= f(x), that is then revolved around the x-axis to form a 3 dimensional figure (the "surface of revolution"), then the surface area is given by that formula:
$$2\pi \int_a^b f(x)\sqrt{1+ f'^2(x)}dx$$.

The "f(x)" has to be given as a description of the curve, y= f(x), being rotated around the x-axis.

You probably know that the surface area of a cylinder, of radius r and length L is given by $2\pi r L$: that's really the circumference of the circle times the length.

If y= f(x) is rotated around the x-axis, then each point describes a circle of radius f(x) and so circumference $2\pi f(x)$. The length of a tiny ("infinitesmal") cylinder is given by the arclength formula: $L= ds= \sqrt{1+ f'^2(x)}dx$. So the differential of area is given by $dA= 2\pi f(x)\sqrt{1+ f'^2(x)}dx$. You integrate that to get the surface area: $A= \int_a^b2\pi f(x)\sqrt{1+ f'^2(x)}dx$.

HOWEVER, and this may be what is causing your confusion, if the graph of y= f(x) is rotated around the y-axis rather than the x-axis, each point follows a circle of radius x, not f(x). That circle has radius x, the distance from the point to the y-axis, not f(x), the distance from the point to the x-axis.

If you are given y= f(x), its graph being rotated around the x-axis, then the surface area is given by
$$2\pi\int_a^b f(x)\sqrt{1+ f '^2(x)}dx$$

If you are given y= f(x), its' graph being rotated around the y-axis, then the surface area is given by
$$2\pi\int_a^b x \sqrt{1+ f '^2(x)}dx$$

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