# Surface areas

1. Jan 22, 2004

### tandoorichicken

I know that the equation for the surface area of any solid of revolution around, say, the x-axis is
$$SA = 2\pi\int_{a}^{b} y\sqrt{1 + (\frac{\,dy}{\,dx})^2} \,dx$$

What I need is the same formula except in parametric terms, like if the problem was given in terms of x(t) and y(t). Any takers?

2. Jan 22, 2004

### KLscilevothma

If it revolves about the x-axis on the closed interval [a,b], then
$$SA = 2\pi\int_{a}^{b} y(t)\sqrt{[x'(t)]^2 + [y'(t)]^2} \,dt$$

For example. if the surface are of the solid generated by revolving the region enclosed by the curve with parametric equations x(t), y(t) from t = 0 to t = pi/2, then the upper limit, b = pi/2, lower limit a = 0.

If it revolves about the y-axis on the closed interval [a,b], then
$$SA = 2\pi\int_{a}^{b} x(t)\sqrt{[x'(t)]^2 + [y'(t)]^2} \,dt$$