Y = sin pi * x Arc Length/Surface Revolution

1. Nov 27, 2008

kevtimc

1. The problem statement, all variables and given/known data

y = sin $$\pi$$x Using arc length and surface revoultion on x-axis 0 <= x <= 1

3. The attempt at a solution

d/dx sin $$\pi$$x = $$\pi$$ cos $$\pi$$x
($$\pi$$ cos$$\pi$$x)^2 = $$\pi$$^2 cos^2$$\pi$$x

$$\int sin pi * x * 2 * pi * \sqrt{1 + pi^2 * cos^2 (pi*x)}$$
u = pi cos (pi * x)
du = -pi^2 * sin (pi * x) dx

-1/2pi$$\int \sqrt{1 + u^2}$$
u = tan $$\alpha$$
du = sec^2 $$\alpha$$

We get the integral of sec^3,

This doesn't seem to be right, and if it is, the limits of integration don't work out . . .

Last edited: Nov 27, 2008
2. Nov 28, 2008

tiny-tim

Welcome to PF!

Welcome to PF!

(have a pi: π )

(and in tex, it's \pi)

… what is the actual question?

3. Nov 28, 2008

kevtimc

Sorry, I'm just getting used to all the quirks of latex.

The question is to find the length using surface area revoultion forumla of y = sin (pi * x) (rotated about the x-axis) from 0 - 1.

$$\int2\pi*sin(\pi*x) * \sqrt{1 + (sin(\pi*x)')^2}$$

I think sec^3 is actually correct. You have to reverse substitute using reverse trigonometric identites. $$u = tan \theta , sec \theta = \sqrt{1 + u^2}$$