Y = sin pi * x Arc Length/Surface Revolution

[kevtimc]

Homework Statement

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

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 . . .

 

[tiny-tim]

Welcome to PF!

Welcome to PF!

(have a pi: π )

(and in tex, it's \pi)

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

… what is the actual question?

 

[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. $$u = tan \theta , sec \theta = \sqrt{1 + u^2}$$