Y = sin pi * x Arc Length/Surface Revolution

[kevtimc]

Homework Statement

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

The Attempt at a Solution

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

\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 \pix 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)&#039;)^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}