How Is the Surface Area of a Rotated Curve Calculated Using Integrals?

[countzander]

Homework Statement

Consider the surface S formed by rotating the graph of y = f(x) around the x-axis between x = a and x = b. Assume that f(x) ≥ 0 for a ≤ x ≤ b. Show that the surface area of S is 2π times integral of f(x)sqrt(1 + f ' (x)^2) dx from a to b.

http://i.imgur.com/qFeGP.png

Homework Equations

The integral of the magnitude of the cross product of the partial derivatives of parameterization vector, r = r(s,t). The region is R.

The Attempt at a Solution

I tried parameterizing the surface with parameters of x and f(x). The surface I set as g(x,f(x)). But when I took the cross product of that thing, I ended up with a useless statement involving partial derivatives which does not lead to the solution.

Homework Statement

Homework Equations

The Attempt at a Solution

 

[haruspex]

Homework Equations

The integral of the magnitude of the cross product of the partial derivatives of parameterization vector, r = r(s,t). The region is R.

That's not an equation. Can you elaborate?

I tried parameterizing the surface with parameters of x and f(x). The surface I set as g(x,f(x)). But when I took the cross product of that thing, I ended up with a useless statement involving partial derivatives which does not lead to the solution.

Please post details of you working.

 

[tiny-tim]

hi countzander!

I tried parameterizing the surface with parameters of x and f(x) …

why so complicated?

just use trig! ​

(and f' = tan)

 

[countzander]

http://i52.photobucket.com/albums/g12/countzander/Untitled-1.png

 

[LCKurtz]

Homework Statement

Consider the surface S formed by rotating the graph of y = f(x) around the x-axis between x = a and x = b. Assume that f(x) ≥ 0 for a ≤ x ≤ b. Show that the surface area of S is 2π times integral of f(x)sqrt(1 + f ' (x)^2) dx from a to b.

http://i.imgur.com/qFeGP.png

Homework Equations

The integral of the magnitude of the cross product of the partial derivatives of parameterization vector, r = r(s,t). The region is R.

The Attempt at a Solution

I tried parameterizing the surface with parameters of x and f(x). The surface I set as g(x,f(x)).

That isn't two parameters. $x$ is free to use as a parameter but then $f(x)$ is determined. You need another parameter if you want to do it that way. I would suggest $\theta$, the angle of rotation as the second parameter. Express your surface as$$
\vec R(x,\theta) = \langle x, ?, ?\rangle$$where the question marks are the appropriate expressions for $y$ and $z$ in terms of $x,\, f(x),\, \theta$.