# Homework Help: Surface area of a rotated curve

1. Nov 28, 2006

### glid02

Here's the question:
Find the area of the surface obtained by rotating the curve

from x=0 to x=9 about the x-axis.

I'm supposed to parametrize the curve, using rcos(theta) as x and rsin(theta) as y, at least I think I am.

That would give f(x,y) = 3rcos^3(theta),rsin(theta)

Then find the partial derivatives with respect to r and theta and find their cross product. Then find the magnitude of the cross product and integrate with limits int[0-2pi] int[0-9].

Is this right? I can't find any information on the internet to do it this way and the book isn't much help either.

Thanks.

Last edited by a moderator: May 2, 2017
2. Nov 29, 2006

### HallsofIvy

No. That is a circle in the xy-plane. Your curve, in the xy-plane, is y= 3x3, not at all a circle. When you "rotate" around the x-axis, every point, with coordinates (x0, y) in the xy-plane, traces out a circle of points in the (x0, y, z) plane.
Since the radius of the circle is y= 3x^3, that would have parametric equations $x= x_0, y= 3x_0^3cos(\theta ), z= 3x_0^3 sin(\theta)$.

Using the parametric equations I gave, with x0 and $\theta$ as parameters, that should work. However, it looks awfully complicated to me. The area of the curved surface of a cylinder with radius y and height ds is $2\pi y ds$. For y= 3x3,
$$ds= \sqrt{1+ (9x^2)^2}dx= \sqrt{1+ 81x^4}dx$$.
The integral for surface area would be given by
$$2\pi \int_0^9(3x^3)\sqrt{1+81x^4}dx$$

Last edited by a moderator: May 2, 2017