Surface area of a rotated curve

[glid02]

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

http://ada.math.uga.edu/webwork2_files/tmp/equations/18/d733a6e52ad8ca260230969bdc3f401.png

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.

 

[HallsofIvy]

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

http://ada.math.uga.edu/webwork2_files/tmp/equations/18/d733a6e52ad8ca260230969bdc3f401.png

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.

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).

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.

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