area of surface of revolution: y=3x from x=1 to x=4 is rotated about the y axis.

[hndalama]

I am learning to find the area of a surface of revolution. The method I am being taught is to divide the surface into frustrums and then calculate the lateral area of the frustrums and then integrate to add up the areas of the frustrums. While learning this I had an idea of a different method. Instead of dividing the surface into frustrums, what about dividing it into cylinders and then the lateral surface area of the cylinders will approximate the area of the surface. And when I integrate, as the height approaches 0 the lateral surface area should equal the area of the surface. as far as I can tell the logic is the same so both should work. The problem is mine is not working and I don't know why.

Take this question for example.

The line segment y=3x from x=1 to x=4 is rotated about the y axis. Find the area of the surface of revolution.

so since the surface is formed by rotating around the y-axis, I divide it into horizontal cylinders with radius=x and height delta y. Because my delta variable is y I rewrite x in terms of y, to get x=y/3. My endpoints in terms of y become y=3 and y=12.

The lateral surface area of a cylinder is 2pi*r*h . in terms of my variables it is 2pi*y/3*delta y . I have an equation in delta y so integrating this should give me the sum of the areas as delta y approaches zero.

The antiderivative is pi*y²/3 , evaluating this at y= 12 and 3 I get that the area is 45pi.

The answer by the frustrum method is 15pi(10)^0.5.

Why is it not working?

 

[mmm4444bot]

While learning this I had an idea of a different method. Instead of dividing the surface into frustrums, what about dividing it into cylinders and then the lateral surface area of the cylinders will approximate the area of the surface ... Why is it not working?

We can't use cylinders, for determining the surface area of a conical shape. The height of the cylinder is less than the slant height of the frustum, so the surface area of a cylinder will be less than the surface area of a frustum having the same vertical height.

If you "unwrap" a cylinder, you get a rectangle. With a frustum, you get part of a washer (i.e., circular segment) whose width is the slant height. The slant height is greater than the rectangle's height, so the rectangle has smaller area; that's why the cylinder method under-counts surface area of a conical shape. :cool:

 

[mmm4444bot]

The cylinder height is h.

The frustum's slant height is H.

 

[MarkFL]

Yes, by using cylinders instead of conical frustums, this would be analogous to saying the arc-length of a curve from f(a) to f(b) (where a < b) is simply s = b - a (this is only true for a horizontal line). The magnitude of the instantaneous slope of the curve must be taken into account all along the way, which the frustums do.

 

[hndalama]

thank you