Integral to find the Surface area

1. Jun 22, 2012

c77793

For a given function f(x)=x, if we rotate this function from [0,1] around the x-axis, we'll have a cone.... why can't I find its surface area by adding the perimeter of all the circunferences from [0,1] with radius = f(x)?Something like 2*pi∫xdx? The formulas says we have to do 2*pi∫x*√(i+y'²)dx .....

2. Jun 22, 2012

mathman

Since the function f(x) is not parallel to the x axis the area differential has to take into account that effect, which for your problem is √2.

In general, you need the differential of arc length of the revolving figure which is
√(1+y'²)dx.

Last edited: Jun 22, 2012
3. Jun 22, 2012

c77793

But to find the volume of the cone, for exemple, the fact that the function isn't parallel to the x-axis doens't influence that "volume is the sum of all areas of all circles".... What changes?

4. Jun 23, 2012

mathman

Think of the area of the cone's surface as the sum of a bunch of thin circular strips. The extra factor is the width of the strip.

5. Jun 24, 2012

HallsofIvy

Consider this fallacious argument: If we divide the interval from x= 0 to x= 1 into n equal subintervals, each of length 1/n, we can "approximate" the line from (0, 0) to (1, 1) by a broken line, the horizontal line from (0, 0) to (1/n, 0), the vertical line from I(1/n, 0) to (1/n, 1/n), the horizontal line from (1/n, 1/n) to (2/n, 1/n), the vertical line from (2/n, 1/n) to (2/n, 2/n), etc.

The area under the horizontal line from (k/n, k/n) to ((k+1)/n, k/n) is (k/n)(1/n)= k/n^2 so the total area is $\sum_{k=0}^n k/n^2= (1/n^2)\sum_{k=0}^n k= (1/n^2)(n(n+1)/2= (n^2+ n)/2n= n/2+ 1/2. And the limit, as n goes to infinity, is 1/2, the area under the straight line. But if we, instead, use the length of the broken line to approximate the length of the line, each segment has length 1/n and there are 2n such segments so that we get a length of 2 which is wrong. The length of the line segment from (k/n, k/n) to ((k+1)/n, (k+1)/n) simply cannot be approximated by the length of those two horizontal and vertical segments. We have to use the Pythagorean theorem to get the length of each segment: [itex]\sqrt{(1/n)^2+ (1/n)^2}= (1/n)\sqrt{2}$. And that leads to the arclength formula.