# Integral to find the Surface area

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 .....

## Answers and Replies

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:
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?

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.