# How to calculate surface area (surfaces of revolution)

You may have seen equations for the surface area of certain shapes like cones, spheres etc. but where did they come from? It is a simple idea. Take an equation for a line, y = something, making sure that the axis does not intersect the line:

then you rotate it around the y axis between x1 and x2:

split the shape up in to bits (resembling small slices of a cone):

and calculate the surface area of each bit and you have the area.
The circumference of a circle is 2πr. The surface area of this small part of the shape is the circumference of the circle multiplied by the length of it:
ΔA  ≈ 2πy•Δs
the '2πy' part is like 2πr and the Δs part is a small length of the line.
Dividing both sides by Δx gives:
 ΔA Δx = 2πy Δs Δx

As Δx gets smaller (Δx →0) we can now say:
 dA dx = 2πy ds dx
I am going to use a previous equation that calculates ds / dx which you can read about here.
 ds dx = √(1 + (dy / dx)2)
so I can put this in to the equation and get:
 dA dx = 2πy√(1 + (dy / dx)2)
All we need to do now is to add all these dA / dx parts up which we can do with integration giving:
 Surface area = ∫ x2 2πy√(1 + (dy / dx)2) dx x1

You can see an example of the surface area of a sphere which I've written here.

## The curved surface of a cone

Let's do an example of the surface area of a cone.
We start by drawing a line from 0 to h (h is the height of the cone) and 'r' is the radius of the base of the cone:
 y = x r h

Now using pythagoras' theorem, the length of s is √(r2 + h2) and is a constant. Let's look at the surface area equation again:
 Surface area of the cone = ∫ h 2πy√(1 + (dy / dx)2) dx 0
I shall break this up in to parts
 y = x r h so we can see that dy dx = r h
I can put these back in to the equation and taking the constant 2π and r / h out, giving us:
 Surface area of the cone = 2πr / h ∫ h (x√(1 + (r / h)2) dx 0
I can modify the 1 + (r / h) by making 1 = h / h and simplifying it to (h2 + r2)/h2 which is another constant I can place outside of the integrand:
 Surface area of the cone = 2πr / h((h2 + r2)/h2) ∫ h x dx 0
which is very simple to integrate, especially considering that the 0 in the integrand removes a part of the equation.
Solving this and simplifying gives us:
The surface area of the cone = πr√(r2 + h2)