Area of a Circle

 

`y=rsin(theta)` `x=rcos(theta)`

The area under the curve, which is clearly one quarter the area of a circle, can be calculated as the integral `int F(x)dx`.

`A/4=int_0^(x=r) ydx`

Where we can express `y` in terms of the radius and the angle subtending between the radius and the `x`-axis.

`A/4=int_0^(x=r) rsin(theta)dx`

We can calculate the change term, `dx`, through differentiation,

`x=rcos(theta) rarr (dx)/(d theta)=r(dcos(theta))/(d theta)=-rsin(theta)`

and so,

`dx=-rsin(theta)d(theta)`

Substituting this into the integral we obtain

`A/4=-r^2int_0^(x=r) sin^2(theta) d(theta)`

Next the limits of integration must be changed from x to theta. We can see from the diagram that when `x=0`, the angle `theta=pi/2` and that when `x=r`, `theta=0`.

`A/4=-r^2int_(pi/2)^(theta=0) sin^2(theta) d(theta)`

To evaluate this integral, we first make use of the half-angle formula, viz.

`sin^2(theta)=(1-cos(2theta))/2`

This makes the integral easier to solve,

`A/4=-r^2/2int_(pi/2)^(theta=0) (1-cos(2theta)) d(theta)`

Applying the distributive property of integration, we can expand the brackets,

`A/4=-r^2/2(int_(pi/2)^(theta=0)d theta - int_(pi/2)^(theta=0) cos(2theta)d(theta))`

The firt integral is simply `[theta]_(pi/2)^(0)` and the second can be calculated using u-substitution. Using the substitution,

`u=2theta` we have `(du)/(d theta)=2` and `d theta = (du)/2`

Therefore,

`int cos(2theta)d(theta)=1/2 int cos(u)du=-1/2 sin(u) = -1/2 sin(2theta)`

Using this result the equation for the area becomes,

`A/4=-r^2/2[theta + 1/2 sin(2theta)]_(pi/2)^(0)=-r^2/2[0-pi/2 + 1/2 sin(0)-1/2 sin(pi)]`

Both sine terms equal zero (see the table of special values for the trig functions), we cancel the minus signs, remove the brackets, and (recalling that the area calculated is for one quarter of a circle) cancel the denominators to obtain,

`A=pir^2`