# Area Enclosed by Parametric Curves

We know that area under the curve `y=F(x)` is `A=int_a^b F(x)dx` where `f(x)>=0`.

If curve is given by parametric equations `x=f(t)` and `y=g(t)` then using substitution rule with `x=f(t)` we have that `dx=f'(t)dt` and since `x` is changing from `a` to `b` then `t` is changing from `alpha=f^(-1)(a)` to `beta=f^(-1)(b)`. It is not always a case, sometimes `t` is changing from `beta` to `alpha`.

So, `A=int_a^bF(x)dx=int_alpha^beta F(f(t))f'(t)dt=int_alpha^beta g(t)f'(t)dt`.

**Area Enclosed by Parametric Curves**: `A=int_(alpha)^(beta)g(t)f'(t)dt` (or `int_beta^alpha g(t)f'(t)dt`).

**Example**. Find the area under one arch of cycloid. `x=r(t-sin(t))`, `y=r(1-cos(t))`.

One arch of the cycloid is given by `0<=t<=2pi`.

We have that `x_t'=r(1-cos(t))`.

Therefore, `A=int_0^(2pi)r(1-cos(t))r(1-cos(t))dt=`

`=r^2 int_0^(2pi)(1-cos(t))^2dt=r^2 int_0^(2pi) (1-2cos(t)+cos^2(t))dt`.

Now, we need to use double angle formula and integral becomes:

`Ar^2 int_0^(2pi)(1-2cos(t)+1/2(1+cos(2t)))dt=r^2 int_0^(2pi) (3/2 -2cos(t)+1/2cos(2t))dt=`

`=r^2(3/2 t-2sin(t)+1/4sin(2t))|_0^(2pi)=`

`=r^2((3/2 2pi-2sin(2pi)+1/4sin(2* 2pi))-(3/2 *0-2sin(0)+1/4sin(2*0)))=`

`=r^2((3pi-2*0+1/4*0)-(0-2*0+1/4*0))=3pir^2`.