Calculations of areas in the plane using Green's theorem

A very powerful tool in integral calculus is Green's theorem. Let's consider a vector field $F(x,y)=(P(x,y),Q(x,y))$, $C$ being a closed curve in the plane and $S$ the interior surface delimited by the curve.

Then: $$\int_C F \ dr=\iint_S\big( Q_x-P_y \big) \ dx \ dy$$

The application in the calculation of areas is the following one. We will think such a field being $Q_x-P_y=1$.Then the term on the right is only the area of the enclosure $S$. Therefore, we will be able to calculate it by doing one line integral on the border of the enclosure.

There are many fields that satisfy the property $Q_x-P_y=1$, but the most used are:

• $F(x,y)=(0,x)$
• $F(x,y)=(-y,0)$
• $F(x,y)=(-y,x)$

For example, we are going to calculate the area delimited by the parametric curve: $$\alpha(\theta)=(3\sin(2\theta)\cdot\cos(\theta),3\sin(2\theta)\cdot\sin(\theta))$$

with $\theta\in\big[0,\dfrac{\pi}{2}\big]$.

Now we take the vector field $F(x,y)=(0,x)$ and integrate the field along the curve $\alpha(\theta)$. Let's calculate: $$\alpha'(\theta)=(6\cos(2\theta)\cdot\cos(\theta)-3\sin(2\theta)\cdot\sin(\theta),6\cos(2\theta)\cdot\sin(\theta)-3\sin(2\theta)\cdot\cos(\theta))$$

and we have:

$$\begin{array}{rl} \text{Area}=& \iint_D 1 \ dx \ dy=\int_C F \ dr = \int_0^{\frac{\pi}{2}} F(\alpha(t))\cdot\alpha'(t) \ dt \\ =& \int_0^{\frac{\pi}{2}} (0,3\sin(2t)\cdot\sin(t)) \cdot (6\cos(2t)\cdot\cos(t)-3\sin(2t)\cdot\sin(t), \\ & \quad \quad \quad \quad 6\cos(2t)\cdot\sin(t)-3\sin(2t)\cdot\cos(t)) \ dt \\ =& \int_0^{\frac{\pi}{2}} 3\sin(2t)\cos(t)\cdot (6\cos(2t)\cdot\sin(t)-3\sin(2t)\cdot\cos(t)) \ dt \\ = & 18\int_0^{\frac{\pi}{2}} \cos(t)\cos(2t)\sin(t)\sin(2t) \ dt + 9\int_0^{\frac{\pi}{2}} \sin^2(2t)\cos^2(2t) \ dt \\ =& 9\int_0^{\frac{\pi}{2}} \sin^2(2t)\cos(2t) \ dt + 9\int_0^{\frac{\pi}{2}} \sin^2(2t)\Big(\dfrac{1+\cos^2(2t)}{2}\Big) \ dt \\ =& \dfrac{9}{2}\Big[\dfrac{sin^3(2t)}{3})\Big]_0^{\frac{\pi}{2}}+ \dfrac{9}{2}\int_0^{\frac{\pi}{2}} \dfrac{1-\cos(4t)}{2} \ dt +\dfrac{9}{2}\int_0^{\frac{\pi}{2}} \sin^2(2t)\cos(2t) \ dt \\ =& \dfrac{9}{8}\cdot\pi \end{array}$$