Calculations of areas in the plane using Green's theorem

Example

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 $\theta \in [0,{\displaystyle \frac{\pi }{2}}]$.

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

and we have:

$$\begin{array}{rl}\text{Area}=& {\iint }_{D}1dxdy={\int }_{C}Fdr={\int }_0^{\frac{\pi }{2}}F(\alpha (t))\cdot {\alpha }^{\prime }(t)dt\\ =& {\int }_0^{\frac{\pi }{2}}(0,3\sin (2t)\cdot \sin (t))\cdot (6\cos (2t)\cdot \cos (t)-3\sin (2t)\cdot \sin (t),\\ & 6\cos (2t)\cdot \sin (t)-3\sin (2t)\cdot \cos (t))dt\\ =& {\int }_0^{\frac{\pi }{2}}3\sin (2t)\cos (t)\cdot (6\cos (2t)\cdot \sin (t)-3\sin (2t)\cdot \cos (t))dt\\ =& 18{\int }_0^{\frac{\pi }{2}}\cos (t)\cos (2t)\sin (t)\sin (2t)dt+9{\int }_0^{\frac{\pi }{2}}{\sin }^2(2t){\cos }^2(2t)dt\\ =& 9{\int }_0^{\frac{\pi }{2}}{\sin }^2(2t)\cos (2t)dt+9{\int }_0^{\frac{\pi }{2}}{\sin }^2(2t)({\displaystyle \frac{1+{\cos }^2(2t)}{2}})dt\\ =& {\displaystyle \frac{9}{2}}[{\displaystyle \frac{si{n}^3(2t)}{3}}){]}_0^{\frac{\pi }{2}}+{\displaystyle \frac{9}{2}}{\int }_0^{\frac{\pi }{2}}{\displaystyle \frac{1-\cos (4t)}{2}}dt+{\displaystyle \frac{9}{2}}{\int }_0^{\frac{\pi }{2}}{\sin }^2(2t)\cos (2t)dt\\ =& {\displaystyle \frac{9}{8}}\cdot \pi \end{array}$$