I have to prove that the area of $$D=\{(x,y)\in \mathbb{R²}|x\leq y\leq 3x, 2-y \leq x \leq 4-y \}$$ is $\frac{3}{2}$.
The picture is easy, We have 4 lines with intersection points: $(1,3),(2,2),(1,1),(\frac{1}{2},\frac{3}{2})$. With the Gauss-Green identity we know that $$A(D)=\frac{1}{2} \int_{\partial D} xdy - ydx $$ Now, whit the parametrization
- $(4-t,t)$ with $t \in [2,3]$
- $(t,t)$ with $t \in [1,2]$
- $(2-t,t)$ with $t \in [1,\frac{3}{2}]$
- $(4-t,t)$ with $t \in [\frac{3}{2},3]$
We have $$A(D)=-\frac{1}{2}\Big[\int_2 ^ 3 (-t,4-t)(-1,1)dt+ \int_1^ 2 (-t,t)(1,1)dt+ \int_1^{\frac{3}{2}}(-t,2-t)(1,-1)dt + \int_{\frac{3}{2}}^ 3 (-t,4-t)(1,-1)dt\Big]=2$$ Where am I doing wrong?
Your first two parameterizations of line segments on the boundary are correct and oriented in the positive direction. Your third parameterization is "backwards": it is moving in the negative direction around the boundary. Your final parameterization is incorrect: it does the first one but with the wrong domain.