Compute the area of the set $T\subset R^2$of the points lying within the trace of the closed curve $γ = γ_1 ∪ γ_2 ∪ γ_3$

Question

Hi i found this exercise in the solved exercise session in my university portal and there was something there that i did not understand and i was hoping i could receive some answers here if possible.

So the exercise goes like this :

Compute the area of the set $T ⊂ R^2$ of the points lying within the trace of the closed curve $γ = γ_1 ∪ γ_2 ∪ γ_3$, counterclockwise oriented, where

$γ_1(t) =(sin^3(t), cos^3(t))$ , $t\in [0,\frac \pi2]$

$γ_2$ is the arc of the ellipse $x^ 2 +\frac 1 4 y^2=1 $ contained in the first quadrant and

$γ_3$ is the half-circle, with centre at $(0, \frac 32 $) and radius $\frac 12$ , contained in the first quadrant

What he does is he uses th Green Theorem if $F:R^2\to R^2$ 2 is a vector field of class $C^ 1 , F = (f1, f2)$, such that

$(1.1)$ $\forall(x,y)\in R^2: \frac {\partial f_2(x,y)} {\partial x}-\frac {\partial f_1(x,y)} {\partial y}=1$ then

$Area(T)=\int _T 1dxd = \int _T (\frac {\partial f_2(x,y)} {\partial x}-\frac {\partial f_1(x,y)} {\partial y})dxdy=\int_T F dP=\int_{\gamma_1}F dP+\int_{\gamma_2}F dP+\int_{\gamma_3}F dP$

Then he parametrizes all the $\gamma1,\gamma2,\gamma3$ and then he finds the derivative of each

Up until now i think i understand everything

Here is the part i dont understand

He writes: If we consider the vector field $F(x, y) = (f1(x, y), f2(x, y)) = (0, x) $satisfying $(1.1)$(is referencing $(1.1)$ above), we have

$\int_{\gamma_1}F dP=\int^{\frac {\pi}2}_0F(\gamma_1(t))\gamma_1 '(t)=\int^{\frac {\pi}2}_0 (0,sin^3(t)(\gamma_1'(t)))$

From here on he does the same thing with all the others and from there on is just computation

Why did he decide to consider consider the vector field $F(x, y) = (f1(x, y), f2(x, y)) = (0, x)$?

Answer 1

The short answer is because $2D$ vector field $\vec F = (0, x)$ has the scalar curl of $1$. In other words,

For vector field $ \vec F = (P, Q)$ and a counterclockwise oriented closed curve, Green's Theorem states that,

$\displaystyle \iint_T \left (\cfrac{\partial Q}{\partial x} - \cfrac{\partial P}{\partial y}\right) dA = \int_{C} \vec F(\lambda(t)) \cdot \vec\lambda'(t) \ dt$

So if we want to find area bound by a closed curve, we choose a vector field which has constant scalar curl, for example, $(-y, x)$ or $(0, x)$ or $(-y, 0)$.

In this case, $\vec F = (P, Q) = (0, x)$

So, $\cfrac{\partial Q}{\partial x} - \cfrac{\partial P}{\partial y} = 1$

So LHS is $ \displaystyle \iint_T 1 \ dA \ , \ $ which clearly gives the area.

Here is the diagram of the region.

You can see the region is bound by $\lambda_1, \lambda_2$, and $\lambda_3$.

So it comes down to line integral of the vector field over $\lambda_1$, $\lambda_2$ and $\lambda_3$. Make sure you are choosing the right orientation for line integral over each curve so it gives the area bound by them.