Line integral

Let $P(x,y) = y^2$ and $Q(x,y) = 5xy$ . By Green’s theorem, we have $\displaystyle \oint_C (P(x,y)dx + Q(x,y)dy) = \iint_A \left(\frac{\partial Q(x,y)}{\partial x} - \frac{\partial P(x,y)}{\partial y}\right)$ , where $C$ is the path and $A$ is the area enclosed by the path. So, we have $\frac{\partial Q(x,y)}{\partial x} = 5y$ , $\frac{\partial P(x,y)}{\partial y} =2y$ and $\frac{\partial Q(x,y)}{\partial x} - \frac{\partial P(x,y)}{\partial y} = 5y - 2y = 3y$ . Since the path is enclosed by two circles, we work in polar coordinates, where $x = r\cos \theta$ , $y = r\sin \theta$ with the limits $1 \leq r \leq 2$ and $0 \leq \theta \leq \pi$ . Substituting this in to the double integral, we have

$\begin{aligned} \displaystyle \oint_C (y^2\ dx + 5xy\ dy) &= \int_0^{\pi}\int_1^2 (3r\sin\theta)r\ dr\ d\theta \\ &= \int_0^{\pi}\sin\theta\ d\theta \int_1^2 3r^2\ dr \\ &= \left[-\cos\theta\right]_0^{\pi} \cdot \left[r^3\right]_1^2 \\ &= 2 \cdot 7 \\ &= 14 \\ \end{aligned}$

Here's for peeps like me who don't know nothing about Green's theorem .

Break down the integral into four parts, starting from (2,0), going in the anti-clockwise direction.

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$I_1:$

Parametrize as $x=2\cos\theta$ and $y=2\sin\theta$ .

We have $\mathrm dx=-2\sin\theta\ \mathrm d\theta$ and $\mathrm dy=2\cos\theta\ \mathrm d\theta$ .

$I_1=\oint_C(y^2\ \mathrm dx+5xy\ \mathrm dy)$

$\quad\,=\int_0^\pi[(2\sin\theta)^2(-2\sin\theta\ \mathrm d\theta)+5(2\cos\theta)(2\sin\theta)(2\cos\theta\ \mathrm d\theta)]$

$\quad\,=\int_0^\pi(-8\sin^3\theta+40\cos^2\theta\sin\theta)\ \mathrm d\theta$

$\quad\,=\int_0^\pi(40\sin\theta-48\sin^3\theta)\ \mathrm d\theta$

$\quad\,=16$

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$I_2:$

Since $y$ is always $0$ , so is the integral.

$I_2=0$

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$I_3:$

Parametrize similarly as $I_1$ to obtain:

$I_3=\int_\pi^0[\sin^2\theta(-\sin\theta\ \mathrm d\theta)+5(\cos\theta)(\sin\theta)(\cos\theta\ \mathrm d\theta)]$

Compare with $I_1$ :

$I_3=-\frac18I_1=-2$

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$I_4:$

Since $y$ is always $0$ , so is the integral.

$I_4=0$

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Therefore $\oint_C(y^2\ \mathrm dx+5xy\ \mathrm dy)=16+0-2+0=14$ .

By Green's theorem we get, $\displaystyle\oint_{C}(y^2dx+5xydy)=\int\int_{A}3ydA={3}*(\text{Area of figure enclosed})*{\bar{y}}$ where $\bar{y}$ is the center of mass of the enclosed curve ${A}$ .Given that center of mass of a semicircular disc is $\frac{4R}{3\pi}$ we can evaluate the center of mass of given closed curve by the principle of superposition. It turns out that $\bar{y}=\frac{28}{9\pi}$ . Plugging in this value and the area of figure(which can be easily found out by subtracting the area of smaller semicircle from the bigger semicircle) we get the desired result.