Evaluating area of $\frac{3}{4}$ of a disc 
Evaluate $$\int_{\gamma_1\cup \gamma_2}xdx+x^2ydy$$
Where
$\gamma_1(t)=(2\cos t,2\sin t),t\in [-\frac{\pi}{2},\pi]$
$\gamma_2(t)=(\cos t,\sin t),t\in [-\frac{\pi}{2},\pi]$
Using Green's theorem

Visualising the curve it is a there quarters disc
So the curves will be as followed
$\gamma_a(t)=(0,1+t),t\in[0,1]$
$\gamma_1(t)=(2\cos t,2\sin t),t\in [-\frac{\pi}{2},\pi]$
$\gamma_{b}(t)=(2-t,0),t\in[0,1]$
$\gamma_2(t)=(\cos t,\sin t),t\in [-\frac{\pi}{2},\pi]$
Using Green's theorem
In the case of $\gamma_a(t)$ and $\gamma_b(t)$ both vanishes as $Q_y=2xy$ and once $x=0$ and once $y=0$ so the integrand vanishes
$$I_1=\int_{\gamma_1}xdx+x^2ydy=\int\int_{D} 2xy dx dy=\int_{0}^2\int_{-\frac{\pi}{2}}^{\pi}(2\rho \cos \theta *\rho \sin \theta *\rho )d \theta d\rho=\\=\int_{0}^2\int_{-\frac{\pi}{2}}^{\pi}(2\rho^3 \cos \theta\sin \theta)d \theta d\rho=\int_{0}^2\int_{-\frac{\pi}{2}}^{\pi}(\rho^3 2\cos \theta\sin \theta)d \theta d\rho=\\=\int_{0}^2\int_{-\frac{\pi}{2}}^{\pi}(\rho^3 \sin 2\theta)d \theta d\rho=\int_{0}^2\rho^3*(-\frac{\cos2\theta}{2}|_{-\frac{\pi}{2}}^{\pi})d\rho=-\int_{0}^2\rho^3 d\rho=-4$$
The same happen to
$$I_2=-\int_{\gamma_2}xdx+x^2ydy=-\int_{0}^1\int_{-\frac{\pi}{2}}^{\pi}(2\rho^3 \cos \theta\sin \theta)d \theta d\rho=\int_{0}^1\rho^3 d\rho=1$$
So overall we have $-4+1=-3$ which is obviously wrong, what have I missed?
Which is obviously wrong, what have I missed?
 A: Hint: Consider follwoing paths to close the area with positive directions 
$\gamma_a(t)=(0,-1-t),t\in[0,1]$,
$\gamma_1(t)=(2\cos t,2\sin t),t\in [-\dfrac{\pi}{2},\pi]$,
$\gamma_{b}(t)=(-2+t,0),t\in[0,1]$,
$\gamma_2(t)=(\cos t,\sin t),t\in [-\dfrac{\pi}{2},\pi]$.
Now use direct way with path $\gamma_a+\gamma_1+\gamma_b-\gamma_2$. With Green's theorem
$$\int_1^2\int_{-\frac{\pi}{2}}^{\pi}2 r^3\ \sin t\cos t\ dt\ dr=\color{blue}{-\dfrac{15}{4}}$$

Edit: Details in direct way are
$$\displaystyle\int_{\gamma_a}=\int_0^10d0+0dt=0$$
$$\displaystyle\int_{\gamma_1}=\int_{-\frac{\pi}{2}}^{\pi}-4\sin t\cos t+16\sin t\cos^3 t\ dt=-2$$
$$\displaystyle\int_{\gamma_b}=\int_0^1(-2+t)\ dt=-\dfrac32$$
$$\displaystyle\int_{\gamma_2}=\int_{-\frac{\pi}{2}}^{\pi}-\sin t\cos t+\sin t\cos^3 t\ dt=\dfrac14$$
these conclude $0+(-2)+(-\dfrac32)-(\dfrac14)=\color{blue}{-\dfrac{15}{4}}$
A: Let define


*

*$\gamma_3(t)=(-2+2t,0),t\in[0,1]$

*$\gamma_4(t)=(0,-2t),t\in[0,1]$

*$\gamma_5(t)=(-1+t,0),t\in[0,1]$

*$\gamma_6(t)=(0,-t),t\in[0,1]$


then we have
$$\int_{\gamma_1\cup \gamma_2}xdx+x^2ydy=\int_{\gamma_1\cup \gamma_3\cup \gamma_4}xdx+x^2ydy+\int_{\gamma_2\cup \gamma_5\cup \gamma_6}xdx+x^2ydy-\int_{\gamma_3}xdx+x^2ydy-\int_{\gamma_4}xdx+x^2ydy-\int_{\gamma_5}xdx+x^2ydy-\int_{\gamma_6}xdx+x^2ydy$$
and by Green theorem
$$\int_{\gamma_1\cup \gamma_3\cup \gamma_4}xdx+x^2ydy+\int_{\gamma_2\cup \gamma_5\cup \gamma_6}xdx+x^2ydy=\iint_{D_1} 2xy\,dx\,dy+\iint_{D_2}2xy\,dx\,dy$$
and by polar coordinates
$$\iint_{D_1}2xy\,dx\,dy=\int_{-\pi/2}^\pi d\theta \int_0^2 2\rho^3\cos \theta \sin \theta\, d\rho=-\frac14$$
$$\iint_{D_2}2xy\,dx\,dy=\int_{-\pi/2}^\pi d\theta \int_0^1 2\rho^3\cos \theta \sin \theta\, d\rho=-4$$
and
$$\int_{\gamma_3}xdx+x^2ydy=\int_{0}^{1} 2(-2+2t)dt=\left[-4t+2t^2\right]_{0}^{1}=-2$$
$$\int_{\gamma_4}xdx+x^2ydy=0$$
$$\int_{\gamma_5}xdx+x^2ydy=\int_{0}^{1} (-1+t)dt=\left[-t+\frac12 t^2\right]_{0}^{1}=-\frac12$$
$$\int_{\gamma_6}xdx+x^2ydy=0$$
therefore
$$\int_{\gamma_1\cup \gamma_2}xdx+x^2ydy=-\frac14-4+2+\frac12=\frac{-1-16+8+2}{4}=-\frac74$$
which is consistent with direct evaluation for


*

*integral 1 $\int_{\gamma_1}xdx+x^2ydy=-2$

*integral 2 $\int_{\gamma_1}xdx+x^2ydy=\frac14$

A: You are wrong in the limits of the outer integral. It should be from 1 to 2 therefore:$$\int_{1}^{2}\int_{-\frac{\pi}{2}}^{\pi}r^3\sin2\theta d\theta dr=\int_{1}^{2}-r^3dr=-\dfrac{15}{4}$$
