Problem with integral calculation I am supposed to calculate the following integral:
$\int_{M}^{}x\left ( y-1 \right )dA$, where: $M=\left \{ \left ( x,y \right )\in\mathbb{R}^{2}:x^{2}+y^{2}\leq 1\wedge y\leq x+1\wedge y\geq 0 \right \}$
I know how to calculate it, when I have two inequations. I guess it would be by using Polar coordinates and Jabocian, but I do not know, how to start here. 
Can anyone please help me?
 A: I presume you have sketched the circle and two lines so you know the region $M$ looks like

You can straightforwardly integrate over this region as
$$ \int_{-1}^0\int_0^{x+1} \dots \,\mathrm{d}y \,\mathrm{d}x + \int_{0}^1\int_0^{\sqrt{1-x^2}} \dots \,\mathrm{d}y \,\mathrm{d}x  \text{,}  $$
$$  \int_0^1 \int_{y-1}^{\sqrt{1-y^2}} \dots \mathrm{d}x \,\mathrm{d}y  \text{,}  $$
$$  \int_{-1}^1 \int_{0}^{\sqrt{1-x^2}} \dots \,\mathrm{d}x \,\mathrm{d}y - \int_{0}^{1} \int_{\sqrt{1-y^2}}^{y-1} \,\mathrm{d}y  \text{,}  $$
or
$$  \int_{-1}^1 \int_0^{x+1} \dots \,\mathrm{d}y \,\mathrm{d}x - \int_{0}^1 \int_{\sqrt{1-x^2}}^{x+1} \dots \,\mathrm{d}y \,\mathrm{d}x  \text{.}  $$
And there may be other easy ways to break this up that I'm not seeing immediately.
A: I would apply Stokes theorem, and integrate along the boundary. That is, for $\alpha = \tfrac{1}{2} x^2 (y - 1) dy$ we have $d \alpha = x (y - 1) dx \wedge dy$, so by Stokes we find $\int_M x(y - 1) dx \wedge dy = \int_{\partial M} \alpha$. Now $\partial M$ consists of three parts (mind the direction of the curves):
(1) part of $y = 0$, over which $\alpha$ will integrate to zero,
(2) the arc $\gamma_1(\theta) = (\cos \theta, \sin \theta)$ with $\theta \in [0, \pi/2]$,
(3) the line $\gamma_2(t) = (-t, 1 - t)$ with $t \in [0, 1]$.
We compute
$$\int_{\gamma_1} \alpha = \int_0^{\pi/2} \frac{1}{2} \cos^3 \theta (\sin \theta - 1) d \theta = - 5 / 24$$
and 
$$\int_{\gamma_2} \alpha = \int_0^1 \frac{1}{2} t^2 (-t) (- dt) = \int_0^1 \frac{1}{2} t^3 dt = 1/8$$
Therefore, $\int_M x (y - 1) dx \wedge dy = \int_{\partial M} \alpha = 0 - 5/24 + 1/8 = - 1 / 12$.
