Let \(\gamma\) be the closed path which goes counterclockwise around the circle \(C(0,2) = \{(x,y) \in \mathbb{R}^2 : x^2+y^2 = 2^2\}\) from the point \((2,0)\) to the point \((0,2)\), then goes to the origin along the y-axis, and then back to \((2,0)\) along the x-axis.
Compute \(\int_\gamma 2x\,dy + y\,dx\) in one of two ways:
Directly, using the definition of a line integral; or
By using Green’s theorem or the general Stokes’ theorem.
<1>3. By Green’s theorem, \[ \int_\gamma y\,dx+2x\,dy =\iint_D 1\,dA. \]
<1>4. Therefore \[ \boxed{\int_\gamma y\,dx+2x\,dy=\pi}. \]
Solution
1. The path \(\gamma\) is the positively oriented boundary of the quarter disk\[ D=\{(x,y):x\ge0,\ y\ge0,\ x^2+y^2\le4\}. \]
The circular arc runs counterclockwise from \((2,0)\) to \((0,2)\), then the path follows the positive \(y\)-axis down to the origin and the positive \(x\)-axis back to \((2,0)\). Thus the enclosed region lies to the left of the traversal.
2. Write the integrand as\[ P\,dx+Q\,dy \quad\text{with}\quad P(x,y)=y,\qquad Q(x,y)=2x. \]Then\[ \frac{\partial Q}{\partial x}-\frac{\partial P}{\partial y}=2-1=1. \]
The region \(D\) is one quarter of a disk of radius \(2\), so \[ \operatorname{area}(D)=\frac14\pi(2)^2=\pi. \] Combining this with 3 gives the result.