Evaluate the line integral of (e^y + x) dx + (x e^y - 2y) dy around the closed triangular path through the points (0,0), (0,1) and (1,2).

Step-by-step solution

1. 1

   Prefer Green's theorem to direct parametrisation. The path is closed and piecewise smooth, and parametrising three segments separately would mean three messy integrals involving $e^y$. Green's theorem converts the whole thing into one double integral over the enclosed triangle $D$:

   $\oint_L P\,dx+Q\,dy=\iint_D\left(\frac{\partial Q}{\partial x}-\frac{\partial P}{\partial y}\right)dA$

2. 2

   Identify P and Q.

   $P(x,y)=e^{y}+x,\qquad Q(x,y)=xe^{y}-2y$

3. 3

   Compute the two partial derivatives. Differentiating $Q$ with respect to $x$ treats $y$ as constant, so $e^y$ is just a constant multiplier and $-2y$ dies:

   $\frac{\partial Q}{\partial x}=e^{y}$

   Differentiating $P$ with respect to $y$, the $x$ term dies:

   $\frac{\partial P}{\partial y}=e^{y}$

4. 4

   Subtract and integrate.

   $\frac{\partial Q}{\partial x}-\frac{\partial P}{\partial y}=e^{y}-e^{y}=0$

   so

   $\oint_L P\,dx+Q\,dy=\iint_D 0\,dA=0$

   $\boxed{0}$

5. 5

   Explain why the triangle's shape was irrelevant. Because $\partial Q/\partial x=\partial P/\partial y$ everywhere on the plane (a simply connected domain), the field is conservative: it is the gradient of the potential

   $\varphi(x,y)=xe^{y}+\tfrac{x^2}{2}-y^2$

   since $\varphi_x=e^y+x=P$ and $\varphi_y=xe^y-2y=Q$. Any closed loop therefore returns to the same potential value, giving $0$ — the three given vertices could have been replaced by any others.

6. 6

   Check with the potential directly. Going $(0,0)\to(0,1)\to(1,2)\to(0,0)$, the values of $\varphi$ are $0$, $-1$, $e^2+\tfrac12-4$, and back to $0$. Summing the three differences telescopes to $0$, confirming the Green's-theorem result.