Evaluate the line integral of (e^x sin y - xy) dx + (xy + e^x cos y) dy along the right half of the circle x^2 + y^2 = 4, traversed in the direction of increasing y.

Step-by-step solution

1. 1

   Fix the path and its endpoints. The right half of $x^2+y^2=4$ runs between $(0,-2)$ and $(0,2)$ through $(2,0)$. Increasing $y$ means we start at $(0,-2)$ and end at $(0,2)$. Knowing the endpoints matters because part of the integrand will turn out to be path-independent.

2. 2

   Test for exactness before parametrising. With $P=e^x\sin y-xy$ and $Q=xy+e^x\cos y$,

   $P_y=e^x\cos y-x,\qquad Q_x=y+e^x\cos y.$

   These are not equal, so the field is not conservative and the whole integral is not path independent. But the mismatch is only in the $-xy$ and $xy$ terms.

3. 3

   Split off the exact part. Notice that

   $d\!\left(e^x\sin y\right)=e^x\sin y\,dx+e^x\cos y\,dy,$

   so those two terms form a perfect differential and depend only on the endpoints:

   $\int_L e^x\sin y\,dx+e^x\cos y\,dy=\Big[e^x\sin y\Big]_{(0,-2)}^{(0,2)}=\sin 2-\sin(-2)=2\sin 2.$

4. 4

   Parametrise the remaining piece. For $\displaystyle\int_L -xy\,dx+xy\,dy$ use $x=2\cos t$, $y=2\sin t$ with $t$ from $-\tfrac{\pi}{2}$ to $\tfrac{\pi}{2}$ (this keeps $x\ge 0$ and makes $y$ increase, matching the required direction). Then $dx=-2\sin t\,dt$, $dy=2\cos t\,dt$ and $xy=4\sin t\cos t$:

   $\int_{-\pi/2}^{\pi/2}\left(8\sin^2 t\cos t+8\sin t\cos^2 t\right)dt.$

5. 5

   Use symmetry, then substitute. $\sin t\cos^2 t$ is odd on a symmetric interval, so its integral is $0$. For the other term put $u=\sin t$:

   $8\int_{-1}^{1}u^2\,du=8\cdot\frac{2}{3}=\frac{16}{3}.$

6. 6

   Add the two contributions. The line integral is

   $\frac{16}{3}+2\sin 2\approx 7.1519,$

   which matches a direct numerical evaluation of the parametrised integral to five decimal places.