\begin{aligned} I=& \int_{L+L_{1}}\left(e^{x} \sin y-b(x+y)\right) d x+\left(e^{x} \cos y-a x\right) d y \\ &-\int_{L_{1}}\left(e^{x} \sin y-b(x+y)\right) d x+\left(e^{x} \cos y-a x\right) d y \end{aligned}

$$I_{1}=\iint_{D}\left(\frac{\partial Q}{\partial x}-\frac{\partial P}{\partial y}\right) d x d y=\iint_{D}(b-a) d x d y=\frac{\pi}{2} a^{2}(b-a)$$

$$\left\{\begin{array}{l} x=x \\ y=0 \end{array},(0 \leq x \leq 2 a),\right.$$

\begin{aligned} I &=\int_{L}\left(e^{x} \sin y-b(x+y)\right) d x+\left(e^{x} \cos y-a x\right) d y \\ &=\int_{L} e^{x} \sin y d x+e^{x} \cos y d y-\int_{L} b(x+y) d x+a x d y \end{aligned}

$$\int_{L} e^{x} \sin y d x+e^{x} \cos y d y=\left.e^{x} \sin y\right|_{(2 a, 0)} ^{(0,0)}=0$$

\begin{aligned}
&\left\{\begin{array} { c }
{ x = a + a \operatorname { c o s } t } \\
{ y = a \operatorname { s i n } t }
\end{array} , \text { 则 } \left\{\begin{array}{c}
d x=-a \sin t d t \\
d y=a \cos t d t
\end{array}, t \text { 从 } 0 \text { 到 } \pi,\right.\right. \text { 得 }\\
\int_{L} b(x+y) d x+a x d y \\
&=\int_{0}^{\pi}\left(-a^{2} b \sin t-a^{2} b \sin t \cos t-a^{2} b \sin ^{2} t+a^{3} \cos t+a^{3} \cos ^{2} t\right) d t \\
&=-2 a^{2} b-\frac{1}{2} \pi a^{2} b+\frac{1}{2} \pi a^{3} \\
\text { 从而 } \quad I &=0-\left(-2 a^{2} b-\frac{1}{2} \pi a^{2} b+\frac{1}{2} \pi a^{3}\right)=\left(\frac{\pi}{2}+2\right) a^{2} b-\frac{\pi}{2} a^{3}
\end{aligned}