Compute $\int_{C} \vec F\cdot d\vec r$, where
- $\vec F(x,y)=xy \vec i +ye^x \vec j$ and $C$ is a rectangle with vertices $(0,0),(2,0),(2,1),(0,1)$
- $\vec F(x,y)=xy \vec i +(y-x)\vec j$ and $C$ is the line described by $y=2x-4$ from $(1,-2)$ to $(2,0)$
The side $\;(0,0) \to (2,0)\;$ can be parametrized
$$C_1:\;\;r(t)=(t,0)\;,\;\;t\in[0,2]\implies F(r(t))=(0,0)\;,\;\;r'(t)=(1,0)\implies$$
$$\int_{C_1} \vec F\cdot d\vec r=\int_0^2 (0,0)\cdot (1,0)\,dt=\int_0^2 0\,dt=0$$
For the side $\;(2,0)\to(2,1)\;$ we have
$$C_2:\;\;r(t)=(2,t)\;,\;\;t\in[0,1]\implies F(r(t))=(2t,te^2)\;,\;\;r'(t)=(0,1)\implies$$
$$\int_{C_2}\vec F\cdot d\vec r=\int_0^1(2t,te^2)\cdot(0,1)\,dt=e^2\int_0^1t\,dt$$
$$=e^2\frac{t^2}{2}|_0^1=\frac{e^2}{2}$$
For the side $\;(2,1)\to(0,1)\;$ we have
$$C_3:\;\;r(t)=(t,1)\;,\;\;t\in[0,2]\implies F(r(t))=(t,e^t)\;,\;\;r'(t)=(1,0)\implies$$
$$\int_{C_3}\vec F\cdot d\vec r=\int_0^2(t,e^t)\cdot(1,0)\,dt=\int_0^2t\,dt$$ $$=\frac{t^2}{2}|_0^2=2$$
For the side $\;(0,1)\to(0,0)\;$ we have
$$C_4:\;\;r(t)=(0,t)\;,\;\;t\in[0,1]\implies F(r(t))=(0,t)\;,\;\;r'(t)=(0,1)\implies$$
$$\int_{C_4}\vec F\cdot d\vec r=\int_0^1(0,t)\cdot(0,1)\,dt=\int_0^1t\,dt$$ $$=\frac{t^2}{2}|_0^1=\frac{1}{2}$$
So the answer should be :
$$\int_{C} \vec F\cdot d\vec r=\int_{C_1} \vec F\cdot d\vec r+\int_{C_2} \vec F\cdot d\vec r+\int_{C_3} \vec F\cdot d\vec r+\int_{C_4} \vec F\cdot d\vec r=0+\frac{e^{2}}{2}+2+\frac{1}{2}=\frac{e^{2}+5}{2}$$
The line is indeed : $$C: r(t)=(t,2t-4)\;,\;\;t\in[1,2]\implies F(r(t))=(2t^2-4t,t-4)\;,\;\;r'(t)=(1,2)\implies$$
$$\int_{C}\vec F\cdot d\vec r=\int_1^2(2t^2-4t,t-4)\cdot(1,2)\,dt=2\int_1^2 (t^2-t-4)\,dt$$ $$=2(\frac{t^3}{3}-\frac{t^2}{2}-4t)|_1^2=-\frac{19}{3}$$
I want to know how much of my work is correct.
Your answer for the first one has mistakes. The mistakes are in evaluation of line integral over $C3$ and $C4$. For $C1$ and $C2$, you are going counter-clockwise but for $C3$ and $C4$, you are going clockwise. Assuming it is counterclockwise orientation, you need to fix $C3$ and $C4$
$C3$ is straight line from $\;(2,1)\to(0,1)\;$. So if you are parametrizing it as $(t, 1)$, you must go from $t = 2$ to $t = 0$. So the integral should be,
$ \displaystyle \int_{C_3}\vec F\cdot d\vec r=\int_2^0(t,e^t)\cdot(1,0)\,dt = \int_2^0 t\,dt = - \int_0^2 t\,dt$
Similarly for $C4$, it should be -
$ \displaystyle \int_{C_4}\vec F\cdot d\vec r=\int_1^0(0,t)\cdot(0,1)\,dt=\int_1^0 t\,dt = -\int_0^1 t\,dt$