Line integral along a parametrised path proof

Question

I need some help starting the following proof, where $\vec F=x(\hat \imath+\hat k)+2y\hat \jmath$.

Prove that for all paths $\Gamma$ running from $\hat \imath$ to $\hat \jmath$ and lying in the $xy$-plane $\{z=0\}$, the equality $\int_\Gamma \vec F \cdot d \vec r = \int_{\Gamma_S} \vec F \cdot d \vec r$ where $\Gamma_S$ is the straight segment from $\hat \imath$ to $\hat \jmath$.

Answer 1

If the given integrand is an "exact differential" (That is, there exist a function F(x,y,z) such that the differential is $\frac{\partial F}{\partial x}dx+ \frac{\partial F}{\partial y} dy+ \frac{\partial f}{\partial z}dz$) then the integral along any path from $(x_0, y_0, z_0)$ to $(x_1, y_1, z1)$ is $F(x_1, y_1, z_1)- F(x_0, y_0, z_0)$ and so is independent of the path. To show that a integrand is an "exact differential" remember that, for nice functions, the mixed partial derivatives are equal: $\frac{\partial}{\partial x}\left(\frac{\partial F}{\partial y}\right)$$= \frac{\partial}{\partial y}\left(\frac{\partial F}{\partial x}\right)$ etc. so that, if p(x,y,z)dx+ g(x,y,z)dy+ h(x,y,z)dz is an "exact differential" then $\frac{\partial p}{\partial y}= \frac{\partial q}{\partial x}$, etc.

Answer 2

$$\forall \Gamma, \exists (\gamma_1,\gamma_2) \in C^0 / \gamma_1(0) =1,\gamma_1(1) =0,\gamma_2(0)=0, \gamma_2(1)=1, \\\Gamma = \{ (\gamma_1(t),\gamma_2(t),0), t\in[0,1] \} $$ $$\int_\Gamma \vec{F}.d\vec{r} =\int_\Gamma (x\vec{i}+2y\vec{j}).d\vec{r} \\ =\int_0^1 (\gamma_1(t)\vec{i}+2\gamma_2(t)\vec{j}). (\gamma_1'(t)\vec{i}+\gamma_2'(t)\vec{j})dt \\ =\int_0^1 (\gamma_1'(t)\gamma_1(t)+2\gamma_2'(t)\gamma_2(t))dt \\ = \frac{1}{2}[\gamma_1^2(t)+2\gamma_2^2(t)]_0^1 \\ =\frac{1}{2}$$

Answer 3

Thanks for your help, I think I have proved it for the general case.

Let $\Gamma$ have parametrisation $\vec a(t)=a_1(t)\hat \imath+a_2(t)\hat \jmath + a_3(t)\hat k \:$ (where $\: a_3(t)=0$ as $\Gamma$ lies in the $xy$-plane) with $t\in[t_I,t_F]$. Also, due to the fact that $\Gamma$ runs from $\hat \imath$ to $\hat \jmath$, $a_1(t_I)=1, a_2(t_I)=0, a_1(t_F)=0, a_2(t_F)=1$. Now,

$$\int_\Gamma \vec F \cdot d \vec r =\int_{t_I}^{t_F}\vec F(\vec a(t))\cdot\frac{d\vec a}{dt}(t) \: dt$$

$$=\int_{t_I}^{t_F} \left ( \left( a_1(t)\left(\hat \imath+\hat k\right) \right)+2a_2(t)\hat \jmath \right) \cdot \left( a_1^{'}(t)\hat \imath + a_2^{'}(t) \hat \jmath \right) \: dt$$

$$= \int_{t_I}^{t_F} a_1(t)a_1^{'}(t)+2a_2(t)a_2^{'}(t) \: dt$$

$$= \left[ \frac12 a_1^2(t)+a_2^2(t) \right]_{t_I}^{t_F}$$

$$= \frac12a_1^2(t_F)+a_2^2(t_F)- \left( \frac12 a_1^2(t_I)+a_2^2(t_I) \right)$$

$$=1-\frac12 = \frac12$$