Line integral of 3 segments, Green not applicable...

Question

Let $\mathcal{C}$ be the 3 segments successively going from $(0,0,0)$ to $(2,4,6)$ to $(3,6,2)$ and to $(0,0,1)$. I need to calculate the work made by the vector field : $$\vec{F}=(6zx^2e^{x^3}+yz)\vec{i}+(xz+y)\vec{j}+(xy+2e^{x^3})\vec{k}$$

to move a particle along $\mathcal{C}$. So, I need to calculate :

$$W=\int_{\mathcal{C}}{\vec{F}\cdot{d\vec{r}}}$$

We know that $\vec{F}$ is conservative $\iff\vec{F}=\nabla f= \begin{pmatrix} \partial f/\partial x\\ \partial f/\partial y\\ \partial f/\partial z\\ \end{pmatrix}$, but that condition isn't met.

We choose to close $\mathcal{C}$ using segment $\mathcal{C_4}$ going from point $(0,0,1)$ to $(0,0,0)$.

Let $\mathcal{C_1}$ be the segment from $(0,0,0)$ to $(2,4,6)$, $\mathcal{C_2}$ the segment from $(2,4,6)$ to $(3,6,2)$ and $\mathcal{C_3}$ the segment from $(3,6,2)$ to $(0,0,1)$, then we have :

$$W=\int_{\mathcal{C_1}\cup\mathcal{C_2}\cup\mathcal{C_3}}{\vec{F}\cdot{d\vec{r}}}+\int_{\mathcal{C_4}}{\vec{F}\cdot{d\vec{r}}}-\int_{\mathcal{C_4}}{\vec{F}\cdot{d\vec{r}}}=\oint_{\mathcal{C_1}\cup\mathcal{C_2}\cup\mathcal{C_3}\cup\mathcal{C_4}}{\vec{F}\cdot{d\vec{r}}}-\int_{\mathcal{C_4}}{\vec{F}\cdot{d\vec{r}}}$$

Parameterizing $\mathcal{C_4}$ : $$\vec{r_4}(t)=(1-t) \begin{pmatrix} 0\\ 0\\ 1\\ \end{pmatrix}+t \begin{pmatrix} 0\\ 0\\ 0\\ \end{pmatrix}= \begin{pmatrix} 0\\ 0\\ 1-t\\ \end{pmatrix}$$ with $t\in[0,1]$.

Using Green theorem, we have :

$$\oint_{\mathcal{C_1}\cup\mathcal{C_2}\cup\mathcal{C_3}\cup\mathcal{C_4}}{\vec{F}\cdot{d\vec{r}}}=!!!$$

Green's theorem doesn't apply in 3D space !

I'm getting kinda lost here. Anyone would like to share a clue?

Thanks !

Answer 1

Use Stokes' Theorem:

We have $$\tag 1 \int_{\vec C} \vec F\cdot d\vec r=\int _{S}(\nabla \times \vec F)\cdot \vec ndS$$ where $\vec C=\vec C_{1}+\vec C_{2}$ is your closed curve, as you have generated it, and $S$ is the surface bounded by $\vec C$.

You are in luck here because $\nabla \times \vec F=\vec 0$, so you can write

$\int_{\vec C} \vec F\cdot d\vec r=0$. Then $$\tag 2 \int _{\vec C_{1}}\vec F\cdot d\vec r+\int _{\vec C_{2}}\vec F\cdot d\vec r=0$$ You want $\int _{\vec C_{1}}\vec F\cdot d\vec r$, which is, using $(2)$, $$-\int _{\vec C_{2}}\vec F\cdot d\vec r$$.

On $\vec C_{2}, x=y=0$ and $0\leq z\leq 1$ so $\vec F\cdot d\vec r=2dz$ so that finally, $$-\int _{\vec C_{2}}\vec F\cdot d\vec r=-2\int_{1}^{0}dz=2$$.

Answer 2

Stokes' Theorem is the generalization of Green's Theorem in the plane. Stokes' Theorem states

$$\oint_C \vec A\cdot \vec dr=\int_S \nabla \times \vec A\cdot \hat n \,dS$$

where $C$ is the closed contour that bounds the surface $S$, holds for sufficiently smooth vector fields $\vec A$ and sufficiently regular surface $S$.

Here, we can show that $\nabla \times \vec F=0$ either by direct evaluation (this is straightforward) or by recalling that for any smooth scalar field $\phi$, we have $\nabla \times \nabla \phi=0$ (i.e, the curl of the gradient is zero).

Here, we find that

$$\vec F=\nabla \phi$$

where $\phi = 2ze^{x^3}+xyz+\frac12 y^2$ (we may add an arbitrary constant to $\phi$). Inasmuch as $\vec F =\nabla \phi$, then $\nabla \times \vec F=\nabla \times \nabla \phi =0$.

Therefore, applying Stokes' Theorem reveals that

$$\oint_C \vec F\cdot \vec dr=\int_S \nabla \times \vec F\cdot \hat n \,dS=0$$

which in turn implies

$$\int_{C_1+C_2+C_3} \vec F\cdot \vec dr=-\int_{C_4} \vec F\cdot \vec dr$$

Can you finish?