Integral of vector field to a curl

Question

Consider the function $$g(x,y,z)= \left(\frac {x^3} {2(z^2 +1)(x^4 +4y^6)^2},\frac {3y^5} {(z^2 +1)(x^4 +4y^6)^2}, \frac z {4(z^2 +1)2(x^4 +4y^6)}\right)$$

defined on the set $U:= \{(x,y,z) \in R^3:(x,y)\not=(0,0)\}$

Calculate $\int_γ g$, where $γ$ is the oriented piece wise $C_1$ curve parametrizing the boundary of the γ square $[−1, 1]\times[−1, 1]\times 0$. The orientation is counter-clockwise, following the right-hand rule.

So what I have done is proven that the function is irrotational or curl-free. But this does not guarantees conservativeness so i am not sure how to proceed; in addition, how can i write down the function of the boundary? I am a little confused about only three points for a square.

Answer 1

Given the vector field $\mathbf{F} : \mathcal{D} \to \mathbb{R}^3$ of law:

$$ \mathbf{F}(x,\,y,\,z) := \left\{ \frac{x^3}{2\left(x^4+4\,y^6\right)^2\left(z^2+1\right)}, \; \frac{3\,y^5}{\left(x^4+4\,y^6\right)^2\left(z^2+1\right)}, \; \frac{z}{4\left(x^4+4\,y^6\right)\left(z^2+1\right)^2} \right\} $$

of natural domain:

$$ \mathcal{D} = \left\{ (x,\,y,\,z) \in \mathbb{R}^3 : x^4 + 4\,y^6 \ne 0 \right\} $$

and the support curve $\gamma := \gamma_1 \cup \gamma_2 \cup \gamma_3 \cup \gamma_4$, naturally parameterizable as follows:

$$ \begin{aligned} & \gamma_1 : (x,\,y,\,z) = \mathbf{r}_1(t) := (t,\,-1,\,0)\,, \; \; \; \text{for} \; t \in [-1,\,1]\,; \\ & \gamma_2 : (x,\,y,\,z) = \mathbf{r}_2(t) := (1,\,t,\,0)\,, \; \; \; \; \; \, \text{for} \; t \in [-1,\,1]\,; \\ & \gamma_3 : (x,\,y,\,z) = \mathbf{r}_3(t) := (t,\,1,\,0)\,, \; \; \; \; \; \, \text{for} \; t \in [-1,\,1]\,; \\ & \gamma_4 : (x,\,y,\,z) = \mathbf{r}_4(t) := (-1,\,t,\,0)\,, \; \; \; \text{for} \; t \in [-1,\,1]\,; \\ \end{aligned} $$

by definition, the work $\mathcal{W}$ of $\mathbf{F}$ along $\gamma$ is equal to:

$$ \small \begin{aligned} \mathcal{W}_{\gamma}(\mathbf{F}) := \int\limits_{\gamma^+} \mathbf{F} \cdot \text{d}\mathbf{r} & = \sum_{i = 1}^4 \int\limits_{\gamma_i^+} \mathbf{F}(\mathbf{r}_i(t)) \cdot \mathbf{\dot{r}}_i(t)\,\text{d}t \\ & = \int_{-1}^1 \frac{t^3}{2\left(t^4+4\right)^2}\,\text{d}t + \int_{-1}^1 \frac{3\,t^5}{\left(4\,t^6+1\right)^2}\,\text{d}t + \int_1^{-1} \frac{t^3}{2\left(t^4+4\right)^2}\,\text{d}t + \int_1^{-1} \frac{3\,t^5}{\left(4\,t^6+1\right)^2}\,\text{d}t \\ & = \int_{-1}^1 \frac{t^3}{2\left(t^4+4\right)^2}\,\text{d}t + \int_{-1}^1 \frac{3\,t^5}{\left(4\,t^6+1\right)^2}\,\text{d}t - \int_{-1}^1 \frac{t^3}{2\left(t^4+4\right)^2}\,\text{d}t - \int_{-1}^1 \frac{3\,t^5}{\left(4\,t^6+1\right)^2}\,\text{d}t \\ & = 0\,. \end{aligned} $$

That's all.

Answer 2

One should recognize that

$g(x,y,z) = \nabla\left( \frac{-1}{8(z^2+1)(x^4+4y^6)}\right)$

That notation $[-1,1]\times[-1,1]\times\{0\}$ is not three points, it describes a square in the $z=0$ plane. Since $z$ does not change $(dz=0)$, let's consider the equivalent $2$D problem:

$$ \int_\gamma \nabla f \cdot dr \hspace{24 pt} f(x,y) = \frac{-1}{8(x^4+4y^6)} $$

One would be tempted to use the fundamental theorem of line integrals, but we do not have that the vector field is necessarily conservative everywhere because the domain of integration is not simply connected (we cannot include the origin). Instead let's use Green's theorem (for non simply connected regions) to encompass the origin by a different curve inside the "unit square"

$$x^4+4y^6 = 1$$

And parametrize it by an inspired substitution

$$x^4 = \sin^2 t \hspace{16 pt} 4y^6 = \cos^2 t$$

$x$ depends on the sign but $y$ will not because it's a cube root, so we'll have two curves

$$\gamma_1(t) = (\sqrt{\sin t}, \sqrt[3]{\cos t/2}) \hspace{12 pt} t\in [0,\pi] \implies \gamma'_1(t) = \left(\frac{\cos t}{2x}, \frac{-\sin t}{6y^2}\right)$$

$$\gamma_2(t) = (-\sqrt{-\sin t}, \sqrt[3]{\cos t/2}) \hspace{12 pt} t\in [-\pi,0] \implies \gamma'_2(t) = \left(\frac{\cos t}{2x}, \frac{-\sin t}{6y^2}\right)$$

Let's compute these two integrals first:

$$\int_{\gamma_1} \nabla f \cdot dr = \int_0^\pi \frac{\frac{1}{2}x^2\cos t - y^3\sin t}{2(1)^2}dt = \frac{1}{4} \int_0^\pi \sin t \cos t - \cos t \sin t dt = 0$$

The $\gamma_2$ integral will give us the same result, so we have that

$$\int_\gamma \nabla f \cdot dr = 0$$

How does this help us with the integral on the unit square? Denoting $D$ as the region in between the two curves, we can use Green's theorem to say

$$\int_{\text{square}} \nabla f \cdot dr - \int_\gamma \nabla f \cdot dr = \iint_{D} Q_x - P_y dA = 0$$

Thus the integral along the square is $0$. This in fact proves that all line integrals that surround the origin evaluate to the same thing. Since these and the other closed loop integrals that don't surround the origin (conservative by simply connectedness) all evaluate to $0$, this proves the vector field was conservative.