Prove that the work
$$\mathcal{W}=- \oint \limits_{\gamma} \frac{(x, y, z)}{\left ( x^2+y^2+z^2 \right )^{3/2}} \cdot \, {\rm d}(x, y, z)$$
produced along a $\mathcal{C}^1$ oriented curve $\gamma$ of $\mathbb{R}^3 \setminus \{(0, 0, 0) \}$ depends only on the distances of starting and ending point of $\gamma$ about the origin.
Solution
This question also comes from a mutlivariable examination and for me it is a bit tricky as to what the examiner asks. Basically the exercise wants us to prove that integrand function is actually a conservative field.
Based on the above description let us consider the function
$$f:\mathbb{R}^3 \setminus \big\{\overline{0}\big\}\longrightarrow\mathbb{R}\,; \quad f(x,y,z)=-\dfrac{1}{\sqrt{x^2+y^2+z^2}}$$
which is continuously differentiable in $\mathbb{R}^3 \setminus \{0\}$ and it holds that:
$$({\rm{grad}}\, f)(x,y,z)= \frac{(x,y,z)}{( x^2+y^2+z^2)^{3/2}}\,,\quad (x,y,z)\in\mathbb{R}^3\setminus\big\{\overline{0}\big\}\,.$$
Hence , the vector field
$$ :\mathbb{R}^3\setminus\big\{\overline{0}\big\}\longrightarrow\mathbb{R}\,; \quad \overline{F}(x,y,z)= \frac{(x,y,z)}{( x^2+y^2+z^2)^{3/2}}\,,\quad (x,y,z)\in\mathbb{R}^3\setminus\big\{\overline{0}\big\}$$
is a conservative one. This , in return, means that $\mathcal{W}$ is actually independent of the road we choose , meaning that it only depends on distances of starting and ending point of $\gamma$ about the origin.
The exercise was migrated from mathimatikoi.org .
$$\mathcal{W}=- \oint \limits_{\gamma} \frac{(x, y, z)}{\left ( x^2+y^2+z^2 \right )^{3/2}} \cdot \, {\rm d}(x, y, z)$$
produced along a $\mathcal{C}^1$ oriented curve $\gamma$ of $\mathbb{R}^3 \setminus \{(0, 0, 0) \}$ depends only on the distances of starting and ending point of $\gamma$ about the origin.
Solution
This question also comes from a mutlivariable examination and for me it is a bit tricky as to what the examiner asks. Basically the exercise wants us to prove that integrand function is actually a conservative field.
Based on the above description let us consider the function
$$f:\mathbb{R}^3 \setminus \big\{\overline{0}\big\}\longrightarrow\mathbb{R}\,; \quad f(x,y,z)=-\dfrac{1}{\sqrt{x^2+y^2+z^2}}$$
which is continuously differentiable in $\mathbb{R}^3 \setminus \{0\}$ and it holds that:
$$({\rm{grad}}\, f)(x,y,z)= \frac{(x,y,z)}{( x^2+y^2+z^2)^{3/2}}\,,\quad (x,y,z)\in\mathbb{R}^3\setminus\big\{\overline{0}\big\}\,.$$
Hence , the vector field
$$ :\mathbb{R}^3\setminus\big\{\overline{0}\big\}\longrightarrow\mathbb{R}\,; \quad \overline{F}(x,y,z)= \frac{(x,y,z)}{( x^2+y^2+z^2)^{3/2}}\,,\quad (x,y,z)\in\mathbb{R}^3\setminus\big\{\overline{0}\big\}$$
is a conservative one. This , in return, means that $\mathcal{W}$ is actually independent of the road we choose , meaning that it only depends on distances of starting and ending point of $\gamma$ about the origin.
The exercise was migrated from mathimatikoi.org .
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