Integrating the cross product of a gradient

Question

Is it possible to compute the integral, $$\displaystyle\int_{\mathbf{r_1}(t)}^{\mathbf{r_1}(t)}\nabla(f)\times \mathrm{d}\mathbf{r}$$ Without resorting to using parametric definitions and then integrating component-wise? Thank you.

Answer 1

$\newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,} \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack} \newcommand{\dd}{\mathrm{d}} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,} \newcommand{\ic}{\mathrm{i}} \newcommand{\mc}[1]{\mathcal{#1}} \newcommand{\mrm}[1]{\mathrm{#1}} \newcommand{\pars}[1]{\left(\,{#1}\,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,} \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}} \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$ \begin{align} \int_{\mathbf{r}_{1}\pars{t}}^{\mathbf{r}_{2}\pars{t}} \nabla\mrm{f}\pars{\mathbf{r}}\times\dd\mathbf{r} & = \int_{\mathbf{r}_{1}\pars{t}}^{\mathbf{r}_{2}\pars{t}} \verts{\begin{array}{ccc} \ds{\hat{x}} & \ds{\hat{y}} & \ds{\hat{z}} \\[1mm] \ds{\mrm{f}_{x}\pars{\mathbf{r}}} & \ds{\mrm{f}_{y}\pars{\mathbf{r}}} & \ds{\mrm{f}_{z}\pars{\mathbf{r}}} \\[1mm] \ds{\dd x} & \ds{\dd y} & \ds{\dd z} \end{array}} \\[5mm] & = \hat{x}\int_{\mathbf{r}_{1}\pars{t}}^{\mathbf{r}_{2}\pars{t}} \bracks{\mrm{f}_{y}\pars{\mathbf{r}}\,\dd z - \mrm{f}_{z}\pars{\mathbf{r}}\,\dd y} + \hat{y}\int_{\mathbf{r}_{1}\pars{t}}^{\mathbf{r}_{2}\pars{t}} \bracks{\mrm{f}_{z}\pars{\mathbf{r}}\,\dd x - \mrm{f}_{x}\pars{\mathbf{r}}\,\dd z} \\[2mm] & + \hat{z}\int_{\mathbf{r}_{1}\pars{t}}^{\mathbf{r}_{2}\pars{t}} \bracks{\mrm{f}_{x}\pars{\mathbf{r}}\,\dd y - \mrm{f}_{y}\pars{\mathbf{r}}\,\dd x} \end{align}