In my physics book, the author writes that line integral of a scalar function $\phi$ over a curve $C$ can be written as the following: $$\int_C\phi d{\textbf r}=\textbf{i}\int_C\phi(x,y,z)dx\,+\textbf{j}\int_C\phi(x,y,z)dy\,+\textbf{k}\int_C\phi(x,y,z)dz\,$$
It seemed to me that the output would be a value in $\mathbb R$ rather than in $\mathbb R^3$. Could you please explain?
He provided an example too, which I will add if you think it would help!
Any help is greatly appreciated :)
Given a domain $\Omega\subset{\mathbb R}^3$, a scalar field $$\phi:\>\Omega\to{\mathbb R},\quad (x,y,z)\mapsto\phi(x,y,z)\ ,$$ or a vector field $${\bf F}:\>\Omega\to{\mathbb R}^3,\quad (x,y,z)\mapsto\bigl(u(x,y,z), v(x,y,z), w(x,y,z)\bigr)\ ,$$ and a curve $$C:\quad t\mapsto{\bf r}(t)=\bigl(x(t),y(t),z(t)\bigr)\in\Omega\qquad(a\leq t\leq b)$$ in $\Omega$ there are various line integrals of these fields along $C$ possible, and the values of the integrals will be real numbers or vectors. The integral in your example is vector-valued, albeit I could not think of a concrete physical example where such an integral occurs.
We can integrate $\phi$ with respect to arc length on $C$: $$\int_C\phi \>ds=\int_a^b \phi\bigl({\bf r}(t)\bigr)\>\bigl|\dot{\bf r}(t)\bigr|\>dt\quad\in{\mathbb R} ;$$ then in your way: $$\int_C\phi\> d{\bf r}=\int_a^b \phi\bigl({\bf r}(t)\bigr)\>\dot{\bf r}(t)\>dt\quad\in{\mathbb R}^3\ .$$ Similarly we can with the vector field ${\bf F}$ form the line integrals $$\int_C{\bf F}\> ds\in{\mathbb R}^3\>,\qquad\int_C{\bf F}\cdot d{\bf r}\in{\mathbb R},\quad \int_C{\bf F}\times d{\bf r}\in{\mathbb R}^3\>,$$ and maybe others, depending on the geometrical or physical purpose.