Answer
(a) See the explanation below.
(b) See the explanation below.
(c) See the explanation below.
(d) See the explanation below.
(e) See the explanation below.
Work Step by Step
a)
The line integral along a smooth curve $C$ of a scalar function $f(x,y,z)$with respect to arc length can be shown as: $\int_Cf(x,y,z)ds$.
where, $ds=\sqrt {(dx)^2+(dy)^2+(dz)^2}$
Thus, $\int_Cf(x,y,z)\sqrt {(dx)^2+(dy)^2+(dz)^2}$
Here, $ds$ represents an infinitesimally small part arc length.
b)
A line integral for a space curve $C$ is defined as for an interval $[m,n]$
We have $\int_Cf(x,y,z)ds=\int_m^n\sqrt {(\frac{dx}{dt})^2+(\frac{dy}{dt})^2+(\frac{dz}{dt})^2}dt$
c)
A thin wire can be considered as a one dimensional straight wire whose mass is given as: $m= \int_C \rho (x,y) ds$
Center of mass $(\bar {x},\bar {y})$ is defined as: $\bar {x}=\frac{1}{m}\int_C x \rho (x,y) ds$ ;
$\bar {y}=\frac{1}{m}\int_C y \rho (x,y) ds$
d)
The line integral of a scalar function with respect to $x$ is defined as:
$\int_Cf(x,y,z)dx$
The line integral of a scalar function with respect to $y$ is defined as:$\int_Cf(x,y,z)dy$
The line integral of a scalar function with respect to $z$ is defined as: $\int_Cf(x,y,z)dz$
e)
From part (a), we have that the line integral along a smooth curve $C$ of a scalar function $f(x,y,z)$with respect to arc length can be shown as: $\int_Cf(x,y,z)ds$.
where, $ds=\sqrt {(dx)^2+(dy)^2+(dz)^2}$
Thus, $\int_Cf(x,y,z)\sqrt {(dx)^2+(dy)^2+(dz)^2}$
Here, $ds$ represents an infinitesimally small part arc length.
A line integral for a space curve $C$ is defined as for an interval $[m,n]$
We have $\int_Cf(x,y,z)ds=\int_m^n\sqrt {(\frac{dx}{dt})^2+(\frac{dy}{dt})^2+(\frac{dz}{dt})^2}dt$
$\int_Cf(x,y,z)dx=\int_m^nf(r(t))|x'(t)|dt$;
$\int_Cf(x,y,z)dy=\int_m^nf(r(t))|y'(t)|dt$;
$\int_Cf(x,y,z)dz=\int_m^nf(r(t))|z'(t)|dt$.
Here, $r(t)$ represents $r(t)=x(t)i+y(t)j+z(t)k=\lt x(t),y(t),z(t) \gt$; $m \leq t \leq n$.
