Answer
(a) A scalar function is a function of the form $f(x,y,z)$ , whose range is scalar.
The line integral along $C$ of a scalar function with respect to arc length is defined as: $\int_Cf(x,y,z)ds$.
where, $ds=\sqrt {(dx)^2+(dy)^2+(dz)^2}$
Thus, $\int_Cf(x,y,z)\sqrt {(dx)^2+(dy)^2+(dz)^2}$
$ds$ is the arc length of an infinitesimally small part of the curve $C$.
(b) The line integral along $C$ of a scalar function with respect to arc length is defined as: $\int_Cf(x,y,z)ds$.
where, $ds=\sqrt {(dx)^2+(dy)^2+(dz)^2}$
Thus, $\int_Cf(x,y,z)\sqrt {(\frac{dx}{dt})^2+(\frac{dy}{dt})^2+(\frac{dz}{dt})^2}dt$
A vector function for a space curve $C$ is
$r(t)=x(t)i+y(t)j+z(t)k=\lt x(t),y(t),z(t) \gt$; $a \leq t \leq b$
$ds=\sqrt {(dx)^2+(dy)^2+(dz)^2}$
So, $f(x,y,z)=f(x(t),y(t),z(t))=f(r(t))$
Compute the integral, we get $\int_Cf(x,y,z)ds=\int_a^bf(r(t))|r'(t)|dt$
(c) Since, this is a thin wire, so we only need to think about one dimensional object and integrate the function along the curve $C$.
Thus, $m= \int_C \rho (x,y) ds$
Center of mass can be calculated as:
$\bar {x}=\frac{1}{m}\int_C x \rho (x,y) ds$
and
$\bar {y}=\frac{1}{m}\int_C y \rho (x,y) ds$
(d) Line integral of a scalar function with respect to $x$ is,
$\int_Cf(x,y,z)dx$
Line integral of a scalar function with respect to $y$ is,$\int_Cf(x,y,z)dy$
Line integral of a scalar function with respect to $z$ is, $\int_Cf(x,y,z)dz$
(e) The line integral along $C$ of a scalar function with respect to arc length is defined as: $\int_Cf(x,y,z)ds$.
where, $ds=\sqrt {(dx)^2+(dy)^2+(dz)^2}$
Thus, $\int_Cf(x,y,z)\sqrt {(\frac{dx}{dt})^2+(\frac{dy}{dt})^2+(\frac{dz}{dt})^2}dt$
A vector function for a space curve $C$ is
$r(t)=x(t)i+y(t)j+z(t)k=\lt x(t),y(t),z(t) \gt$; $a \leq t \leq b$
$ds=\sqrt {(dx)^2+(dy)^2+(dz)^2}$
So, $f(x,y,z)=f(x(t),y(t),z(t))=f(r(t))$
Compute the integral, we get
Line integral of a scalar function with respect to $x$ is,
$\int_Cf(x,y,z)dx=\int_a^bf(r(t))|x'(t)|dt$
Line integral of a scalar function with respect to $y$ is,$\int_Cf(x,y,z)dy=\int_a^bf(r(t))|y'(t)|dt$
Line integral of a scalar function with respect to $z$ is, $\int_Cf(x,y,z)dz=\int_a^bf(r(t))|z'(t)|dt$
Work Step by Step
(a) A scalar function is a function of the form $f(x,y,z)$ , whose range is scalar.
The line integral along $C$ of a scalar function with respect to arc length is defined as: $\int_Cf(x,y,z)ds$.
where, $ds=\sqrt {(dx)^2+(dy)^2+(dz)^2}$
Thus, $\int_Cf(x,y,z)\sqrt {(dx)^2+(dy)^2+(dz)^2}$
$ds$ is the arc length of an infinitesimally small part of the curve $C$.
(b) The line integral along $C$ of a scalar function with respect to arc length is defined as: $\int_Cf(x,y,z)ds$.
where, $ds=\sqrt {(dx)^2+(dy)^2+(dz)^2}$
Thus, $\int_Cf(x,y,z)\sqrt {(\frac{dx}{dt})^2+(\frac{dy}{dt})^2+(\frac{dz}{dt})^2}dt$
A vector function for a space curve $C$ is
$r(t)=x(t)i+y(t)j+z(t)k=\lt x(t),y(t),z(t) \gt$; $a \leq t \leq b$
$ds=\sqrt {(dx)^2+(dy)^2+(dz)^2}$
So, $f(x,y,z)=f(x(t),y(t),z(t))=f(r(t))$
Compute the integral, we get $\int_Cf(x,y,z)ds=\int_a^bf(r(t))|r'(t)|dt$
(c) Since, this is a thin wire, so we only need to think about one dimensional object and integrate the function along the curve $C$.
Thus, $m= \int_C \rho (x,y) ds$
Center of mass can be calculated as:
$\bar {x}=\frac{1}{m}\int_C x \rho (x,y) ds$
and
$\bar {y}=\frac{1}{m}\int_C y \rho (x,y) ds$
(d) Line integral of a scalar function with respect to $x$ is,
$\int_Cf(x,y,z)dx$
Line integral of a scalar function with respect to $y$ is,$\int_Cf(x,y,z)dy$
Line integral of a scalar function with respect to $z$ is, $\int_Cf(x,y,z)dz$
(e) The line integral along $C$ of a scalar function with respect to arc length is defined as: $\int_Cf(x,y,z)ds$.
where, $ds=\sqrt {(dx)^2+(dy)^2+(dz)^2}$
Thus, $\int_Cf(x,y,z)\sqrt {(\frac{dx}{dt})^2+(\frac{dy}{dt})^2+(\frac{dz}{dt})^2}dt$
A vector function for a space curve $C$ is
$r(t)=x(t)i+y(t)j+z(t)k=\lt x(t),y(t),z(t) \gt$; $a \leq t \leq b$
$ds=\sqrt {(dx)^2+(dy)^2+(dz)^2}$
So, $f(x,y,z)=f(x(t),y(t),z(t))=f(r(t))$
Compute the integral, we get
Line integral of a scalar function with respect to $x$ is,
$\int_Cf(x,y,z)dx=\int_a^bf(r(t))|x'(t)|dt$
Line integral of a scalar function with respect to $y$ is,$\int_Cf(x,y,z)dy=\int_a^bf(r(t))|y'(t)|dt$
Line integral of a scalar function with respect to $z$ is, $\int_Cf(x,y,z)dz=\int_a^bf(r(t))|z'(t)|dt$
