## Calculus: Early Transcendentals 8th Edition

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 $r(t)=x(t)i+y(t)j+z(t)k=\lt x(t),y(t),z(t) \gt$; $m \leq t \leq n$.