## Calculus 8th Edition

(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, $\int_Cf(x,y,z)dz=\int_a^bf(r(t))|z'(t)|dt$
(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, $\int_Cf(x,y,z)dz=\int_a^bf(r(t))|z'(t)|dt$