Answer
(a) See the explanation below.
(b) See the explanation below.
(c) See the explanation below.
(d) See the explanation below.
Work Step by Step
a) The surface integral of a scalar function $f(x,y,z)$ over the surface $S$ can be defined as: $\iint_Sf(x,y,z)dS$
b) A parametric surface, let us say $S$, is known to be a surface in $R^3$ with two parameters $(m,n)$ represented as:
$r(m,n)=p(m,n)i+q(m,n)j+r(m,n)k$
Here, $p,q,r$ are scalar functions.
Area of a surface parametric surface $S$ can be defined as: $\iint_Sf(x,y,z)dS=\iint_Df(r(m,n))|r_m \times n_v|$ dA
where $(m,n) \in D$
c) Area of the surface $S$ of equation $z=g(x,y)$ can be calculated as:
$\iint_Sf(x,y,z)dS=\iint_Sf(x,y,g(x,y))\sqrt {1+(g_x)^2+(g_y)^2}dA=\iint_Sf(x,y,g(x,y))\sqrt {1+(\dfrac{\partial z}{\partial x})^2+(\dfrac{\partial z}{\partial y})^2}dA$
d) Mass of the thin sheet has the shape of a surface $S$ can be expressed as:
$m=\iint_S \rho(x,y,z) dS$
Center of mass of the sheet can be defined as:$(\bar{x},\bar{y},\bar{z})$
Here,
$\bar{x}=\frac{1}{m}\iint_Sx \rho(x,y,z) dS$;
$\bar{y}=\frac{1}{m}\iint_Sy \rho(x,y,z) dS$;
$\bar{z}=\frac{1}{m}\iint_Sz \rho(x,y,z) dS$
