91Ó°ÊÓ

The surface area of the smooth curve is defined by the integral${∫}_{S}1dS$.

(a) If $S$is given by $z=f(x,y)$, where to be a differentiable function defined over a region $R^2$, then

$dS=\sqrt{{\left(\frac{∂z}{∂x}\right)}^2+{\left(\frac{∂z}{∂y}\right)}^2+1}dA$

Example: The portion of the plane $z=6x+2y$ that lies above the rectangle $[0,2]×[2,7]$ in the

$xy$-plane.

Find $\frac{∂z}{∂x}$and $\frac{∂z}{∂y}$.

$\begin{array}{rcl}\frac{∂z}{∂x}& =& 6\\ \frac{∂z}{∂y}& =& 2\end{array}$

Find $dS$.

$\begin{array}{rcl}dS& =& \sqrt{6^2+2^2+1}dA\\ & =& \sqrt{36+4+1}dA\\ & =& \sqrt{41}dA\end{array}$

The surface area is as follows:

$\begin{array}{rcl}{∫}_{S}1dS& =& {∫}_2^7{∫}_0^2\sqrt{41}dA\\ & =& 10\sqrt{41}\end{array}$

If $S$is given by $r(u,v)$defined over a region $D$, then $dS$ will be as follows:

$\begin{array}{rcl}dS& =& \left|\left|r_u×r_v\right|\right|dA\\ & =& \left|\left|r_u×r_v\right|\right|dudv\end{array}$

Example: Find the surface area of the portion of the sphere of radius $4$.

The sphere of radius $4$is parametrized by $r(u,v)=(4\cos u\sin v,4\sin u\sin v,4\cos v)$for $u∈[0,2\mathrm{π}]$and $v∈[0,\mathrm{π}]$.

Thus, $\left|\left|r_u×r_v\right|\right|$.

$\left|\left|r_u×r_v\right|\right|=16\sin  v$

So, the surface area is as follows:

$\begin{array}{rcl}{∫}_{S}1dS& =& {∫}_0^{2\mathrm{π}}{∫}_0^{\mathrm{π}}16\sin vdvdu\\ & =& {∫}_0^{2\mathrm{π}}16{[-\cos v]}_0^{\mathrm{π}}=-16{∫}_0^{2\mathrm{π}}[-1-1]du\\ & =& 32{∫}_0^{2\mathrm{π}}du\end{array}$

Simplify further.

$\begin{array}{rcl}{∫}_{S}1dS& =& 32{\left[u\right]}_0^{2\mathrm{π}}\\ & =& 32[2\mathrm{π}-0]\\ & =& 64\mathrm{π}\end{array}$