4.9 Parametric Surfaces and their Areas
Surface Area
If a smooth parametric surface \(S\) is given by the equation
\[\overline {r}(u,v) = x(u,v)\,\textbf {i} + y(u,v)\,\textbf {j} + z(u,v)\,\textbf {k}\hspace {0.2cm},\hspace {0.5cm} (u,v)\in D\] and \(S\) is covered just once as \((u,v)\) ranges throughout the parameter domain \(D\), then \[A(S) = \iint \limits _D\left |\overline {r}_u \times \overline {r}_v\\\right | dA\hspace {0.2cm}, \hspace {0.5cm} \text {where}\]
\[\overline {r}_u = \frac {\partial x}{\partial u}\,\textbf {i} + \frac {\partial y}{\partial u}\,\textbf {j} + \frac {\partial z}{\partial u}\,\textbf {k}\]
\[\overline {r}_v = \frac {\partial x}{\partial v}\,\textbf {i} + \frac {\partial y}{\partial v}\,\textbf {j} + \frac {\partial z}{\partial v}\,\textbf {k}\]
Solution.
For a sphere, we get
\[ x = a\sin \phi \cos \theta \hspace {0.3cm}, \hspace {1cm} y = a\sin \phi \sin \theta \hspace {0.3cm}, \hspace {1cm} z = a\cos \phi \]
\[D = \big \{\big (\phi , \theta \big ) : 0\leq \phi \leq \pi \hspace {0.4cm} , \hspace {0.4cm} 0 \leq \theta \leq 2\pi \big \}\]
\[\overline {r}\big (\phi , \theta \big ) = a \sin \phi \cos \theta \, \textbf {i} + a \sin \phi \sin \theta \, \textbf {j} + a\cos \phi \,\textbf {k}\]
\[\overline {r}_{\phi } = a \cos \phi \cos \theta \, \textbf {i} + a \cos \phi \sin \theta \, \textbf {j} - a\sin \phi \,\textbf {k}\]
\[\overline {r}_{\theta } \big (\phi ,\theta \big ) = -a\sin \phi \sin \theta \,\textbf {i} + a\sin \phi \cos \theta \,\textbf {j}\]
\begin {align*} \overline {r}_{\phi } \times \overline {r}_{\theta } & = \begin {vmatrix} \textbf {i} & \textbf {j} & \textbf {k}\\ a\cos \phi \cos \theta & a\cos \phi \sin \theta & -asin\phi \\ -a\sin \phi \sin \theta & a\sin \phi \cos \theta & 0\\ \end {vmatrix}\\\\ & = a^2 \sin ^2\phi \cos \theta \textbf {i} + a^2\sin ^2 \phi \sin \theta \textbf {j} + a^2\sin \phi \cos \theta \textbf {k}\\ \end {align*}
\begin {align*} \left |\overline {r}_{\phi } \times \overline {r}_{\theta }\\\right | & = \sqrt {a^4\sin ^4\phi \cos ^2\theta + a^4\sin ^4\phi \sin ^2\theta + a^4\sin ^2\phi \cos ^2\theta }\\ & = \sqrt {a^4\sin ^4\phi ( \cos ^2\theta + \sin ^2 \theta ) + a^4 \sin ^2 \phi \cos ^2\theta }\\ & = a^2 \sin \phi \sqrt {\sin ^2 \phi + \cos ^2\phi } \end {align*}
Thus the surface Area of sphere is \begin {align*} A & = \iint \limits _D \left |\overline {r}_{\phi } \times \overline {r}_{\theta }\\\right |dA\\\\ & = \int ^{2\pi }_0\int ^{\pi }_0 a^2\sin \phi d \phi d \theta \\\\ & = 4\pi a^2\\\\\\ \end {align*}
Surface Area of the Graph of a Function
For the specific case of a sphere \(S\) given by \(Z = f(x,y)\), where \((x,y)\in D\) and \(f\) has continuous partial derivatives, we take \(X\) and \(Y\) as parameters and have \[X=x\hspace {0.3cm}, \hspace {1cm} Y = y\hspace {0.3cm} , \hspace {1cm} Z = f(x,y).\hspace {0.4cm} \text {So}\] \[\overline {r}_x = \textbf {i} + \frac {\partial f}{\partial x}\,\textbf {k}\hspace {0.3cm} ,\hspace {1cm} \overline {r}_y = \textbf {j} + \frac {\partial f}{\partial y}\,\textbf { k}\]
\begin {align*} \overline {r}_x\times \overline {r}_y & = \begin {vmatrix} \textbf {i} & \textbf { j} & \textbf {k}\\\\ 1 & 0 & \dfrac {\partial f}{\partial x}\\\\ 0 & 1 & \dfrac {\partial f}{\partial y}\\ \end {vmatrix}\\\\ & = -\frac {\partial f}{\partial x}\,\textbf {i} - \frac {\partial f}{\partial y}\,\textbf {j} +\textbf {k}\\ \end {align*}
\[\left |\overline {r}_x \times \overline {r}_y\\\right | = \sqrt {\Bigg (\dfrac {\partial f}{\partial x}\Bigg )^2 + \Bigg (\dfrac {\partial f}{\partial y}\Bigg )^2 + 1}\]
So surface area is \[A(S) = \iint \limits _D \sqrt {1 +\Bigg (\dfrac {\partial f}{\partial x}\Bigg )^2 + \Bigg (\dfrac {\partial f}{\partial y}\Bigg )^2 }\,\,dA\]
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