4.9 Parametric Surfaces and their Areas
Surface Area
Recall from the section on multiple integration the area of a parametric surface.
Definition 4.9.1. If a smooth parametric surface \(S\) is given by \[\mathbf r(u,v) = x(u,v)\,\mathbf i + y(u,v)\,\mathbf j + z(u,v)\,\mathbf k , \qquad (u,v)\in D ,\] and \(S\) is covered just once as \((u,v)\) ranges throughout the parameter domain \(D\), then \[A(S) = \iint \limits _D|\mathbf r_u \times \mathbf r_v|\,dA ,\] where \(\mathbf r_u = \frac {\partial x}{\partial u}\,\mathbf i + \frac {\partial y}{\partial u}\,\mathbf j + \frac {\partial z}{\partial u}\,\mathbf k\) and \(\mathbf r_v = \frac {\partial x}{\partial v}\,\mathbf i + \frac {\partial y}{\partial v}\,\mathbf j + \frac {\partial z}{\partial v}\,\mathbf k\).
For the sphere of radius \(a\), parametrised by spherical coordinates, \[\mathbf r(\phi ,\theta ) = \langle a\sin \phi \cos \theta ,\ a\sin \phi \sin \theta ,\ a\cos \phi \rangle ,\qquad (\phi ,\theta )\in D = [0,\pi ]\times [0,2\pi ] ,\] the calculation of Example 3.14.2 gives \[\mathbf r_\phi \times \mathbf r_\theta = a^2\sin ^2\phi \cos \theta \,\mathbf i + a^2\sin ^2\phi \sin \theta \,\mathbf j + a^2\sin \phi \cos \phi \,\mathbf k ,\] with \(|\mathbf r_\phi \times \mathbf r_\theta | = a^2\sin \phi \), so \(A = \int _0^{2\pi }\int _0^\pi a^2\sin \phi \,d\phi \,d\theta = 4\pi a^2\). The vector \(\mathbf r_\phi \times \mathbf r_\theta \) is \(a\sin \phi \) times the position vector, so it points outward; this is used below.
Surface Area of the Graph of a Function
For a surface \(S\) given by \(z = f(x,y)\), \((x,y)\in D\), with \(f\) having continuous partial derivatives, take \(x\) and \(y\) as parameters. Then \(\mathbf r_x = \mathbf i + f_x\,\mathbf k\), \(\mathbf r_y = \mathbf j + f_y\,\mathbf k\), \[\mathbf r_x\times \mathbf r_y = -\frac {\partial f}{\partial x}\,\mathbf i - \frac {\partial f}{\partial y}\,\mathbf j +\mathbf k ,\qquad |\mathbf r_x \times \mathbf r_y| = \sqrt {1 + \left (\frac {\partial f}{\partial x}\right )^2 + \left (\frac {\partial f}{\partial y}\right )^2} ,\] and the surface area is \[A(S) = \iint \limits _D \sqrt {1 +\left (\frac {\partial f}{\partial x}\right )^2 + \left (\frac {\partial f}{\partial y}\right )^2 }\,dA .\]
Example 4.9.2. Find the area of the part of the hyperbolic paraboloid \(z = y^2 - x^2\) that lies between the cylinders \(x^2 + y^2 = 1\) and \(x^2 + y^2 = 4\).
Solution. Here \(f_x^2 + f_y^2 = 4x^2 + 4y^2 = 4r^2\), so \[A = \int _0^{2\pi }\int _1^2\sqrt {1 + 4r^2}\,r\,dr\,d\theta = 2\pi \cdot \frac {1}{12}\left [(1 + 4r^2)^{3/2}\right ]_1^2 = \frac {\pi }{6}\left (17\sqrt {17} - 5\sqrt 5\right )\approx 30.85 .\]
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