Chapter 6: Applications of Double Integration
Section 6.3: Surface Area
Example 6.3.15
Derive the expression for when the surface is given implicitly. See Table 6.3.2.
Solution
In principle, the surface defined implicitly by the equation can be obtained explicitly by solving for , in which case , that is, the equation becomes an identity in and .
Implicit differentiation of this identity leads to
and
Use of , valid for the explicitly given surface , leads to
= = =
<< Previous Example Section 6.3 Next Example >>
© Maplesoft, a division of Waterloo Maple Inc., 2025. All rights reserved. This product is protected by copyright and distributed under licenses restricting its use, copying, distribution, and decompilation.
For more information on Maplesoft products and services, visit www.maplesoft.com
Download Help Document