Chapter 6: Applications of Double Integration
Section 6.3: Surface Area
Calculate the surface area of the surface defined by the function F=5−3 x2−2 y2 whose domain is the plane region R, the interior of the triangle whose vertices are 0,0,1,0,0,1. See Example 6.2.5.
The surface is Fx,y=5−3 x2−2 y2, so the surface-area element is
dσ=1+Fx2+Fy2dA=1+36 x2+16 y2dA
The surface area is then given by the iterated integral
∫01∫01−x1+36 x2+16 y2ⅆy ⅆx = 1.469876015
The inner integral can be evaluated in closed form, but then the outer integral must be evaluated numerically.
Maple Solution - Interactive
Table 6.3.5(a) contains a solution via a task template that implements the SurfaceArea command from the Student MultivariateCalculus package.
Calculus - Multivariate≻Integration≻Surface Area
5−3 x2−2 y2
Inert integral: dy dx
Table 6.3.5(a) Solution by SurfaceArea command implemented in a task template
The upper limit for the inner integral is obtained by determining the equation of the line representing the hypotenuse of the right triangle whose vertices are given.
The upper limit for the outer integral is expressed in floating-point form. This causes Maple to evaluate the integral numerically. The alternative would be to use a symbolic "1" and find that Maple can evaluate the inner integral exactly, but not evaluate the outer integral. An extra step would then be needed to evaluate the outer integral numerically.
The most direct calculation of the surface area over a triangular domain is implemented with the task template contained in Table 6.3.5(b).
Calculus - Vector≻Integration≻Surface Integration≻Surface Defined over a Triangle
Surface Integral on a Surface Defined over a Triangle
Table 6.3.5(b) Task-template implementation of the SurfaceInt command
A solution from first principles begins with the equation of the hypotenuse of the right triangle defining the region R, found in Table 6.3.5(c).
Tools≻Load Package: Student Precalculus
Write a sequence of two lists, each list representing an endpoint of the hypotenuse.
Context Panel: Student Precalculus≻Lines And Segments≻Line≻Equation
0,1,1,0→equation of liney=1−x
Table 6.3.5(c) Obtaining the equation of the hypotenuse for the triangle defining region R
The calculation of λ=1+Fx2+Fy2 appears in Table 6.3.5(d).
Context Panel: Assign Name
F=5−3 x2−2 y2→assign
Expression palette: Square-root template
Calculus palette: Partial-derivative operator
Context Panel: Evaluate and Display Inline
1+∂∂ x F2+∂∂ y F2 = 36⁢x2+16⁢y2+1
Table 6.3.5(d) Calculation of λ=1+Fx2+Fy2
The iterated integral by means of which the surface area over R can be found is given in Table 6.2.5(e).
Iterate in the order dy dx via the template in the Calculus palette
Calculus palette: Iterated double-integral template
Context Panel: 2-D Math≻Convert To≻Inert Form
Press the Enter key.
Context Panel: Approximate≻10 (digits)
∫01∫01−x1+36 x2+16 y2ⅆy ⅆx
→at 10 digits
Table 6.2.5(e) Iterated double-integral for finding the surface area of F over R
Once again, the inner integral is evaluated in closed form, but the outer integral must be evaluated numerically.
Finally, having obtained λ=1+36 x2+16 y2, the Context Panel provides access to the MultiInt command in the Student MultivariateCalculus package. The requisite iterated integral can be formulated interactively in the ensuing dialog.
Maple Solution - Coded
Install the Student MultivariateCalculus package.
Define the integrand F.
F≔5−3 x2−2 y2:
Obtain the equation of the hypotenuse in region R
Use the Line and GetRepresentation commands from the Student MultivariateCalculus package.
Use the diff command to obtain the requisite partial derivatives.
Top-level, using the Int and evalfcommands
Use the MultiInt command from the Student MultivariateCalculus package
Use the MultiInt command with a pre-defined domain option
MultiIntλ,x,y=Triangle0,0,1.0,0,0,1 = 1.469876015
MultiIntλ,x,y=Region0..1.0,0.. 1−x = 1.469876015
Use the SurfaceInt command in the Student VectorCalculus package
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