Chapter 6: Applications of Double Integration
Section 6.3: Surface Area
Calculate the surface area of the surface defined by the function F=2−x2−4 y2 whose domain is the plane region R, the interior of the ellipse x2+4 y2=1. See Example 6.2.8.
The surface is defined by Fx,y=2−x2−4 y2, so the surface-area element is
dσ=1+Fx2+Fy2dA=12⁢y2+22− x2− 4 y2dA
Iterating in the order dy dx, results in the integral
∫−11∫−1−x2/21−x2/212⁢y2+22− x2− 4 y2 dy dx = 2.155517778
As a function of y, the integrand is sufficiently complicated that Maple cannot find an antiderivative for it. Hence, the iterated integral is evaluated numerically.
Maple Solution - Interactive
Table 6.3.8(a) contains a solution via a task template that implements the SurfaceArea command from the Student MultivariateCalculus package.
Calculus - Multivariate≻Integration≻Surface Area
Inert integral: dy dx
Table 6.3.8(a) Solution by SurfaceArea command implemented in a task template
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 calculation of λ=1+Fx2+Fy2 appears in Table 6.3.8(b).
Context Panel: Assign Name
Expression palette: Square-root template
Calculus palette: Partial-derivative operator
Press the Enter key.
Context Panel: Simplify≻Simplify
Context Panel: Assign to a Name≻f
1+∂∂ x F2+∂∂ y F2
→assign to a name
Table 6.3.8(b) Calculation of λ=1+Fx2+Fy2
An alternative approach is to employ the task template in Table 6.2.8(c). Note the use of the lower limit in floating-point form, and note that the "Floating-Point Approximation" button is pressed to obtain a numeric answer.
Calculus - Vector≻Integration≻Multiple Integration≻2-D≻Over an Ellipse
Integrate fx,y over an Ellipse
Equation of ellipse:
From θ= to θ=
Table 6.2.8(c) Task template for integration over an ellipse
Maple elects to implement the integration in polar coordinates, representing the ellipse as
which is what would be obtained by the following "direct" conversion to polar coordinates.
Expression palette: Evaluation template
Context Panel: Solve≻Obtain Solutions for≻r
x2+4 y2=1x=a|f(x)x=r cosθ,y=r sinθ
→solutions for r
A solution from first principles is given in Table 6.2.8(d).
Solve for y=yx
Control-drag (or type) the equation of the ellipse.
Context Panel: Solve≻Obtain Solutions for≻y
Context Panel: Assign to a Name≻Y
x2+4 y2=1→solutions for y12⁢−x2+1,−12⁢−x2+1→assign to a nameY
Write an appropriate iterated integral and evaluate
Calculus palette: Iterated double-integral template
Context Panel: 2-D Math≻Convert To≻Inert Form
Context Panel: Approximate≻10 (digits)
∫−11.∫Y2Y1f ⅆy ⅆx→at 10 digits2.155517778
Table 6.2.8(d) Solution from first principles
Maple Solution - Coded
Install the Student MultivariateCalculus package.
Define the surface F.
Use the diff command to obtain the requisite partial derivatives.
Apply the simplify command for a more compact expression.
Form the integral via the MultiInt command with a pre-defined domain option
Evaluate numerically with the evalf command.
evalfq = 2.155517778
Use the SurfaceInt command from the Student VectorCalculus package
evalfQ = 2.155517778
The inner integral is sufficiently complicated as to justify an immediate numeric evaluation of the complete iterated integral.
Solution from first principles
Apply the solve command to the equation of the ellipse, obtaining the separate branches y=yx.
Top-level, using the Int and evalf
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