Chapter 6: Applications of Double Integration
Section 6.3: Surface Area
Obtain the surface integral of gx,y=x y2 over the part of the top half of the ellipsoid 3 x2+5 y2+7 z2=1 that sits above the region bounded by the ellipse 3 x2+5 y2=1/2. See Example 6.2.9.
The surface is defined by Fx,y=7−21 x2−35⁢y2/7 , the top half of the ellipsoid 3 x2+5 y2+7 z2=1, so the surface-area element is
Iterating in the order dy dx results in the integral
∫−1616⁢∫−110⁢10−60 x2110⁢10−60 x2x2⁢y2⁢17⁢12⁢x2+10⁢y2−73⁢x2+5⁢y2−1ⅆyⅆx ≐ 0.0003283936762
where 3 x2+5 y2=1/2, the equation of the ellipse, has been solved for y=yx, and x±=±1/6.
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
Solve 3 x2+5 y2+7 z2=1 for z=zx,y
Context Panel: Assign to a Name≻q
3 x2+5 y2+7 z2=1→assign to a nameq
Type the name q and press the Enter key.
Context Panel: Solve≻Obtain Solutions for≻z
Context Panel: Assign to a Name≻Z
→solutions for z
→assign to a name
The simplest approach is to employ the task template in Table 6.3.11(a), after noting that the plane region R is an ellipse.
Calculus - Vector≻Integration≻Surface Integration≻Surface Defined over an Ellipse
Surface Integral on a Surface Defined over an Ellipse
Equation of Ellipse:
From θ= to θ=
Table 6.3.11(a) Task template for surface integration over an ellipse
A solution from first principles is given in Table 6.3.11(b).
Solve the equation of the ellipse for y=yx
Write the equation of the ellipse.
Context Panel: Solve≻Obtain Solutions for≻y
Context Panel: Assign to a Name≻Y
3 x2+5 y2=1/2→solutions for y110⁢−60⁢x2+10,−110⁢−60⁢x2+10→assign to a nameY
Obtain x=xL and x=xR for the ellipse
Write the equation yT=0 and press the Enter key.
Context Panel: Solve≻Obtain Solutions for≻x
→solutions for x
Calculus palette: Partial-derivative template
Context Panel: Assign Name
λ=1+∂∂ x Z12+∂∂ y Z12→assign
Write an appropriate iterated integral and evaluate numerically
Calculus palette: Iterated double-integral template
Context Panel: 2-D Math≻Convert To≻Inert Form
Context Panel: Approximate≻10 (digits)
∫−1/61/6∫Y2Y1x y2 λ ⅆy ⅆx
→at 10 digits
Table 6.3.11(b) Solution from first principles
Maple Solution - Coded
Install the Student MultivariateCalculus package.
Define the bounding surfaces zx,y=Z± and the bounds of the region R as yx=Y±, and X=±1/6. Note the use of the solve and eval commands.
q≔3 x2+5 y2=1/2:Z≔solve3 x2+5 y2+7 z2=1,z:Y≔solveq,y:X≔solveevalq,y=0, x:
Use the diff command to obtain the partial derivatives with respect to x and y.
Apply the simplify command.
Form the integral via the MultiInt command with a pre-defined domain option
Evaluate the integral with the evalf command
evalfS = 0.0003283936762
Use the SurfaceInt command from the Student VectorCalculus package
evalfQ = 0.0003283936762
Because the region of integration is an ellipse, Maple has automatically switched to polar coordinates. Without the list of coordinates r,θ included in the definition of the circle, Maple would use x for r and y for θ, writing a deceptive iterated integral. The integrand of this surface integral is so complicated that any attempt to find a closed-form for the inner integral has been abandoned in favor of a numeric evaluation of the double integral.
A solution from first principles that uses the top-level Int command is given below.
Intx y2 λ,y=Y2..Y1,x=X2..X1=evalfIntx y2 λ,y=Y2..Y1,x=X2..X1
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