Chapter 6: Applications of Double Integration
Section 6.3: Surface Area
Obtain the surface integral of gx,y=x y over the part of the top half of the ellipsoid 3 x2+5 y2+7 z2=1 that sits above the disk x−1/62+y−1/52≤1/52. 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
∫−1301130∫15−130⁢11−900 x2+300⁢x15+130⁢11−900 x2+300⁢xx⁢y⁢17⁢12⁢x2+10⁢y2−73⁢x2+5⁢y2−1ⅆyⅆx ≐ 0.00589739046
where x−1/62+y−1/52=1/52, the equation of the circle, has been solved for y=yx, and xL=−1/30, xR=11/30.
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 6.3.9(a), after noting that the plane region R is the circle whose center is 1/6,1/5 and whose radius is 1/5.
Calculus - Vector≻Integration≻Surface Integration≻Surface Defined over a Disk
Surface Integral on a Surface Defined over a Disk
From θ= to θ=
Table 6.3.9(a) Task template for surface integration over a disk
A solution from first principles is given in Table 6.3.9(b).
Solve the equation of the circle for y=yx
Write the equation of the circle.
Context Panel: Solve≻Obtain Solutions for≻y
Context Panel: Assign to a Name≻Y
x−1/62+y−1/52=1/52→solutions for y15+130⁢−900⁢x2+300⁢x+11,15−130⁢−900⁢x2+300⁢x+11→assign to a nameY
Obtain x=xL and x=xR for the circle
Write the equation yT=1/5 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
Press the Enter key
Context Panel: Approximate≻10 (digits)
∫−1/3011/30∫Y2Y1x y λ ⅆy ⅆx
→at 10 digits
Table 6.3.9(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±1/5. Note the use of the solve and eval commands.
q≔x−1/62+y−1/52=1/25:Z≔solve3 x2+5 y2+7 z2=1,z:Y≔solveq,y:X≔solveevalq,y=1/5,x:
Use the diff command to obtain the partial derivatives with respect to x and y.
Apply the simplify command.
Use the MultiInt command with a pre-defined domain option
S≔MultiIntx y λ,x,y=Circle1/6,1/5,1/5,r,θ,output=integral
evalfS = 0.005897390462
Use the SurfaceInt command from the Student VectorCalculus package
evalfQ = 0.005897390462
Because the region of integration is a disk, 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 y λ,y=Y2..Y1,x=X2..X1=evalfIntx y λ,y=Y2..Y1,x=X2..X1
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