Chapter 6: Applications of Double Integration
Section 6.3: Surface Area
Example 6.3.10
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 rectangle 0≤x≤3/10,0≤y≤1/10. See Example 6.2.9.
Solution
Mathematical Solution
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
dσ=1+Fx2+Fy2dA=17⁢12⁢x2+10⁢y2−73⁢x2+5⁢y2−1dA
Iterating in the order dy dx results in the integral
∫0310∫0110x⁢y⁢17⁢12⁢x2+10⁢y2−73⁢x2+5⁢y2−1ⅆyⅆx ≐ 0.0002353725944
As a function of y, the integrand is sufficiently complicated that, although Maple can find an antiderivative for it, this antiderivative fills several pages. 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
q
3⁢x2+5⁢y2+7⁢z2=1
→solutions for z
17⁢−21⁢x2−35⁢y2+7,−17⁢−21⁢x2−35⁢y2+7
→assign to a name
Z
The simplest approach is to employ the task template in Table 6.3.10(a).
Tools≻Tasks≻Browse:
Calculus - Vector≻Integration≻Surface Integration≻Surface Defined over a Rectangle
Surface Integral on a Surface Defined over a Rectangle
Integrand
fx,y,z=
Surface
Rectangle
zx,y=
x2=
y2=
x1=
y1=
Table 6.3.10(a) Task template for surface integration over a rectangle
A solution from first principles is given in Table 6.3.10(b).
Obtain λ=1+fx2+fy2
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)
∫03/10∫01/10x y λ ⅆy ⅆx
∫0310∫0110x⁢y⁢1+9⁢x2−21⁢x2−35⁢y2+7+25⁢y2−21⁢x2−35⁢y2+7ⅆyⅆx
→at 10 digits
0.0002353725944
Table 6.3.10(b) Solution from first principles
Maple Solution - Coded
Initialize
Install the Student MultivariateCalculus package.
withStudent:-MultivariateCalculus:
Define the bounding surfaces zx,y=Z± by invoking the solve and eval commands.
q≔x−1/62+y−1/52=1/25:Z≔solve3 x2+5 y2+7 z2=1,z:
Use the diff command to obtain the partial derivatives with respect to x and y.
Apply the simplify command.
λ≔simplify1+diffZ1,x2+diffZ1,y2
17⁢7⁢12⁢x2+10⁢y2−73⁢x2+5⁢y2−1
Form the integral via the MultiInt command with a pre-defined domain option
Evaluate the integral with the evalf command
S≔MultiIntx y λ,x,y=Rectangle0..3/10,0..1/10,output=integral
∫0110∫031017⁢x⁢y⁢7⁢12⁢x2+10⁢y2−73⁢x2+5⁢y2−1ⅆxⅆy
evalfS = 0.0002353725944
Use the SurfaceInt command from the Student VectorCalculus package
Q≔Student:-VectorCalculus:-SurfaceIntx y,x,y,z=Surfacex,y,Z1,x,y=Rectangle0..3/10,0..1/10,output=integral
∫0110∫0310x⁢y⁢1+9⁢x2−21⁢x2−35⁢y2+7+25⁢y2−21⁢x2−35⁢y2+7ⅆxⅆy
evalfQ = 0.0002353725944
A solution from first principles that uses the top-level Int command is given below.
Intx y λ,y=0..1/10,x=0..3/10=evalfIntx y λ,y=0..1/10,x=0..3/10
∫0310∫011017⁢x⁢y⁢7⁢12⁢x2+10⁢y2−73⁢x2+5⁢y2−1ⅆyⅆx=0.0002353725944
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