Chapter 6: Applications of Double Integration
Section 6.3: Surface Area
Example 6.3.7
Calculate the surface area of the surface defined by the function F=2−y−1/22 whose domain is the plane region R, the region bounded by the graphs of 1, cosx, and y=x on 0≤x≤1. See Example 6.2.7.
Solution
Mathematical Solution
The surface is defined by Fx,y=2−y−1/22, so the surface-area element is
dσ=1+Fx2+Fy2dA=1+−2⁢y+12dA
The simplest iteration of the double integral that gives the surface area takes the integrand as λ=1+−2⁢y+12 and uses the order dx dy, with Q≐0.74:
∫Q1∫arccosyyλ dx dy = 0.1300942772
If the order of integration is taken as dy dx, then the iterated integral would be the more tedious
∫0Q∫cosx1λ dy dx+∫Q1∫x1λ dy dx = 0.1300942772
Because the integration limit Q is a floating-point number, Maple immediately evaluates the integrals numerically.
Maple Solution - Interactive
Initialize
Context Panel: Assign Name
F=2−y−1/22→assign
XL=arccosy→assign
Solve cosx=x for Q
Write the equation cosx=x.
Context Panel: Solve≻Numerically Solve
Context Panel: Assign to a Name≻Q
cosx=x→solve0.7390851332→assign to a nameQ
Table 6.3.7(a) provides a solution by task template.
Tools≻Task≻Browse:
Calculus - Vector≻Integration≻Surface Integration≻Surface Defined over a 2-D Region
Surface Integral on a Surface Defined over a General 2-D Region
Integrand
fx,y,z=
Surface
zx,y=
∫∫Sf ⅆσ =∫x=ax=b∫y=uxy=vxfx,y,zx,y 1+zx2+zy2 ⅆy ⅆx
∫∫Sf ⅆσ=∫y=ay=b∫x=uyx=vyfx,y,zx,y 1+zx2+zy2 ⅆx ⅆy
=
b=
a=
Table 6.3.7(a) Surface area via task template
A solution from first principles requires the calculation of λ=1+−2⁢y+12.
Obtain λ=1+−2⁢y+12
Expression palette: Square-root template
Calculus palette: Partial-derivative operator
Context Panel: Assign to a Name≻lambda
1+∂∂ x F2+∂∂ y F2 = 1+−2⁢y+12→assign to a nameλ
Iterate in the order dx dy via the template in the Calculus palette
Calculus palette: Iterated double-integral template
Context Panel: Evaluate and Display Inline
∫Q1∫arccosyyλ ⅆx ⅆy = 0.1300942772
Iterate in the order dy dx via the template in the Calculus palette
∫0Q∫cosx1λ ⅆy ⅆx+∫Q1∫x1λ ⅆy ⅆx = 0.1300942771
Maple Solution - Coded
Install the Student MultivariateCalculus package.
withStudent:-MultivariateCalculus:
Define the surface F.
F≔2−y−1/22:
Calculate the value of x=Q, the solution of cosx=x
Use the fsolve command for a numeric solution.
Q≔fsolvex=cosx,x
0.7390851332
For a solution from first principles, obtain λ=1+Fx2+Fy2
Use the diff command to obtain the requisite partial derivatives.
λ≔sqrt1+diffF,x2+diffF,y2
1+−2⁢y+12
Top-level, using the Int and int commands
Intλ,x=arccosy..y,y=Q..1=intλ,x=arccosy..y,y=Q..1
∫0.73908513321∫arccos⁡yy1+−2⁢y+12ⅆxⅆy=0.1300942772
Intλ,y=cosx..1,x=0..Q+Intλ,y=x..1,x=Q..1=intλ,y=cosx..1,x=0..Q+intλ,y=x..1,x=Q..1
∫00.7390851332∫cos⁡x11+−2⁢y+12ⅆyⅆx+∫0.73908513321∫x11+−2⁢y+12ⅆyⅆx=0.1300942772
Use the MultiInt command from the Student MultivariateCalculus package
Student:-MultivariateCalculus:-MultiIntλ,x=arccosy..y,y=Q..1 = 0.1300942772
Student:-MultivariateCalculus:-MultiIntλ,x=arccosy..y,y=Q..1,output=integral
∫0.73908513321∫arccos⁡yy1+−2⁢y+12ⅆxⅆy
Student:-MultivariateCalculus:-MultiIntλ,x=arccosy..y,y=Q..1,output=steps
∫0.73908513321∫arccos⁡yy1+−2⁢y+12ⅆxⅆy=∫0.739085133211+−2⁢y+12⁢xx=arccos⁡y..y|1+−2⁢y+12⁢xx=arccos⁡y..yⅆy=∫0.739085133211+−2⁢y+12⁢y−arccos⁡yⅆy=∫1+−2⁢y+12⁢y−arccos⁡yⅆyy=0.7390851332..1|∫1+−2⁢y+12⁢y−arccos⁡yⅆyy=0.7390851332..1
Use the SurfaceInt command from the Student VectorCalculus package
Student:-VectorCalculus:-SurfaceInt1,x,y,z=Surfacex,y,F,y,x=RegionQ..1, arccosy..y
0.1300942772
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