Chapter 6: Applications of Double Integration
Section 6.3: Surface Area
Example 6.3.2
Calculate the surface area of the surface defined by the function F=2 x2+3 y2 whose domain is the plane region R, the finite region bounded by the graphs of x=y2 and y=3−2 x. See Example 6.2.2.
Solution
Mathematical Solution
The surface is Fx,y=2 x2+3 y2, so the surface-area element is
dσ=1+Fx2+Fy2dA=1+16 x2+36 y2dA
The surface area is then given by the iterated integral
∫−3/21∫y23−y/21+16 x2+36 y2 dx dy ≐ 13.91521369
The inner integral can be evaluated in closed form, but then the resulting outer integral must be evaluated numerically.
Maple Solution - Interactive
Table 6.3.2(a) contains a solution via a task template that implements the SurfaceInt command from the Student VectorCalculus package. By setting the integrand to 1, the template computes surface area.
Tools≻Tasks≻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 6.3.2(a) Solution by task template
Clicking the "Value" button will produce a partly evaluated integral - the inner integral evaluates to a cumbersome expression that the outer integral cannot handle in closed form. The solution given here is a result of pressing the "Floating-Point Approximation" button.
Table 6.3.2(b) contains a solution from first principles.
Initialize
Context Panel: Assign Name
F=2 x2+3 y2→assign
Obtain λ=1+Fx2+Fy2
Expression palette: Square-root template
Calculus palette: Partial derivative template
Context Panel: Evaluate and Display Inline
1+∂∂ x F2+∂∂ y F2 = 16x2+36y2+1
Write and evaluate the relevant iterated integral
Calculus palette: Template for iterated double integral Context Panel: 2-D Math≻Convert To≻Inert Form Press the Enter key.
Context Panel: Approximate≻10 (digits)
∫−3/21∫y23−y/21+16 x2+36 y2ⅆx ⅆy
∫−321∫y232−y216x2+36y2+1ⅆxⅆy
→at 10 digits
13.91521368
Table 6.3.2(b) Solution from first principles
Maple Solution - Coded
Install the Student MultivariateCalculus package.
withStudent:-MultivariateCalculus:
Define the surface.
F≔2 x2+3 y2:
Apply the SurfaceArea command from the Student MultivariateCalculus package
The inner integral can be evaluated in closed form, but then the outer integral must be evaluated numerically by applying the evalf command to the result of the inner integral.
SurfaceAreaF,y=−3/2..1,x=y2..3−y/2,output=integral
q≔SurfaceAreaF,y=−3/2..1,x=y2..3−y/2
q≔∫−321−y216y4+36y2+12−9ln4y2+16y4+36y2+1y22−ln4y2+16y4+36y2+18+340y2−24y+374−40y2−24y+37y4+9ln6−2y+40y2−24y+37y22+ln6−2y+40y2−24y+378ⅆy
evalfq = 13.91521369
Alternate approach: Include a floating-point number as one of the arguments
SurfaceAreaF,y=−3/2..1.0,x=y2..3−y/2 = 13.91521369
Apply the SurfaceInt command from the Student VectorCalculus package
Student:-VectorCalculus:-SurfaceInt1,y,x,z=Surfacex,y,F,y=−3/2..1, x=y2..3−y/2,output=integral
A solution from first principles must necessarily begin with the calculation of
λ=1+Fx2+Fy2=1+16 x2+36 y2
The top-level Int command returns the inert iterated integral
Q≔Int1+16 x2+36 y2,x=y2..3−y/2,y=−3/2..1
Q≔∫−321∫y232−y216x2+36y2+1ⅆxⅆy
Apply the evalf command to obtain a numeric evaluation of the iterated integral
evalfQ = 13.91521368
The slight difference in the numeric values obtained is a reminder that the first value results from the numeric evaluation of a single integral; the second, from a double integral.
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