Chapter 6: Applications of Double Integration
Section 6.3: Surface Area
Example 6.3.3
Calculate the surface area of the surface defined by the function F=2 x+3 y+1 whose domain is the plane region R, the region bounded by the graphs of fx=sinx and gx=sin2 x on 0≤x≤π.
See Example 6.2.3.
Solution
Mathematical Solution
The surface is Fx,y=2 x+3 y+1, so the surface-area element is
dσ=1+Fx2+Fy2dA=1+22+32dA=14dA
The surface area is then given by the iterated integral
∫0π/3∫sinxsin2 x14ⅆy ⅆx+∫π/3π∫sin2 xsinx14ⅆy ⅆx = 514/2
Because λ=14, the integral can be evaluated in closed form.
Maple Solution - Interactive
Table 6.3.3(a) contains a solution from first principles.
Initialize
Context Panel: Assign Name
F=2 x+3 y+1→assign
Obtain dσ′=1+Fx2+Fy2
Expression palette: Square-root template
Calculus palette: Partial derivative template
Context Panel: Evaluate and Display Inline
1+∂∂ x F2+∂∂ y F2 = 14
Write and evaluate the relevant iterated integral
Calculus palette: Template for iterated double integral
∫0π/3∫sinxsin2 x14ⅆy ⅆx+∫π/3π∫sin2 xsinx14ⅆy ⅆx = 52⁢14
Table 6.3.3(a) Solution from first principles
Maple Solution - Coded
Install the Student MultivariateCalculus package.
withStudent:-MultivariateCalculus:
Define the surface.
F≔2 x+3 y+1:
Apply the SurfaceArea command from the Student MultivariateCalculus package
Because λ=14, the surface integral can be evaluated in closed form.
SurfaceAreaF,x=0..π/3,y=sinx..sin2 x,output=integral+SurfaceAreaF,x= π/3..π,y=sin2 x..sinx,output=integral
∫013⁢π∫sin⁡xsin⁡2⁢x14ⅆyⅆx+∫13⁢ππ∫sin⁡2⁢xsin⁡x14ⅆyⅆx
SurfaceAreaF,x=0..π/3, y=sinx..sin2 x+SurfaceAreaF,x= π/3..π,y=sin2 x..sinx
52⁢14
A solution from first principles must necessarily begin with the calculation of
λ=1+Fx2+Fy2=1+22+32=14
The top-level Int command returns the inert iterated integral
Int14,y=sinx..sin2 x,x=0..π/3+Int14,y=sin2 x..sinx,x= π/3..π
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