Chapter 6: Applications of Double Integration
Section 6.3: Surface Area
Calculate the surface area of the surface defined by the function F=3 x2+2 y2+1 whose domain is the plane region R, the region bounded by the graphs of fx=arctanx+1−1/2 and gx=sinx on the interval x∈0,π/2. See Example 6.2.4.
The surface is Fx,y=2 x2+3 y2, so the surface-area element is
dσ=1+Fx2+Fy2dA=1+36 x2+16 y2dA
The surface area is then given by the iterated integral
∫0X∫gxfx1+36 x2+16 y2 ⅆy ⅆx+∫Xπ/2∫fxgx1+36 x2+16 y2 ⅆy ⅆx ≐ 1.898369230
where X≐0.5059163080 is the intersection of the graphs of fx and gx.
The inner integral can be evaluated in closed form, but then the resulting outer integral must be evaluated numerically. However, since one limit, namely X, is a floating-point number, Maple immediately evaluates the integrals numerically.
Maple Solution - Interactive
Table 6.3.4(a) contains a solution from first principles.
Context Panel: Assign Name
F=3 x2+2 y2+1→assign
Context Panel: Assign name
Obtain the intersections of the curves bounding R
Equate fx and gx, then press the Enter key.
Context Panel: Solve≻Numerically Solve
Context Panel: Assign to a Name≻X
→assign to a name
Expression palette: Square-root template
Calculus palette: Partial derivative template
Context Panel: Evaluate and Display Inline
1+∂∂ x F2+∂∂ y F2 = 36⁢x2+16⁢y2+1
Iterate in the order dy dx via the template in the Calculus palette
Calculus palette: Template for iterated double integral
∫0X∫gf1+36 x2+16 y2 ⅆy ⅆx+∫Xπ/2∫fg1+36 x2+16 y2 ⅆy ⅆx = 1.898369230
Table 6.3.4(a) Solution from first principles
Since one limit of integration, namely X, is a floating-point number, Maple immediately evaluates the integrals numerically.
Maple Solution - Coded
Install the Student MultivariateCalculus package.
Define the surface.
F≔3 x2+2 y2+1:
Define the two curves that bound the region R.
Solve the equation fx=gx for x
Obtain a numeric solution with the fsolve command.
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. However, since one limit, namely X, is a floating-point number, Maple immediately evaluates the integrals numerically.
SurfaceAreaF,x=0..X,y=g..f+SurfaceAreaF,x=X..π/2,y=f..g = 1.898369230
Apply the SurfaceInt command from the Student VectorCalculus package
A solution from first principles must necessarily begin with the calculation of
λ=1+Fx2+Fy2=1+36 x2+16 y2
The top-level Int command returns the inert iterated integral; the top-level int command immediately evaluates.
Int1+36 x2+16 y2,y=g..f,x=0..X+Int1+36 x2+16 y2,y=f..g,x=X..π/2=int1+36 x2+16 y2,y=g..f,x=0..X+int1+36 x2+16 y2,y=f..g,x=X..π/2
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