Chapter 6: Applications of Double Integration
Section 6.3: Surface Area
Example 6.3.13
Obtain the surface integral of gx,y=x y on the surface z1=7−x2−y2defined over the plane region R that is bounded by the curves y=x, y=2−x2, and x=0. See Example 6.2.10.
Solution
Mathematical Solution
The surface is defined by Fx,y=7−x2−y2, so the surface-area element is
dσ=1+Fx2+Fy2dA=⁢1+4 x2+4 y2dA
Iterating in the order dy dx results in the integral
∫01⁢∫x2−x2x⁢y 1+4 x2+4 y2 ⅆyⅆx ≐ 1.324515296
Maple Solution - Interactive
Initialize
Control-drag the equation z1=….
Context Panel: Assign Name
z1=7−x2−y2→assign
The simplest approach is to employ the task template in Table 6.3.13(a).
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 6.3.13(a) Solution by task template
A solution from first principles is given in Table 6.3.13(b).
Obtain λ=1+fx2+fy2
Calculus palette: Partial-derivative template
λ=1+∂∂ x z12+∂∂ y z12→assign
Write an appropriate iterated integral and evaluate
Calculus palette: Iterated double-integral template
Context Panel: Evaluate and Display Inline
∫01∫x2−x2x y λ ⅆy ⅆx = −799960−14⁢ln⁡2+14⁢ln⁡3+17−14⁢ln⁡−3+17+2964⁢17
Table 6.3.13(b) Solution from first principles
Maple Solution - Coded
Install the Student MultivariateCalculus package.
withStudent:-MultivariateCalculus:
Define the bounding surface z1.
z1≔7−x2−y2:
Use to diff command to obtain the partial derivatives with respect to x and y.
λ≔1+diffz1,x2+diffz1,y2
4⁢x2+4⁢y2+1
Form the integral via the MultiInt command with a pre-defined domain option
Evaluate the exact answer with the evalf command
MultiInt⁡x⁢y⁢λ,x,y=Region⁡0..1,x..2−x2,output=integral
∫01∫x−x2+2x⁢y⁢4⁢x2+4⁢y2+1ⅆyⅆx
S≔MultiInt⁡x⁢y⁢λ,x,y=Region0..1,x..2−x2
−799960−14⁢ln⁡2+14⁢ln⁡3+17−14⁢ln⁡−3+17+2964⁢17
evalfS = 1.324515296
Use the SurfaceInt command from the Student VectorCalculus package
Student:-VectorCalculus:-SurfaceIntx y,x,y,z=Surfacex,y,z1,x,y=Region0..1,x..2−x2,output=integral
q≔Student:-VectorCalculus:-SurfaceIntx y,x,y,z=Surfacex,y,z1,x,y=Region0..1,x..2−x2
evalfq = 1.324515296
A solution from first principles that uses the top-level Int command is given below.
Intx y λ,y=x..2−x2,x=0..1=evalfIntx y λ,y=x..2−x2,x=0..1
∫01∫x−x2+2x⁢y⁢4⁢x2+4⁢y2+1ⅆyⅆx=1.324515295
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