Chapter 9: Vector Calculus
Section 9.6: Surface Integrals
Example 9.6.9
Integrate the scalar fx,y,z=x y z on the surface z=x2+y2 defined over the first-quadrant part of the interior of the ellipse x2+4 y2=1 bounded by the lines y=0 and y=x.
Solution
Mathematical Solution
Figure 9.6.9(a) shows the portion of the ellipse over which integration is to take place.
The region shaded in red lies under the line y=x for x∈0,1/5. The region shaded in green lies under the ellipse where y=1−x2/2, for x∈1/5,1.
The integrand of the surface integral is
f=x y x2+y2 1+4 x2+4 y2
so that the surface integral itself is given from first principles in Cartesian coordinates by
use plots, plottools in module() local p1,p2,p3; p1:=plot(sqrt(1-x^2)/2,x=1/sqrt(5)..1,color=black,filled=[color=green]): p2:=plot(x,x=0..1/sqrt(5),color=black,filled=[color=red]); p3:=display(p2,p1,scaling=constrained,labels=[x,y]); print(p3); end module: end use:
Figure 9.6.9(a) Sector of ellipse
∫01/5∫0xF dy dx +∫1/51∫01−x2/2F dy dx = 5504⁢5−1691260000⁢65−13360 ≐ 0.0208
It is also possible to make the change of variables x=r cosθ,y=r sinθ, that is, to polar coordinates. To this end, write the surface as the position vector
R=r cos(θ)r sin(θ)r2(cos2(θ)+sin2(θ)) = r cos(θ)r sin(θ)r2
so that
Rr×Rθ = cos(θ)sin(θ)2 r×−r sin(θ)r cos(θ)0 = −2 r2cos(θ)−2 r2sin(θ)r
and Rr×Rθ=r 1+4 r2. Since the integrand of the surface integral is given by
r cosθr sinθ(r2cos2θ+sin2θ r 1+4 r2=r5sinθcosθ1+4 r2=G
the surface integral itself is given by
∫0π/4∫01/cos2θ+4 sin2θG dr dθ= 5504⁢5−1691260000⁢65−13360≐0.0208
The upper limit on the inner integral is obtained by expressing the equation of the ellipse in polar coordinates. Thus, write
r2cos2θ+4 sin2θ=1⇒r=1/cos2θ+4 sin2θ
Maple Solution - Interactive
Table 9.6.9(a) provides a solution from first principles implemented in polar coordinates.
Initialize
Tools≻Load Package: Student Multivariate Calculus
Loading Student:-MultivariateCalculus
Context Panel: Assign to a Name≻R
r cosθ,r sinθ,r2→assign to a nameR
Obtain Rr×Rθ
Type the norm bars.
Calculus palette: Partial-derivative operator
Common Symbols palette: Cross-product operator
Context Panel: Evaluate and Display Inline
Context Panel: Simplify≻Assuming Positive
Context Panel: Assign to a Name≻dsig
∂∂ r R×∂∂ θ R = 4⁢r4⁢cos⁡θ2+4⁢r4⁢sin⁡θ2+cos⁡θ2⁢r+sin⁡θ2⁢r2→assuming positiver⁢4⁢r2+1→assign to a namedsig
Obtain the integrand in polar coordinates
Context Panel: Assign to a Name≻G
r cosθ r sinθ r2 dsig = r5⁢cos⁡θ⁢sin⁡θ⁢4⁢r2+1→assign to a nameG
Form and evaluate the surface integral
Calculus palette: Iterated double-integral operator
Context Panel: Approximate≻10 (digits)
∫0π/4∫01/cos2θ+4 sin2θG ⅆr ⅆθ = 5504⁢5−13360−1691260000⁢5⁢13→at 10 digits0.02080422869
Table 9.6.9(a) Solution from first principles implemented in polar coordinates
Maple Solution - Coded
Table 9.6.9(b) provides a solution from first principles implemented in polar coordinates.
Install the Student MultivariateCalculus package.
Define the surface as the position vector R.
R≔r cosθ,r sinθ,r2:
Use the CrossProduct, Norm, and simplify commands to obtain Rr×Rθ
dsig≔simplifyNormCrossProductdiffR,r,diffR,θ assuming r>0
r⁢4⁢r2+1
Except for the differentials, let G be the integrand in polar coordinates.
G≔r cosθ r sinθ r2 dsig
r5⁢cos⁡θ⁢sin⁡θ⁢4⁢r2+1
Form the surface integral with the top-level Int command.
Use the value command to evaluate the integral exactly.
Use the evalf command to approximate the integral numerically.
q≔IntG,r=0..1/cos2θ+4 sin2θ,θ=0..π/4
∫014⁢π∫01cos⁡θ2+4⁢sin⁡θ2r5⁢cos⁡θ⁢sin⁡θ⁢4⁢r2+1ⅆrⅆθ
valueq
5504⁢5−13360−1691260000⁢5⁢13
evalfq
0.02080422869
Table 9.6.9(b) Solution from first principles implemented in polar coordinates
Table 9.6.9(c) provides a solution from first principles implemented in Cartesian coordinates.
Assign E to the equation of the ellipse.
E≔x2+4 y2=1:
Use the solve command to obtain the explicit Cartesian representation of the branches of the equation of the ellipse.
Y≔solveE,y
12⁢−x2+1,−12⁢−x2+1
Use the solve command to obtain the coordinates of the intersection of the ellipse with the line y=x.
solveE,y=x,explicit1
x=15⁢5,y=15⁢5
Except for the differentials in dσ, let G be the integrand in Cartesian coordinates.
G≔x y x2+y2 1+4 x2+4 y2:
Form and evaluate the surface integral (See Figure 9.6.9(a))
q≔IntG,y=0..x,x=0..1/5+IntG,y=0..Y1,x=1/5..1
∫015⁢5∫0xx⁢y⁢x2+y2⁢4⁢x2+4⁢y2+1ⅆyⅆx+∫15⁢51∫012⁢−x2+1x⁢y⁢x2+y2⁢4⁢x2+4⁢y2+1ⅆyⅆx
−13360−1691260000⁢5⁢13+5504⁢5
Table 9.6.9(c) Solution from first principles implemented in Cartesian coordinates
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