Chapter 9: Vector Calculus
Section 9.6: Surface Integrals
Example 9.6.4
Integrate the scalar fx,y,z=x y z on the surface z=x2+y2 defined over the plane region R bounded by x=0,x=π/4,y=sinx,y=cosx.
Solution
Mathematical Solution
Take the element of surface area as
dσ=1+zx2+zy2 dA=1+2 x2+2 y2 dA=1+4 x2+4 y2 dA
with dA=dy dx, so that the surface integral becomes
∫0π/4∫sinxcosxx y x2+y2 1+4 x2+4 y2 dy dx ≐ 0.1006308329
Figure 9.6.4(a) shows the region R and that part of the surface defined over R.
use plots, plottools in module() local p1,p2,f; f:=transform((x,y)->[x,y,0]); p1:=plot3d(x^2+y^2,x=0..Pi/4,y=sin(x)..cos(x)); p2:=display(p1,f(p1),scaling=constrained,labels=[x,y,z],tickmarks=[2,2,[0,1]],orientation=[-130,65,0],axes=frame); print(p2); end module: end use:
Figure 9.6.4(a) Plane region R and the surface defined over R
Maple Solution - Interactive
Table 9.6.4(a) provides a solution by task template.
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 9.6.4(a) Solution via task template
Table 9.6.4(b) contains a solution from first principles.
Initialize
Context Panel: Assign Function
fx,y,z=x y z→assign as functionf
Context Panel: Assign to a Name≻Z
x2+y2→assign to a nameZ
Obtain dsig=1+zx2+zy2
Calculus palette: Partial-derivative operator
Context Panel: Evaluate and Display Inline
Context Panel: Assign to a Name≻dsig
1+∂∂ x Z2+∂∂ y Z2 = 4⁢x2+4⁢y2+1→assign to a namedsig
Write and evaluate the appropriate surface integral
Calculus palette: Iterated double-integral template
Context Panel: Approximate≻10 (digits)
∫0π/4∫sinxcosxfx,y,Z dsig ⅆy ⅆx = ∫014⁢π−120⁢4⁢sin⁡x2+4⁢x2+13/2⁢x3−120⁢4⁢sin⁡x2+4⁢x2+13/2⁢x⁢sin⁡x2+1120⁢4⁢sin⁡x2+4⁢x2+13/2⁢x+120⁢x⁢4⁢cos⁡x2+4⁢x2+13/2⁢cos⁡x2+120⁢4⁢cos⁡x2+4⁢x2+13/2⁢x3−1120⁢x⁢4⁢cos⁡x2+4⁢x2+13/2ⅆx→at 10 digits0.1006308329
Table 9.6.4(b) Solution from first principles
Maple can evaluate just the inner integral exactly; the outer integral is then evaluated numerically. The alternative is to make one of the limits of integration a floating-point number so that the whole calculation is done numerically.
∫0.π/4∫sinxcosxfx,y,Z dsig ⅆy ⅆx = 0.1006308329
Maple Solution - Coded
Table 9.6.4(c) contains a solution based on the SurfaceInt command.
Install the Student VectorCalculus package.
withStudent:-VectorCalculus:
Define the surface.
Z≔x2+y2:
Apply the SurfaceInt command to obtain the unevaluated surface integral
SurfaceIntx y z,x,y,z=Surfacex,y,Z,x=0..π/4,y=sinx..cosx,output=integral
∫014⁢π∫sin⁡xcos⁡xx⁢y⁢x2+y2⁢4⁢x2+4⁢y2+1ⅆyⅆx
Float one number in the SurfaceInt command to obtain a numeric result
SurfaceIntx y z,x,y,z=Surfacex,y,Z,x=0..π/4.,y=sinx..cosx = 0.1006308329
Table 9.6.4(c) Solution via the SurfaceInt command
A closed-form is available for the inner integral, but not for the outer. Hence, evaluation of the iterated integral is done numerically by including one floating-point number in the SurfaceInt command.
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