Chapter 9: Vector Calculus
Section 9.6: Surface Integrals
Example 9.6.3
Integrate the scalar fx,y,z=x y z on the surface defined parametrically by the equations xu,v=u+v,yx,v=u−v,zu,v=u v and the bounds u∈0,1, v∈0,u.
Solution
Mathematical Solution
The element of surface area is Ru×Rv dA, where R is a position vector representation of the surface, Ru and Rv are vectors tangent to the coordinate curves, and dA is either du dv or dv du.
Hence, the vectors
R=u+vu−vu v ⇒ Ru=∂uR=11v and Rv=∂vR=1−1u
lead to the cross product
Ru×Rv=ijk11v1−1u = u+vv−u−2 ⇒ Ru×Rv=4+u+v2+v−u2=4+2 u2+2 v2
so that the surface integral is
∫∫Sf dσ = ∫01∫0uu+vu−v u v 4+2 u2+2 v2 dv du = 128105⁢2−16105−356
Maple Solution - Interactive
Table 9.6.3(a) provides a solution via task template.
Tools≻Tasks≻Browse Calculus - Vector≻Integration≻Surface Integration≻Over a parametrically Defined Surface
Surface Integral on a Surface Defined Parametrically
Integrand
fx,y,z=
xu,v=
yu,v=
zu,v=
Fu,v≡fxu,v,yu,v,zu,v
N≡∂y,z∂u,v2+∂z,x∂u,v2+∂x,y∂u,v2
∫∫Sf ⅆσ =∫u=au=b∫v=guv=huFu,vN ⅆv ⅆu
=
b=
∫∫Sf ⅆσ=∫v=av=b∫u=Gvu=HvFu,vN ⅆu ⅆv
a=
Table 9.6.3(a) Solution by task template
Table 9.6.3(b) contains a solution from first principles.
Initialize
Tools≻Load Package: Student Multivariate Calculus
Loading Student:-MultivariateCalculus
Define the surface as the position vector R
Context Panel: Assign Name
R=u+v,u−v,u v→assign
Obtain vectors Ru and Rv tangent to u- and v-coordinate curves, respectively
Calculus palette: Partial-derivative operator
Ru=∂∂ u R→assign
Rv=∂∂ v R→assign
Obtain dσ (sans differentials): dsig=Ru×Rv
Common Symbols palette: Cross-product operator
Context Panel: Evaluate and Display Inline
Context Panel: Student Multivariate Calculus≻Norm≻Euclidean
Context Panel: Assign to a Name≻dsig
Ru×Rv = →Euclidean-norm2⁢u2+2⁢v2+4→assign to a namedsig
Write and evaluate the relevant surface integral
Calculus palette: Double-integral template
∫01∫0uu+v u−v u v dsig ⅆv ⅆu = −16105−35⁢3⁢2+128105⁢2
Table 9.6.3(b) Solution from first principles
Maple Solution - Coded
Table 9.6.3(c) contains a solution based on the SurfaceInt command.
Install the Student VectorCalculus package.
withStudent:-VectorCalculus:
SurfaceIntx y z,x,y,z=Surfaceu+v,u−v,u v,u=0..1,v=0..u,output=integral=SurfaceIntx y z,x,y,z=Surfaceu+v,u−v,u v,u=0..1,v=0..u
∫01∫0uu+v⁢u−v⁢u⁢v⁢4+u+v2+−u+v2ⅆvⅆu=−16105−35⁢3⁢2+128105⁢2
Table 9.6.3(c) Solution via the Surface option in the SurfaceInt command
Table 9.6.3(d) contains a solution from first principles.
Set the display of vectors with the BasisFormat command.
BasisFormatfalse:
Define the function f.
f≔x,y,z→x y z:
Define the surface as a position vector.
R≔u+v,u−v,u v:
Use the diff command to obtain vectors tangent to the coordinate curves.
Ru,Rv≔diffR,u,diffR,v:
Use the CrossProduct and Norm commands to obtain, except for the differentials du and dv, dσ.
dsig≔NormCrossProductRu,Rv
2⁢u2+2⁢v2+4
Use the top-level Int and int commands to write and evaluate the surface integral
Intfu+v,u−v,u v dsig, v=0..u,u=0..1=:-intfu+v,u−v,u v dsig, v=0..u,u=0..1
∫01∫0uu+v⁢u−v⁢u⁢v⁢2⁢u2+2⁢v2+4ⅆvⅆu=−16105+128105⁢2−35⁢2⁢3
Table 9.6.3(d) Solution from first principles
(Because the Student VectorCalculus package contains its own int command, access to the top-level version of the command requires the colon-dash prefix.)
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