Chapter 9: Vector Calculus
Section 9.10: Green's Theorem
Example 9.10.10
Apply the Stokes-form of Green's theorem to F=x−2 y i+x2y j, and R, the region inside the bounding curves y=x2 and y=x, but outside the rectangle defined by the inequalities 1/5≤x≤1/2 and 3/10≤y≤2/5.
Solution
Mathematical Solution
Figure 9.10.10(a) provides a labeled diagram of the region R and its boundaries. The region is subdivided into four subregions Rk,k=1,…,4, and the bounding curves for the inner loop, drawn in red, are labeled Ck,k=1,…,4.
In the line integral on the right side of the Stokes-form of Green's theorem, the boundary is traversed counterclockwise. A counterclockwise orientation of the outer boundary formed by y1 and y2, induces a clockwise orientation on the inner loop whose bounding curves are Ck,k=1,…,4.
A simple calculation gives ∇×F·k=2x y+1≡Q.
Figure 9.10.10(a) The region R
The integral of Q over the region R is then
∫01/5∫x2xQ ⅆy ⅆx+∫1/51/2∫2/5xQ ⅆy ⅆx+∫1/51/2∫x23/10Q ⅆy ⅆx+∫1/21∫x2xQ ⅆy ⅆx = 4595960000
The line integral of F along the outer boundary consisting of y1 and y2 is then
∫012x5−2x2+xⅆx+∫10x−2x+x2/2ⅆx = 56
The line integral of F along the inner boundary consisting of Ck,k=1,…,4, is then
∫1215−310x2ⅆx+∫3102515−2yⅆy+∫1512−25x2ⅆx+∫2531012−2yⅆy = −134720000
The line integral of F along the boundaries of the region R is then 56−134720000 = 4595960000.
Maple Solution
The integration of ∇×F·k=2x y+1≡Q over the subregions Rk,k=1,…,4, as per Figure 9.10.10(a), is given in Table 9.10.10(a).
Define ∇×F·k=2x y+1 as Q
Context Panel: Assign to a Name≻Q
2x y+1→assign to a nameQ
Integrate Q over the region R
Calculus palette: Iterated definite-integral template
Context Panel: 2-D Math≻Convert To≻Inert Form Press the Enter key.
Context Panel: Evaluate Integral
∫01/5∫x2xQ ⅆy ⅆx+∫1/51/2∫2/5xQ ⅆy ⅆx+∫1/51/2∫x23/10Q ⅆy ⅆx+∫1/21∫x2xQ ⅆy ⅆx
∫015∫x2x2xy+2ⅆyⅆx+∫1512∫25x2xy+2ⅆyⅆx+∫1512∫x23102xy+2ⅆyⅆx+∫121∫x2x2xy+2ⅆyⅆx
=
4595960000
Table 9.10.10(a) Integral of Q over the region R
The line integral of F along the two bounding curves y1 and y2 is obtained in Table 9.10.10(b). The passage around this closed contour is in the counterclockwise direction.
Initialize
Tools≻Load Package: Student Vector Calculus
Loading Student:-VectorCalculus
Tools≻Tasks≻Browse: Calculus - Vector≻ Vector Algebra and Settings≻ Display Format for Vectors
Press the Access Settings button and select "Display as Column Vector"
Display Format for Vectors
Define the vector field F
Write the vector field as a free vector. Context Panel: Evaluate and Display Inline
Context Panel: Student Vector Calculus≻Conversions≻To Vector Field
Context Panel: Assign to a Name≻F
x−2 y,x2y = →to Vector Field →assign to a nameF
Obtain the line integral of F along y1 and y2
Apply the LineInt command and press the Enter key.
LineIntF,Pathx,x2,x=0..1,output=integral+LineIntF,Pathx,x,x=1..0,output=integral
∫012x5−2x2+xⅆx+∫10x−2x+12x2ⅆx
56
Table 9.10.10(b) Line Integral of F along y1 and y2
The line integral of F along the bounding curves Ck,k=1,…,4, is obtained in Table 9.10.10(c). The passage around this inner loop is clockwise, consistent with the orientation of the outer loop.
LineIntF,Pathx,3/10,x=1/2..1/5,output=integral+LineIntF,Path1/5,y,y=3/10..2/5,output=integral+LineIntF,Pathx,2/5,x=1/5..1/2,output=integral+LineIntF,Path1/2,y,y=2/5..3/10,output=integral
∫1215x−35ⅆx+∫31025125yⅆy+∫1512x−45ⅆx+∫2531014yⅆy
−134720000
Table 9.10.10(c) Line integral of F along the bounding curves Ck,k=1,…,4
The line integral along the boundaries of the region R is then 56−134720000 = 4595960000.
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