Chapter 9: Vector Calculus
Section 9.10: Green's Theorem
Example 9.10.9
Apply the divergence-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 9.10.9(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 flux integral on the right side of the divergence-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=1+x2≡Q.
Figure 9.10.9(a) The region R
The integral of ∇·F 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 = 80881210000
The flux of F through the outer boundary consisting of y1 and y2 is then
∫012−2x2+xx−x4ⅆx+∫10x/2−1−x5/2ⅆx = 44105
The flux of F through the inner boundary consisting of Ck,k=1,…,4, is then
∫1215−310x2ⅆx+∫3102515−2yⅆy+∫1512−25x2ⅆx+∫2531012−2yⅆy = −33910000
The net flux through the boundaries of the region R is then 44105−33910000 = 80881210000.
Maple Solution
The divergence of F, which is 1+x2≡Q, is integrated over subregions Rk,k=1,…,4, as per Figure 9.10.9(a), in Table 9.10.9(a).
Define ∇·F as Q
Context Panel: Assign to a Name≻Q
1+x2→assign to a nameQ
Integrate ∇·F 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∫x2xx2+1ⅆyⅆx+∫1512∫25xx2+1ⅆyⅆx+∫1512∫x2310x2+1ⅆyⅆx+∫121∫x2xx2+1ⅆyⅆx
=
80881210000
Table 9.10.9(a) Integral of ∇·F over the region R
The flux of F through the two bounding curves y1 and y2 is obtained in Table 9.10.9(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 flux of F through y1 and y2
Apply the Flux command and press the Enter key.
FluxF,Pathx,x2,x=0..1,output=integral+FluxF,Pathx,x,x=1..0,output=integral
∫012−2x2+xx−x4ⅆx+∫1012x−2xx−x5/2ⅆx
44105
Table 9.10.9(b) Flux of F through y1 and y2
The flux of F through the bounding curves Ck,k=1,…,4, is obtained in Table 9.10.9(c). The passage around this inner loop is clockwise, consistent with the orientation of the outer loop.
FluxF,Pathx,3/10,x=1/2..1/5,output=integral+FluxF,Path1/5,y,y=3/10..2/5,output=integral+FluxF,Pathx,2/5,x=1/5..1/2,output=integral+FluxF,Path1/2,y,y=2/5..3/10,output=integral
∫1215−310x2ⅆx+∫3102515−2yⅆy+∫1512−25x2ⅆx+∫2531012−2yⅆy
−33910000
Table 9.10.9(c) Flux of F through the bounding curves Ck,k=1,…,4
The total flux through the boundaries of the region R is then 44105−33910000 = 80881210000.
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