Surface Integrals of Vector Fields When dealing with surface integrals of vector fields we encounter the surface orientation issue. Our basic method of integration is: | [F(x, y, z)-ñ ds = ± z)-nds: ±] [F(x, y, z) - (FF) A R Where the + indicates that we need to decide which it is. Ideally we'd like to parametrize the surface so that the cross-product generated normal vectors point in the direction of desired orientation but this is quite difficult. Instead we take the approach from class - we look at the cross product XT, and examine whether it agrees with or disagrees with Σ's orientation. If it disagrees we negate. For example suppose we wish to evaluate √ √(x²î + yî + zk) • ñ dS where Σ is the part of the parabolic sheet y = x² with −2 ≤ x ≤2 and 0 ≤ z ≤ 3, oriented Σ outwards, meaning with positive y-component. We can parameterize Σ with F(x, z) = x+(4−x²)ĵ+zk with −2 ≤x≤2 and 0≤≤3. We can find FXF, with Matlab: syms x z; rbar [x,4-x^2,2]; cross(diff(rbar,x),diff(rbar,z)) This yields output: ans = [ -2*x, -1, 0] These vectors have negative y-component and hence they disagree with Σ's orientation. The final integral then needs to be negated. Here is the whole thing: syms x y z; rbar -[x,4-x^2,2]; F = [x^2,y,z]; -int (int(simplify (dot (subs (F, [x,y,z], rbar), cross (diff(rbar, x), diff(rbar, z)))),x,-2,2),z,0,3) Suppose Σ is the portion of the parabolic sheet y = x² bounded by -2≤x≤3 and 0≤z≤5 with orientation in the negative y direction. If F(x, y, z) = xî + xyj + yzk describes the flow of a fluid, find the rate of flow of F through Σ in the direction of orientation. Assign the result to q7.
Surface Integrals of Vector Fields When dealing with surface integrals of vector fields we encounter the surface orientation issue. Our basic method of integration is: | [F(x, y, z)-ñ ds = ± z)-nds: ±] [F(x, y, z) - (FF) A R Where the + indicates that we need to decide which it is. Ideally we'd like to parametrize the surface so that the cross-product generated normal vectors point in the direction of desired orientation but this is quite difficult. Instead we take the approach from class - we look at the cross product XT, and examine whether it agrees with or disagrees with Σ's orientation. If it disagrees we negate. For example suppose we wish to evaluate √ √(x²î + yî + zk) • ñ dS where Σ is the part of the parabolic sheet y = x² with −2 ≤ x ≤2 and 0 ≤ z ≤ 3, oriented Σ outwards, meaning with positive y-component. We can parameterize Σ with F(x, z) = x+(4−x²)ĵ+zk with −2 ≤x≤2 and 0≤≤3. We can find FXF, with Matlab: syms x z; rbar [x,4-x^2,2]; cross(diff(rbar,x),diff(rbar,z)) This yields output: ans = [ -2*x, -1, 0] These vectors have negative y-component and hence they disagree with Σ's orientation. The final integral then needs to be negated. Here is the whole thing: syms x y z; rbar -[x,4-x^2,2]; F = [x^2,y,z]; -int (int(simplify (dot (subs (F, [x,y,z], rbar), cross (diff(rbar, x), diff(rbar, z)))),x,-2,2),z,0,3) Suppose Σ is the portion of the parabolic sheet y = x² bounded by -2≤x≤3 and 0≤z≤5 with orientation in the negative y direction. If F(x, y, z) = xî + xyj + yzk describes the flow of a fluid, find the rate of flow of F through Σ in the direction of orientation. Assign the result to q7.
Advanced Engineering Mathematics
10th Edition
ISBN:9780470458365
Author:Erwin Kreyszig
Publisher:Erwin Kreyszig
Chapter2: Second-order Linear Odes
Section: Chapter Questions
Problem 1RQ
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Question
rbar7 = [x,9-x^2,z];
F7 = [x,x*y,y*z];
q7 = -int(int(simplify(dot(subs(F7,[x,y,z],rbar7),cross(diff(rbar7,x),diff(rbar7,z)))),x,-2,3),z,0,5)
![Surface Integrals of Vector Fields
When dealing with surface integrals of vector fields we encounter the surface orientation issue. Our basic method of integration is:
| [F(x, y, z)-ñ ds = ±
z)-nds:
±] [F(x, y, z) - (FF) A
R
Where the + indicates that we need to decide which it is. Ideally we'd like to parametrize the surface so that the cross-product generated normal vectors point in the
direction of desired orientation but this is quite difficult. Instead we take the approach from class - we look at the cross product XT, and examine whether it
agrees with or disagrees with Σ's orientation. If it disagrees we negate.
For example suppose we wish to evaluate √ √(x²î + yî + zk) • ñ dS where Σ is the part of the parabolic sheet y = x² with −2 ≤ x ≤2 and 0 ≤ z ≤ 3, oriented
Σ
outwards, meaning with positive y-component. We can parameterize Σ with F(x, z) = x+(4−x²)ĵ+zk with −2 ≤x≤2 and 0≤≤3. We can find FXF,
with Matlab:
syms x z;
rbar
[x,4-x^2,2];
cross(diff(rbar,x),diff(rbar,z))
This yields output:
ans =
[ -2*x, -1, 0]
These vectors have negative y-component and hence they disagree with Σ's orientation. The final integral then needs to be negated. Here is the whole thing:
syms x y z;
rbar -[x,4-x^2,2];
F =
[x^2,y,z];
-int (int(simplify (dot (subs (F, [x,y,z], rbar), cross (diff(rbar, x), diff(rbar, z)))),x,-2,2),z,0,3)](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F6fb406b3-dfb5-4d65-a424-9ec71ae4c6de%2F107b2550-3c22-405d-84f0-be49aefa811a%2Fhent0qb_processed.png&w=3840&q=75)
Transcribed Image Text:Surface Integrals of Vector Fields
When dealing with surface integrals of vector fields we encounter the surface orientation issue. Our basic method of integration is:
| [F(x, y, z)-ñ ds = ±
z)-nds:
±] [F(x, y, z) - (FF) A
R
Where the + indicates that we need to decide which it is. Ideally we'd like to parametrize the surface so that the cross-product generated normal vectors point in the
direction of desired orientation but this is quite difficult. Instead we take the approach from class - we look at the cross product XT, and examine whether it
agrees with or disagrees with Σ's orientation. If it disagrees we negate.
For example suppose we wish to evaluate √ √(x²î + yî + zk) • ñ dS where Σ is the part of the parabolic sheet y = x² with −2 ≤ x ≤2 and 0 ≤ z ≤ 3, oriented
Σ
outwards, meaning with positive y-component. We can parameterize Σ with F(x, z) = x+(4−x²)ĵ+zk with −2 ≤x≤2 and 0≤≤3. We can find FXF,
with Matlab:
syms x z;
rbar
[x,4-x^2,2];
cross(diff(rbar,x),diff(rbar,z))
This yields output:
ans =
[ -2*x, -1, 0]
These vectors have negative y-component and hence they disagree with Σ's orientation. The final integral then needs to be negated. Here is the whole thing:
syms x y z;
rbar -[x,4-x^2,2];
F =
[x^2,y,z];
-int (int(simplify (dot (subs (F, [x,y,z], rbar), cross (diff(rbar, x), diff(rbar, z)))),x,-2,2),z,0,3)
Transcribed Image Text:Suppose Σ is the portion of the parabolic sheet y = x² bounded by -2≤x≤3 and 0≤z≤5 with orientation in the negative y direction. If
F(x, y, z) = xî + xyj + yzk describes the flow of a fluid, find the rate of flow of F through Σ in the direction of orientation. Assign the result to q7.
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