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Surface (area) Integral
parametrize
get partials of vector for both vars
cross the partials
get magnitude
integrate over bounds
Flux
parametrize
find n(t) = <y’(t),-x’(t)>
evaluate F(r(t))
dot F(r(t)) with n(t)
integrate over bounds
Green’s Theorem:
If positively oriented, simple, closed curve C bounding region D:
<P,Q> = F
Double integral D of (∂P/∂x + ∂Q/∂y) dA
If integrand becomes constant, just multiply by the area
conservative test
3d:
Construct 3×3 vector matrix with ijk,partials, and then components
Components should be <0,0,0>
2d:
y partial of first component should equal x partial of second component
When to find the potential function
fundamental theorem
computing work
integral F * dr over C
testing path independence
evaluating line integrals quickly
when the curve is complicated but the endpoints are easy
Divergence
Sum of partials
Curl
Vector result from the conservative test
Cross-product determinant
Work
If parameterized: integral of F(r(t)) * r’(t) dt from a to b
(line integral)
If conservative: f(B)-f(A) using the potential function
Line integrals of vector fields
Integral F(r(t)) * r’(t) dt from a to b
Line integrals of scalar functions
Integral of f(r(t)) * ||r’(t)|| dt from a to b
Circulation
If parameterized: integral of F(r(t)) * r’(t) dt from a to b
(line integral)
If conservative: f(B)-f(A) using the potential function
Green’s Theorem:
If positively oriented, simple, closed curve C bounding region D:
<P,Q> = F
Circulation: Double integral D of (∂Q/∂x - ∂P/∂y) dA
Wire mass
Integral of ρ(r(t)) * ||r’(t)|| dt from a to b
Fundamental Theorem
If conservative, use potential function and do f(B)-f(A)
Decision Rule
If you see
integral C of F * dr →
Is conservative? Use potential
No? Parameterize
Closed curve in 2D → Green’s
integral C of f ds → f(r(t))*||r’(t)||
Flux Problem:
2D closed curve: Green’s
3D surface: cross-product method