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Unformatted text preview: 6 Div, grad curl and all that 6.1 Fundamental theorems for gradient, divergence, and curl Figure 1: Fundamental theorem of calculus relates df/dx over[ a,b ] and f ( a ) , f ( b ). You will recall the fundamental theorem of calculus says Z b a df ( x ) dx dx = f ( b ) f ( a ) , (1) in other words it’s a connection between the rate of change of the function over the interval [ a, b ] and the values of the function at the endpoints (boundaries) of that interval. There are equivalent “fundamental theorems” for line integrals, area integrals, and volume integrals. In vector calculus we deal with different types of changes of scalar and vector fields, e.g. ~ ∇ φ , ~ ∇ · ~v , and ~ ∇ × ~v , & each has its own theorem. We’ve discussed line integrals before, mostly in the context of the work done along a path, but let’s remind ourselves of the definition: Z B A ~ F ( ~ r ) · d~ r = lim n →∞ n X i =1 ~ F ( ~ r i ) · d~ r i , (2) in other words we add up the area of all the little “rectangles” ~ F ( ~ r i ) · d~ r i consisting of the vector ~ F at the point ~ r i dotted into the path element d~ r i , see Figure. Remember the point is that although we are doing an integral in a 2D space, we are constrained to move along a path, so there is only one real independent variable. 1 Figure 2: Line integral. Example from test: Consider the triangle in the ( x, y ) plane with vertices at (1,0), (1,0), and (0,1). Evaluate the closed line integral I = I ( y ˆ x + x ˆ y ) · d~ r (3) around the boundary of the triangle in the anticlockwise direction. Since the vector ~ r has components ( x, y ), the measure is d~ r = ˆ xdx + ˆ ydy , so ( y ˆ x + x ˆ y ) · d~ r = ydx + xdy . On leg (1,0) → (1,0) we have y = 0, so integral is R 1 1 ( y ) dx = 0. On the leg (1,0) → (0,1) we have y = x + 1, so integral is  R 1 ( x + 1) dx + R 1 (1 y ) dy = 1 2 + 1 2 = 1. On the path (0,1) → (1,0) we have y = x + 1, so integral is R 1 ( x + 1) dx + R 1 ( y 1) dy = 1 2 + 1 2 = 1. So total line integral is 2. Let’s go back and look at the leg from (1,0) → (0,1) again. We could have “parameterized” this leg as x = 1 t , y = t , 0 < t < 1. Then ~ r = (1 t )ˆ x + t ˆ y , and d~ r = ( ˆ x + ˆ y ) dt . We can therefore write this integral as Z (0 , 1) (1 , 0) ~ F · d~ r = Z 1 ( t ˆ x + (1 t )ˆ y ) · ( ˆ x + ˆ y ) dt = Z 1 ( t + 1 t ) = 1 , (4) same answer as above. 6.1.1 Conservative field. Recall we said that the integral of an exact differential was independent of the path. This result can be expressed in our current particular context, line integrals of vector fields. Suppose you have a vector field ~ F which can be expressed as the 2 gradient of a scalar field, ~ F = ~ ∇ φ . Then we have the fundamental theorem for gradients: Z b a ~ ∇ φ · d~ r = Z b a dφ = φ ( b ) φ ( a ) , (5) in other words the integral R b a ~ F · d~ r doesn’t depend on the path between a and b ....
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 Fall '07
 HIRSHFELD
 Vector Calculus, Vector field, Stokes' theorem, Surface integral

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