Integrals along a curve in space. (Sect. 16.1)

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1 Integrals along a curve in space. (Sect. 6.) Line integrals in space. The addition of line integrals. ass and center of mass of wires. Line integrals in space Definition The line integral of a function f : D R 3 R along a curve associated with the function r : [t, t ] R D R 3 is given b s f ds = f (ˆr(s)) ds, s where ˆr(s) is the arc length parametrization of the function r, and s(t ) = s, s(t ) = s are the arc lengths at the points t, t. r r ( s ) f f ( r ( s ) ) s ( f r )

2 Line integrals in space Remarks: A line integral is an integral of a function along a curved path. The name curved integrals would be a better terminolog. Line integrals originate in the earl 8 to stud new phsical situations. For eample: () To describe the work done b a force along a path. (2) To solve problems involving fluid flows, electricit, magnetism. The line integral is independent of the curve original parametrization. In other words, given two different parametrization of the curve, with values r(t), and r( t), then the value of the line integral is the same for both parametrizations. Because the line integral is computed using the curve arc-length parametrization, which is unique for ever curve. Line integrals in space Theorem (General parametrization formula) The line integral of a continuous function f : D R 3 R along a differentiable curve r : [t, t ] R D R 3 can be epressed as s s f (ˆr(s)) ds = t t f (r(t)) r (t) dt, where ˆr(s) is the arc length parametrization of the function r, and s(t ) = s, s(t ) = s are the arc lengths at the points t, t. Proof: Recall the curve arc-length function s(t) = t t r (τ) dτ. Then ds = r (t) dt. Also, ˆr(s(t)) = r(t). Then, the integration b substitution formula implies s s f [ˆr(s(t)) ] ds = t t f ( r(t) ) r s = s(t ), (t) dt, s = s(t ).

3 Line integrals in space Remarks: When performing a line integral, the curve is alwas parametrized with its arc-length function. In this sense, a line integral is independent of the original parametrization of the curve. Line integrals can be defined on curves on the plane. In this case, the line integral is the area of the curtain under the graph of the function is the figure below. z f (,) r ( s ) f ( r (s ) ) The 2-dim line integral is an area, since the curve arc-length parametrization is used in the line integral computation. Line integrals in space Evaluate the line integral of the function f (,, z) = + + z along the curve r(t) = 2t, t, 2 2t in the interval t [, ]. t Solution: (r, straight line.) Recall: f ds = f (r(t)) r (t) dt. The derivative vector is r (t) = 2,, 2, therefore its magnitude is r (t) = = 3. The values of f along the curve are f (r(t)) = (2t)t + t + (2 2t) f (r(t)) = 2t 2 t + 2. f ds = ) (2t 2 t + 2) 3 dt = 3 [(2 t3 3 t t ]. ( 2 f ds = 3 3 ) = t f ds = 3 2.

4 Line integrals in space Evaluate the line integral of the function f (,, z) = 2 + z 2 along the curve r(t) =, a cos(t), a sin(t), in t [, π/2]. t Solution: (r, half circle.)recall: f ds = f (r(t)) r (t) dt. The derivative vector is r (t) =, a sin(t), a cos(t), therefore its magnitude is r (t) = a 2 sin 2 (t) + a 2 cos 2 (t) = a. The values of f along the curve are f (r(t)) = + a 2 sin 2 (t) f (r(t)) = a sin(t). f ds = π/2 a sin(t) a dt = a 2( cos(t) f ds = a 2. t π/2 ). Integrals along a curve in space. (Sect. 6.) Line integrals in space. The addition of line integrals. ass and center of mass of wires.

5 The addition of line integrals Theorem If a curve D in space is the union of the differentiable curves,, n, then the line integral of a continuous function f : D R 3 R along satisfies f ds = f ds + + f ds. n Remark: This result is useful to compute line integral along piecewise differentiable curves. z = U 2 2 The addition of line integrals Evaluate the line integral of f (,, z) = + z 2 along the path = 2, where is the image of r (t) = t, t 2, for t [, ], and 2 is the image of r 2 (t) =,, t for t [, ]. Solution: f ds = f ds + f ds. z 2 f ds = f ds = 4 5 r (t) =, 2t, r (t) = + 4t 2. 2t + 4t 2 dt, u /2 du = 2 (u 3/2 5) f (r (t)) = t + t = 2t. u = + 4t 2, du = 8t dt. f ds = 6 (5 5 ).

6 The addition of line integrals Evaluate the line integral of f (,, z) = + z 2 along the path = 2, where is the image of r (t) = t, t 2, for t [, ], and 2 is the image of r 2 (t) =,, t for t [, ]. Solution: f ds = f ds + f ds. z r 2(t) =,, r 2(t) =. f (r 2 (t)) = + t 2 = 2 t 2. ( f ds = (2 t 2 ) dt = 2 t ) ( t 3 ) = = 5 3. f ds = 6 (5 5 ) + 5 f ds = 3 6 ( ). Integrals along a curve in space. (Sect. 6.) Line integrals in space. The addition of line integrals. ass and center of mass of wires.

7 ass and center of mass of wires Remark: The total mass, the center of mass, and the moments of inertia of wires with arbitrar shapes in space, given b a curve and having a densit function ρ, can be computed using line integrals. = ρ ds; = ρ ds, = I = ( 2 + z 2 )ρ ds, I = ( 2 + z 2 )ρ ds, I z = ( )ρ ds. ρ ds, z = zρ ds; ass and center of mass of wires Find the moments of inertia of a wheel of radius R and densit ρ. Solution: We place the wheel at the center of the z = plane. The curve for the wheel is r(t) = R cos(t), R sin(t),, t [, 2π]. Therefore, r (t) = R sin(t), R cos(t),, hence r (t) = R. Recall: I = ( 2 + z 2 )ρ ds, I z = ( )ρ ds. I = 2π R 2 sin 2 (t)ρ R dt = R 3 ρ 2π [ I = R 3 ρ π ( sin(2t) 4 B smmetr, I = I. Finall, I z = 2π 2π )] [ ] cos(2t) dt 2 I = πr 3 ρ. R 2 ρ R dt I z = 2πR 3 ρ.