of S oriented counterclockwise as seen from above. Let F be the vector field F(x,y,z) = .
Evaluate integral F*dr with bound C
(a) directly as line integral
(b) as a double integral, by using Stokes' Theorem
Evaluate integral F*dr with bound C
(a) directly as line integral
(b) as a double integral, by using Stokes' Theorem
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(a) We evaluate the line integral one segment at a time.
(i) C₁: (0, 0, 0) to (1, 2, 3) ==> r(t) = for t in [0, 1].
So, ∫c₁ F · dr
= ∫(t = 0 to 1) <2t, -t, 3t> · <1, 2, 3> dt
= ∫(t = 0 to 1) 9t dt
= 9/2.
(ii) C₂: (1, 2, 3) to (2, 1, 3) ==> r(t) = <1+t, 2-t, 3> for t in [0, 1].
So, ∫c₂ F · dr
= ∫(t = 0 to 1) <2 - t, -1 - t, 3> · <1, -1, 0> dt
= ∫(t = 0 to 1) 3 dt
= 3.
(iii) C₃: (2, 1, 3) to (0, 0, 0) ==> r(t) = <2t, t, 3t> for t in [0, 1] with opposite orientation.
So, ∫c₃ F · dr
= -∫(t = 0 to 1) · <2, 1, 3> dt
= -∫(t = 0 to 1) 9t dt
= -9/2.
Note: This is clockwise orientation as done through (i)-(iii).
So by (i)-(iii), ∫c F · dr = -(9/2 + 3 + (-9/2)) = -3.
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(b) Note that the plane passing through P, Q, and R is x + y - z = 0
==> z = x + y.
Projecting this onto the xy-plane yields a triangular region with vertices (0, 0), (1, 2), (2, 1).
==> y = x/2 to y = 2x for x in [0, 1], and y = x/2 to y = 3-x for x in [1, 2].
Since curl F = <0, 0, -2>,
∫∫s curl F · dS
= ∫∫ <0, 0, -2> · <-z_x, -z_y, 1> dA
= ∫∫ <0, 0, -2> · <-1, -1, 1> dA
= ∫∫ -2 dA
= -2 * [∫(x = 0 to 1) ∫(y = x/2 to 2x) dy dx + ∫(x = 1 to 2) ∫(y = x/2 to 3-x) dy dx]
= -2 * [∫(x = 0 to 1) (2x - x/2) dx + ∫(x = 1 to 2) ((3 - x) - x/2) dx]
= -[∫(x = 0 to 1) 3x dx + ∫(x = 1 to 2) (6 - 3x) dx]
= -[(3/2)x^2 {for x = 0 to 1} + (6x - 3x^2/2) {for x = 1 to 2}]
= -3.
I hope this helps!
(i) C₁: (0, 0, 0) to (1, 2, 3) ==> r(t) =
So, ∫c₁ F · dr
= ∫(t = 0 to 1) <2t, -t, 3t> · <1, 2, 3> dt
= ∫(t = 0 to 1) 9t dt
= 9/2.
(ii) C₂: (1, 2, 3) to (2, 1, 3) ==> r(t) = <1+t, 2-t, 3> for t in [0, 1].
So, ∫c₂ F · dr
= ∫(t = 0 to 1) <2 - t, -1 - t, 3> · <1, -1, 0> dt
= ∫(t = 0 to 1) 3 dt
= 3.
(iii) C₃: (2, 1, 3) to (0, 0, 0) ==> r(t) = <2t, t, 3t> for t in [0, 1] with opposite orientation.
So, ∫c₃ F · dr
= -∫(t = 0 to 1)
= -∫(t = 0 to 1) 9t dt
= -9/2.
Note: This is clockwise orientation as done through (i)-(iii).
So by (i)-(iii), ∫c F · dr = -(9/2 + 3 + (-9/2)) = -3.
---------------
(b) Note that the plane passing through P, Q, and R is x + y - z = 0
==> z = x + y.
Projecting this onto the xy-plane yields a triangular region with vertices (0, 0), (1, 2), (2, 1).
==> y = x/2 to y = 2x for x in [0, 1], and y = x/2 to y = 3-x for x in [1, 2].
Since curl F = <0, 0, -2>,
∫∫s curl F · dS
= ∫∫ <0, 0, -2> · <-z_x, -z_y, 1> dA
= ∫∫ <0, 0, -2> · <-1, -1, 1> dA
= ∫∫ -2 dA
= -2 * [∫(x = 0 to 1) ∫(y = x/2 to 2x) dy dx + ∫(x = 1 to 2) ∫(y = x/2 to 3-x) dy dx]
= -2 * [∫(x = 0 to 1) (2x - x/2) dx + ∫(x = 1 to 2) ((3 - x) - x/2) dx]
= -[∫(x = 0 to 1) 3x dx + ∫(x = 1 to 2) (6 - 3x) dx]
= -[(3/2)x^2 {for x = 0 to 1} + (6x - 3x^2/2) {for x = 1 to 2}]
= -3.
I hope this helps!