Using Greene'e Theorem, evaluate the integral under C of ∫ x^2cosy dx - xdy where C is given by |x|+|y|= pi............I've been having some trouble with this problem, and if anyone can help it would be greatly appreciated.
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By Green's Theorem,
∫c [x^2 cos y dx - x dy]
= ∫∫R [(∂/∂x)(-x) - (∂/∂y)(x^2 cos y)] dA
= ∫∫R (-1 + x^2 sin y) dA, where we are integrating over R: |x| + |y| ≤ 1, a square.
The first term ∫∫R -1 dA
= -(Area of square with vertices (±1, 0) and (0, ±1)) = -(1^2) = -1.
As for the second term ∫∫R x^2 sin y dA:
Note that this region is symmetric both the x and y axes.
This region can be broken into four subregions, one per quadrant.
(i) First quadrant: Between x + y = 1, x = 0, and y = 0
(ii) Second quadrant: Between -x + y = 1, x = 0, and y = 0
(iii) Third quadrant: Between -x - y = 1, x = 0, and y = 0
(iv) Fourth quadrant: Between x - y = 1, x = 0, and y = 0
I'll relate the last three regions and their integrals to that of (i).
(ii) Letting u = -x and v = y transforms this region into the first quadrant
with jacobian -1.
So, ∫∫ x^2 sin y dA = ∫∫ (-u)^2 sin v * 1 dA' = ∫∫ u^2 sin v dA'
This is the same integral as in (i) (with variable relabelings).
(iii) Letting u = -x and v = -y transforms this region into the first quadrant
with jacobian 1.
So, ∫∫ x^2 sin y dA = ∫∫ (-u)^2 sin(-v) * 1 dA' = -∫∫ u^2 sin v dA'
This is the same integral as in (i), but multiplied by -1.
(iv) Letting u = x and v = -y transforms this region into the first quadrant
with jacobian -1.
So, ∫∫ x^2 sin y dA = ∫∫ u^2 sin(-v) * 1 dA' = -∫∫ u^2 sin v dA'
This is the same integral as in (i), but multiplied by -1.
Adding (i)-(iv) together yields ∫∫R x^2 sin y dA = 0.
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Hence, ∫∫R (-1 + x^2 sin y) dA = -1 + 0 = -1.
I hope this helps!
∫c [x^2 cos y dx - x dy]
= ∫∫R [(∂/∂x)(-x) - (∂/∂y)(x^2 cos y)] dA
= ∫∫R (-1 + x^2 sin y) dA, where we are integrating over R: |x| + |y| ≤ 1, a square.
The first term ∫∫R -1 dA
= -(Area of square with vertices (±1, 0) and (0, ±1)) = -(1^2) = -1.
As for the second term ∫∫R x^2 sin y dA:
Note that this region is symmetric both the x and y axes.
This region can be broken into four subregions, one per quadrant.
(i) First quadrant: Between x + y = 1, x = 0, and y = 0
(ii) Second quadrant: Between -x + y = 1, x = 0, and y = 0
(iii) Third quadrant: Between -x - y = 1, x = 0, and y = 0
(iv) Fourth quadrant: Between x - y = 1, x = 0, and y = 0
I'll relate the last three regions and their integrals to that of (i).
(ii) Letting u = -x and v = y transforms this region into the first quadrant
with jacobian -1.
So, ∫∫ x^2 sin y dA = ∫∫ (-u)^2 sin v * 1 dA' = ∫∫ u^2 sin v dA'
This is the same integral as in (i) (with variable relabelings).
(iii) Letting u = -x and v = -y transforms this region into the first quadrant
with jacobian 1.
So, ∫∫ x^2 sin y dA = ∫∫ (-u)^2 sin(-v) * 1 dA' = -∫∫ u^2 sin v dA'
This is the same integral as in (i), but multiplied by -1.
(iv) Letting u = x and v = -y transforms this region into the first quadrant
with jacobian -1.
So, ∫∫ x^2 sin y dA = ∫∫ u^2 sin(-v) * 1 dA' = -∫∫ u^2 sin v dA'
This is the same integral as in (i), but multiplied by -1.
Adding (i)-(iv) together yields ∫∫R x^2 sin y dA = 0.
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Hence, ∫∫R (-1 + x^2 sin y) dA = -1 + 0 = -1.
I hope this helps!
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I should mention two minor corrections.
(i) The tilted square has equation |x| + |y| = π, not 1.
(ii) So, the vertices are (±π, 0) and (0, ±π).
==> The sides have length π√2.
So, the integral actually equals -(π√2)^2 = -2π^2.
(i) The tilted square has equation |x| + |y| = π, not 1.
(ii) So, the vertices are (±π, 0) and (0, ±π).
==> The sides have length π√2.
So, the integral actually equals -(π√2)^2 = -2π^2.
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so are you saying the final answer is -2π^2 is the answer? i got (π^2/2)-(π^2), but i may have made an error somewhere
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ahh never mind....i have to multiply what i got by four because i only solved for one subregion......thanks again for the help
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