Evaluating β« M dx + N dy: One Parameter Along the Curve
I Key FormulasKey Insight: the value depends on the curve and its direction, not on the parametrisation: along y = x 2 y = x^2 y = x 2 ( 0 , 0 ) (0,0) ( 0 , 0 ) ( 1 , 1 ) (1,1) ( 1 , 1 ) β« C β y β d x + x β d y = 1 3 \int_C -y\,dx + x\,dy = \tfrac13 β« C β β y d x + x d y = 3 1 β t t t ΞΈ \theta ΞΈ x x x
II Three Parametrisations, One ValueEach parametrisation of the arc gives a different integrand in its own parameter, but every one totals 1 3 \tfrac13 3 1 β
Parameter
x = t, y = tΒ², 0 β€ t β€ 1
x = sin ΞΈ, y = sinΒ²ΞΈ, 0 β€ ΞΈ β€ Ο/2
x itself: y = xΒ², 0 β€ x β€ 1
progress 50%
III The Curve Decides the ValueAlong y = x k y = x^k y = x k ( 0 , 0 ) (0,0) ( 0 , 0 ) ( 1 , 1 ) (1,1) ( 1 , 1 ) x x x β y β d x + x β d y -y\,dx + x\,dy β y d x + x d y ( k β 1 ) x k β d x (k-1)x^k\,dx ( k β 1 ) x k d x
IV Quiz Questions
Problem 1 Β· Substitute Into Everything
Given: C C C y = x 2 y = x^2 y = x 2 ( 0 , 0 ) (0,0) ( 0 , 0 ) ( 1 , 1 ) (1,1) ( 1 , 1 ) find β« C y β d x + x β d y \displaystyle\int_C y\,dx + x\,dy β« C β y d x + x d y
a 1 3 \tfrac13 3 1 β b 2 3 \tfrac23 3 2 β c 5 6 \tfrac56 6 5 β d 1 1 1
β
Correct! x 2 β d x + x β
2 x β d x = 3 x 2 β d x x^2\,dx + x \cdot 2x\,dx = 3x^2\,dx x 2 d x + x β
2 x d x = 3 x 2 d x , and
β« 0 1 3 x 2 β d x = 1 \int_0^1 3x^2\,dx = 1 β« 0 1 β 3 x 2 d x = 1 .
β A term is missing. That is
β« C y β d x \int_C y\,dx β« C β y d x alone; the term
x β d y x\,dy x d y contributes
β« 0 1 2 x 2 β d x \int_0^1 2x^2\,dx β« 0 1 β 2 x 2 d x .
β A term is missing. That is
β« C x β d y \int_C x\,dy β« C β x d y alone; add
β« C y β d x = β« 0 1 x 2 β d x \int_C y\,dx = \int_0^1 x^2\,dx β« C β y d x = β« 0 1 β x 2 d x .
β Substitute d y dy d y On the curve
d y = 2 x β d x dy = 2x\,dx d y = 2 x d x ; replacing
d y dy d y by
d x dx d x gives
β« 0 1 ( x 2 + x ) β d x \int_0^1 (x^2 + x)\,dx β« 0 1 β ( x 2 + x ) d x , which is not the integral along
C C C .
Show solution
Use x x x y = x 2 y = x^2 y = x 2 d y = 2 x β d x dy = 2x\,dx d y = 2 x d x 0 β€ x β€ 1 0 \le x \le 1 0 β€ x β€ 1
β« C y β d x + x β d y = β« 0 1 ( x 2 + x β
2 x ) β d x = β« 0 1 3 x 2 β d x = 1. \int_C y\,dx + x\,dy = \int_0^1 \big(x^2 + x \cdot 2x\big)\,dx = \int_0^1 3x^2\,dx = 1. β« C β y d x + x d y = β« 0 1 β ( x 2 + x β
2 x ) d x = β« 0 1 β 3 x 2 d x = 1.
Problem 2 Β· Two Separate Integrals?
Given: a student evaluates β« C β y β d x + x β d y \int_C -y\,dx + x\,dy β« C β β y d x + x d y y = x 2 y = x^2 y = x 2 ( 0 , 0 ) (0,0) ( 0 , 0 ) ( 1 , 1 ) (1,1) ( 1 , 1 ) β« 0 1 β y β d x + β« 0 1 x β d y = β y + x \int_0^1 -y\,dx + \int_0^1 x\,dy = -y + x β« 0 1 β β y d x + β« 0 1 β x d y = β y + x what went wrong?
a Only the bounds: the y y y 0 0 0 1 1 1 y y y b x x x y y y C C C d x dx d x d y dy d y c The sign: the β y β d x -y\,dx β y d x + y β d x +y\,dx + y d x d Nothing; β y + x -y + x β y + x ( 1 , 1 ) (1,1) ( 1 , 1 ) 0 0 0
β
Correct! M β d x + N β d y M\,dx + N\,dy M d x + N d y is one integral along
C C C ; with
y = x 2 y = x^2 y = x 2 ,
d y = 2 x β d x dy = 2x\,dx d y = 2 x d x it becomes
β« 0 1 x 2 β d x = 1 3 \int_0^1 x^2\,dx = \tfrac13 β« 0 1 β x 2 d x = 3 1 β .
β The bounds are not the problem. Integrating
β y -y β y in
x x x alone leaves
y y y behind;
y y y is not a constant along the curve.
β The sign is right. M = β y M = -y M = β y for the rotation field; the error is splitting the integral.
β Not quite. The correct value is
1 3 \tfrac13 3 1 β , not
0 0 0 : separate integrals never see that
y = x 2 y = x^2 y = x 2 on
C C C .
Show solution
Along C C C x x x y = x 2 y = x^2 y = x 2 d y = 2 x β d x dy = 2x\,dx d y = 2 x d x
β« C β y β d x + x β d y = β« 0 1 ( β x 2 + 2 x 2 ) β d x = β« 0 1 x 2 β d x = 1 3 . \int_C -y\,dx + x\,dy = \int_0^1 \big(-x^2 + 2x^2\big)\,dx = \int_0^1 x^2\,dx = \tfrac13. β« C β β y d x + x d y = β« 0 1 β ( β x 2 + 2 x 2 ) d x = β« 0 1 β x 2 d x = 3 1 β .
Problem 3 Β· A Quarter Circle
Given: C C C x = cos β‘ ΞΈ x = \cos\theta x = cos ΞΈ y = sin β‘ ΞΈ y = \sin\theta y = sin ΞΈ 0 β€ ΞΈ β€ Ο 2 0 \le \theta \le \tfrac{\pi}{2} 0 β€ ΞΈ β€ 2 Ο β find β« C x β d y \displaystyle\int_C x\,dy β« C β x d y
a Ο 4 \tfrac{\pi}{4} 4 Ο β b Ο 2 \tfrac{\pi}{2} 2 Ο β c β 1 2 -\tfrac12 β 2 1 β d 1 1 1
β
Correct! x β d y = cos β‘ ΞΈ β
cos β‘ ΞΈ β d ΞΈ x\,dy = \cos\theta \cdot \cos\theta\,d\theta x d y = cos ΞΈ β
cos ΞΈ d ΞΈ , and
β« 0 Ο / 2 cos β‘ 2 ΞΈ β d ΞΈ = Ο 4 \int_0^{\pi/2} \cos^2\theta\,d\theta = \tfrac{\pi}{4} β« 0 Ο /2 β cos 2 ΞΈ d ΞΈ = 4 Ο β .
β Check the integral of cos β‘ 2 ΞΈ \cos^2\theta cos 2 ΞΈ Its average over
[ 0 , Ο 2 ] [0, \tfrac{\pi}{2}] [ 0 , 2 Ο β ] is
1 2 \tfrac12 2 1 β , so the integral is
1 2 β
Ο 2 \tfrac12 \cdot \tfrac{\pi}{2} 2 1 β β
2 Ο β .
β That used d x dx d x β« 0 Ο / 2 cos β‘ ΞΈ β ( β sin β‘ ΞΈ ) β d ΞΈ = β 1 2 \int_0^{\pi/2} \cos\theta\,(-\sin\theta)\,d\theta = -\tfrac12 β« 0 Ο /2 β cos ΞΈ ( β sin ΞΈ ) d ΞΈ = β 2 1 β is
β« C x β d x \int_C x\,dx β« C β x d x ; here
d y = cos β‘ ΞΈ β d ΞΈ dy = \cos\theta\,d\theta d y = cos ΞΈ d ΞΈ .
β Substitute d y dy d y d y dy d y is
cos β‘ ΞΈ β d ΞΈ \cos\theta\,d\theta cos ΞΈ d ΞΈ , not
d ΞΈ d\theta d ΞΈ ;
β« 0 Ο / 2 cos β‘ ΞΈ β d ΞΈ = 1 \int_0^{\pi/2} \cos\theta\,d\theta = 1 β« 0 Ο /2 β cos ΞΈ d ΞΈ = 1 forgets that factor.
Show solution
d y = cos β‘ ΞΈ β d ΞΈ dy = \cos\theta\,d\theta d y = cos ΞΈ d ΞΈ
β« C x β d y = β« 0 Ο / 2 cos β‘ 2 ΞΈ β d ΞΈ = β« 0 Ο / 2 1 + cos β‘ 2 ΞΈ 2 β d ΞΈ = [ ΞΈ 2 + sin β‘ 2 ΞΈ 4 ] 0 Ο / 2 = Ο 4 . \int_C x\,dy = \int_0^{\pi/2} \cos^2\theta\,d\theta = \int_0^{\pi/2} \frac{1 + \cos 2\theta}{2}\,d\theta = \Big[\frac{\theta}{2} + \frac{\sin 2\theta}{4}\Big]_0^{\pi/2} = \frac{\pi}{4}. β« C β x d y = β« 0 Ο /2 β cos 2 ΞΈ d ΞΈ = β« 0 Ο /2 β 2 1 + cos 2 ΞΈ β d ΞΈ = [ 2 ΞΈ β + 4 sin 2 ΞΈ β ] 0 Ο /2 β = 4 Ο β .
Problem 4 Β· Running the Curve Backward
Given: β« C β y β d x + x β d y = 1 3 \int_C -y\,dx + x\,dy = \tfrac13 β« C β β y d x + x d y = 3 1 β y = x 2 y = x^2 y = x 2 ( 0 , 0 ) (0,0) ( 0 , 0 ) ( 1 , 1 ) (1,1) ( 1 , 1 ) C β² C' C β² ( 1 , 1 ) (1,1) ( 1 , 1 ) ( 0 , 0 ) (0,0) ( 0 , 0 ) x = 1 β t x = 1 - t x = 1 β t y = ( 1 β t ) 2 y = (1-t)^2 y = ( 1 β t ) 2 0 β€ t β€ 1 0 \le t \le 1 0 β€ t β€ 1 find β« C β² β y β d x + x β d y \int_{C'} -y\,dx + x\,dy β« C β² β β y d x + x d y
a 1 3 \tfrac13 3 1 β b 0 0 0 c β 1 3 -\tfrac13 β 3 1 β d 2 3 \tfrac23 3 2 β
β
Correct! Reversing the direction flips every
d x dx d x and
d y dy d y , so the value changes sign.
β Direction is part of the curve. Reparametrising in the same direction keeps the value; running backward is a different oriented curve.
β Not quite. 0 0 0 would be the round trip, out and back; this is only the return leg.
β Not quite. Compute it: the integrand is
β ( 1 β t ) 2 β d t -(1-t)^2\,dt β ( 1 β t ) 2 d t on
[ 0 , 1 ] [0, 1] [ 0 , 1 ] .
Show solution
d x = β d t dx = -dt d x = β d t d y = β 2 ( 1 β t ) β d t dy = -2(1-t)\,dt d y = β 2 ( 1 β t ) d t
β y β d x + x β d y = ( 1 β t ) 2 β d t β 2 ( 1 β t ) 2 β d t = β ( 1 β t ) 2 β d t , -y\,dx + x\,dy = (1-t)^2\,dt - 2(1-t)^2\,dt = -(1-t)^2\,dt, β y d x + x d y = ( 1 β t ) 2 d t β 2 ( 1 β t ) 2 d t = β ( 1 β t ) 2 d t , β« C β² β y β d x + x β d y = β β« 0 1 ( 1 β t ) 2 β d t = β 1 3 . \int_{C'} -y\,dx + x\,dy = -\int_0^1 (1-t)^2\,dt = -\tfrac13. β« C β² β β y d x + x d y = β β« 0 1 β ( 1 β t ) 2 d t = β 3 1 β .
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