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Let $\mathbf{F}$ be a vector field defined on $\mathbb R^2 \setminus\{(0,0)\}$ by $$\mathbf {F } (x,y)=\frac{y}{x^2+y^2}i-\frac{x}{x^2+y^2}j $$ Let $\gamma,\alpha:[0,1]\to\mathbb R^2$ be defined by

$$\gamma (t)=(8\cos 2\pi t,17\sin 2\pi t)$$ and $$\alpha (t)=(26\cos 2\pi t,-10\sin 2\pi t)$$

If $$3\int_{\alpha} \mathbf{F\cdot dr} -4 \int_{\gamma} \mathbf{F\cdot dr}= 2m\pi,$$ then what is $m$?

How should I approach this question?

Progress

I see that the parametrization of ellipses are given already. For evaluating say first integral, I need to substitute given parametrization of ellipse in vector field. The parameter $t$ will vary from $ 0$ to $2\pi$. Am I correct?

Sophie Clad
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2 Answers2

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You are not expected to actually compute these line integrals using the parameterizations; this would be a rather painful procedure. The question appears to be testing your knowledge of the winding number. The curve $\gamma$ has winding number $1$ about the origin, since it travels once counterclockwise. The curve $\alpha$ has winding number $-1$, being clockwise.

Since $\mathbf F$ is irrotational in the punctured plane, the integral of this field over a closed loop depends only on the winding number about the origin. (Alternatively, you can the relation with arctangent pointed out by BaronVT to reach the same conclusion.)

Since the integral of $\mathbf F$ over the unit circle is $-2\pi$ (easy direct calculation), it follows that the integral over a closed curve with winding number $w$ is $-2\pi w$. Hence,
$$ \displaystyle 3\int_{\alpha} \mathbf{F\cdot dr} -4\displaystyle \int_{\gamma} \mathbf{F\cdot dr} = 3(-2\pi)(-1) - 4(-2\pi)(1) $$

  • How do you know that direction of other is clockwise . Is is because of minus sign attached to it ? Also as line integral is 2pi , this is because when i take x = cost y=sint and t goes from 0 to 2pi then i get 2pi .Am i correct ? The line integral doesnot depend upon contour But Why ? Thanks – Sophie Clad Dec 18 '14 at 02:18
  • @SophieClad : what is the difference between $t \mapsto (\cos t, \sin t)$ and $t \mapsto (\cos t, - \sin t)$? (If you don't know the answer, I strongly suggest plotting them yourself) Can you relate this to the ellipses given? – BaronVT Dec 18 '14 at 16:51
  • @user147263,in refference to the statement in your answer " Since F is irrotational in the punctured plane, the integral of this field over a closed loop depends only on the winding number about the origin. " can you please provide me the exact statement of the theorem? – Ibrahim Jan 13 '21 at 06:47
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Hint 1: This vector field is essentially $\nabla \theta$ (i.e. $\nabla \arctan(y/x)$). It has trouble at the origin, so the line integrals aren't zero, but you can at least use the fact that it's "almost" conservative to set up easier contours for integration (you are given ellipses, circles would be much easier though).

Hint 2: You can show that the integral around an ellipse is the same as the integral around a circle (up to the choice of clockwise vs counterclockwise). The integral around a circle is easy to compute (you are integrating the change in angle around a circle).

Hint 3: If you have a line integral of a conservative vector field enclosing a simply connected region, the integral is zero. The region I've shaded in this picture: is simply connected, and so the two "main" integrals must be equal.

BaronVT
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  • Vector field is conservative but what are those gamma and alpha functions of t given in question . Thanks – Sophie Clad Dec 12 '14 at 07:12
  • Can you please ealborate more – Sophie Clad Dec 12 '14 at 07:13
  • the $\alpha$ and $\gamma$ are ellipses – BaronVT Dec 12 '14 at 16:27
  • You can show that the integral around an ellipse is the same as the integral around a circle (up to the choice of clockwise vs counterclockwise). The integral around a circle is easy to compute (hint: you are integrating the change in angle around a circle). – BaronVT Dec 12 '14 at 21:47
  • Ok . So parametrization of ellipses are given already. For evaluating say first integral , i need to substitute given parametrization of ellipse in vector field. T will vary from 0 to 2pi .am i correct ? Thanks – Sophie Clad Dec 13 '14 at 10:48
  • Have you not evaluated a line integral before? You would integrate $F(\alpha(t)) \cdot \alpha'(t)$ (so you not only "substitute given parameterization in vector field", but you also take the dot product with the derivative of the parameterization). But as I have said a few times now, make the work easier for yourself and justify integrating over circles instead of the ellipses. The limits of integration will be $0$ to $2\pi$, yes. – BaronVT Dec 13 '14 at 14:02
  • Ok. But parametrization is over ellipses .how do i do change of variable here ? as you are saying to do over circles ? – Sophie Clad Dec 14 '14 at 04:11
  • If you have a line integral of a conservative vector field enclosing a simply connected region, the integral is zero. The region I've shaded in this picture: http://imgur.com/HbThGVB is simply connected, and so the two "main" integrals must be equal. – BaronVT Dec 14 '14 at 18:26
  • Indeed, this is essentially the same vector field as in : Find the velocity of a flow . – Han de Bruijn Dec 18 '14 at 13:35
  • @BaronVT so value of m should be -1 – Sophie Clad Dec 19 '14 at 14:26
  • @BoronVT I have a question here, can't we say in general that this integral over any closed piecewise smooth curves are equal upto (up to the choice of clockwise vs counterclockwise) ? – Ibrahim Jan 29 '21 at 06:54