0
$\begingroup$

F$(x,y) = (2x^2+2xy, 2y^2+2xy)$ where the curve is $f(\theta) = \theta, \theta \in [0,\pi/2]$.

So i attempted to convert the curve into cartesian parametric coordinates so that

$$\gamma = (\theta cos(\theta), \theta sin(\theta))$$

Hence F$(x,y) = (2\theta^2 cos^2(\theta) + 2\theta^2 cos(\theta)sin(\theta), 2\theta^2 sin^2(\theta) + 2\theta^2 cos(\theta)sin(\theta))$

and $\gamma' = (-\theta sin(\theta), \theta cos(\theta)) $

However when i attempt compute the line integral i get $$\int_0^{\pi/2} F dr = 0$$

and the answer in my book says the answer is $ \approx 3.218$

  • 1
    Using the letter $\;\theta\;$ doesn't change, imo, the fact that $\;f(\theta)=\theta\;$ is the indentity map, and thus the curve is the straight line $\;y=x\;,\;\;0\le x\le\frac\pi2\;$ ...unless some different assumptions or conditions are given. I thus can't understand your parametrization of $\;\gamma\;$ . – 2017-01-06
  • 0
    Thats what i originally tried to do, but i couldnt get the correct answer. I believe the use of $\theta$ implies our lecturer wants us to convert to cartesian. It did state that $p(\theta) = \theta$ in the question, but i guessed this doesnt matter .. although having quickly googled, a lot of people use $p(\theta)$ to refer to polar – 2017-01-06
  • 0
    BTW, are you sure that's the correct answer to the correct exercise in your book? – 2017-01-06
  • 0
    @DonAntonio yes positive – 2017-01-06
  • 0
    Well, using my idea I get a rather different answer (something like thrice what you say), so I'd say some assumption on polar coordinates must be assumed... – 2017-01-06
  • 0
    Surely the function $f(\theta) = \theta)$ is not equivalent to $y = x$... Since the line $y = x$ has the polar equation $\theta = \frac{\pi}{4}$ @DonAntonio – 2017-01-06
  • 0
    It looks like that formula, if indeed given in polars, is not defined correctly. – 2017-01-06
  • 0
    @Alex I think it is correct is we assume polar, and it is then a spiral. But then $\;\gamma'(\theta)\;$ is *not* what is written in the question...not even close. – 2017-01-06

1 Answers 1

0

Imo, the parametrization must be

$$\gamma(t)=(t,t)\;,\;\;t\in\left[0,\,\frac\pi2\right]\;\;,\;\;\;\gamma'(t)=(1,1)\implies$$

$$\int_\gamma (2x^2+2xy,\,2y^2+2xy)\cdot d\vec r =\int_0^\frac\pi2 (2t^2+2t^2,\,2t^2+2t^2)\bullet(1,1)dt=$$

$$\int_0^{\frac\pi2}8t^2\,dt=\frac83\left(\frac\pi2\right)^3$$

But then we get something very different of what you say is the answer , so we must assume the curve is in polar coordinates and thus...do basically the same you did! Though I think there is a mistake in the differentiating the curve:

$$\gamma(\theta):=\left(\theta\,\cos\theta,\,\theta\,\sin\theta\right)\;,\;\;0\le\theta\le\frac\pi2\;,\;\;\gamma'(\theta)=(\cos\theta-\theta\,\sin\theta,\,\sin\theta+\theta\,\cos\theta)\implies$$

$$\int_\gamma (2x^2+2xy,\,2y^2+2xy)\cdot d\vec r=\int_0^{\frac\pi2}F(\gamma(\theta))\bullet\gamma'(\theta)\,d\theta$$

The last expression is really nasty though not actually very hard, but still it will take a loooong while to write it down and solve it...at leat to me.

  • 0
    If it is a spiral then we'll need some function relating $r$ to $\theta$ but it's not given @DonAntonio – 2017-01-06
  • 0
    @Alex Sorry, I thought that was clear from the question and the context: it is $\;r(theta)=\theta\;$ , what the OP called $\;f(\theta)\;$ , which is a spiral in polar coordinates. I shall edit. BTW, according to WA, the integral's value is $\;\approx 3.0127\;$ – 2017-01-06
  • 0
    Well putting the integral into wolfram gives the correct answer. Thankyou! I see that it was my differentiation that was wrong, i thought $\theta$ was a constant..ahh – 2017-01-06
  • 0
    https://www.wolframalpha.com/input/?i=integral+of+(cos(r)+-+rsin(r))(2r%5E2cos%5E2(r)+%2B+2r%5E2cos(r)sin(r))+%2B+(2r%5E2sin%5E2(r)+%2B+2r%5E2cos(r)sin(r))(sin(r)%2Brcos(r))+from+0+to+pi%2F2 – 2017-01-06
  • 0
    Ah yes, that makes sense now @DonAntonio – 2017-01-06
  • 0
    @Conor Great...and thus your interpretation as polar was right, but the lack of this data and the notation were a little misleading. – 2017-01-06