Integral Over CFds: Homework Statement & Eqns

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Homework Statement


For F = (x^2+y)i + (y-x)j, calculate the integral over C of Fds for r = (t, t^2) where t goes from 0 to 1.

Homework Equations

The Attempt at a Solution


I know the integral over C of fds is f*sqrt(r'(t)^2+r^2*theta'(t)^2) dt. But I have no theta in this question, is this the wrong integral?
 
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dylanhouse said:

Homework Statement


For F = (x^2+y)i + (y-x)j, calculate the integral over C of Fds for r = (t, t^2) where t goes from 0 to 1.

Homework Equations

The Attempt at a Solution


I know the integral over C of fds is f*sqrt(r'(t)^2+r^2*theta'(t)^2) dt. But I have no theta in this question, is this the wrong integral?

Yes, that's the wrong integral. This problem has nothing to do with polar coordinates. You have ##\vec r(t) =\langle x(t),y(t)\rangle = \langle t,t^2\rangle##. Write the integral in terms of ##t##.
 
So F would be (2t^2)i + (t^2 - t)j. And dr is just <1, 2t>dt. But how would I integrate with respect to ds if I end up with a dt?
 
dylanhouse said:
So F would be (2t^2)i + (t^2 - t)j. And dr is just <1, 2t>dt. But how would I integrate with respect to ds if I end up with a dt?

Your integral is stated with an Fds. I don't think that is an arc length integral, but then you haven't told us what that notation means. F is a vector. What is ds? I would assume you mean ##\vec F\cdot d\vec s## which might otherwise be written ##\vec F\cdot \hat T~ds## or ##\vec F\cdot d\vec r##. In any case I expect you would evaluate it as ##\int\vec F\cdot \vec r'(t)~dt##.
 
Prove $$\int\limits_0^{\sqrt2/4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx = \frac{\pi^2}{8}.$$ Let $$I = \int\limits_0^{\sqrt 2 / 4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx. \tag{1}$$ The representation integral of ##\arcsin## is $$\arcsin u = \int\limits_{0}^{1} \frac{\mathrm dt}{\sqrt{1-t^2}}, \qquad 0 \leqslant u \leqslant 1.$$ Plugging identity above into ##(1)## with ##u...
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