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Using Stokes theorem calculate the integral:$\int_C x^2y^2 dx+dy+zdz$ where C is curve bounded with $x^2+y^2=2x, y\ge 0, z=1$ oriented from $A(2,0,1)$ to $B(0,0,1)$. So using Stokes formula I should bring this down to solving surface integral od second kind. After finding partial derivates od $P=x^2y^2, Q=1, R=z$, I need to calculate normal vector, and then scalar product of field F and normal n. But not sure how to use given orientation, what is it telling me?

Trevor
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Vector field is $\vec{F} = (x^2y^2,1,z)$. The curve is bounded by $x^2+y^2=2x, y\ge 0, z=1$ and you are going from point $A(2,0,1)$ to $B(0,0,1)$.

You can rewrite the surface as $(x-1)^2+y^2 = 1$ which is a cylinder of radius $1$ with center at $(1,0,z)$. Now coming to the curve, it is a semicircle on this surface from point $A$ to point $B$ at $z=1$. Please note you must go anti-clockwise from $A$ to $B$ as $y \geq 0$.

You can parametrize your curve as
$r(t) = (2\cos^2 t, 2\cos t \sin t, 1) $ or $(1+\cos2t, \sin2t,1), 0 \leq t \leq \frac{\pi}{2}$.

$r'(t) = (-2\sin2t, 2\cos 2t,0)$

$\vec{F} \cdot r'(t) = (\sin^2 2t (1+\cos 2t)^2,1,1) \cdot (-2\sin2t, 2\cos2t,0)$

$ = -2\sin^32t(1+\cos 2t)^2 + 2\cos2t$

Line integral is $ \ \displaystyle \int_0^{\pi/2} \vec{F} \cdot r'(t) \ dt$

Math Lover
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  • Do you mind explaining this part with orientation, not sure what is it telling. – Trevor Jan 30 '21 at 10:54
  • You mean why anticlockwise? – Math Lover Jan 30 '21 at 10:55
  • Yes. Not sure how am I looking that on picture. – Trevor Jan 30 '21 at 10:55
  • Got it. And is it correct to solve this without parametrization, precisely can i calculate this integral: $-2\iint_C x^2ydxdy$ and get bounds for $\phi, r$ using polar coordinates. – Trevor Jan 30 '21 at 11:05
  • I understood the part with anticlockwise direction, and that is telling me only that we have $+$ in front of integral right? – Trevor Jan 30 '21 at 11:06
  • Yes that is correct. How did you get to that integral? I am not clear. You are not supposed to take derivative of the vector field as you cannot apply Green's theorem in 3D. You need to do dot product of the vector field with $r'(t)$ and hence the parametrization. – Math Lover Jan 30 '21 at 11:11
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    That parametrization is very simple which you have done before. In polar coordinates, this circle is written as $r = 2 \cos \phi$ so $x = r\cos\phi = 2\cos^2\phi$. Similarly for $y$. As the curve (See both points A and B) is at $z=1$, we substitute $1$ constant for $z$. That is all. When you differentiate, $z$ component will become zero. Is it clear? – Math Lover Jan 30 '21 at 11:15
  • Yes it is, somehow misunderstood the formula. I think it is OK now. Can you just write expression for the integral after dot product, to check? – Trevor Jan 30 '21 at 11:18
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    OK just added that – Math Lover Jan 30 '21 at 11:39
  • Could we use parametrization $x=1+\cos t, y=\sin t$ it still gives equality? Forgot to ask why you set $2t$ instead of $t$ – Trevor Feb 01 '21 at 17:04
  • In vector field, I am just plugging in the values from the parametrization. You may recall that in polar coordinates, $x^2+y^2=2x$ translates to $r = 2 \cos\theta$ so $x = r\cos\theta = 2\cos^2\theta = 1 + \cos 2\theta$ and similarly $y$. – Math Lover Feb 01 '21 at 17:26
  • Yes you can definitely parametrize as $x = 1 + \cos t, y = \sin t$ but then given $y \geq 0$, $0 \leq t \leq \pi$ instead of $0 \leq t \leq \frac{\pi}{2}$. You can choose whichever works out easier. – Math Lover Feb 01 '21 at 17:27