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Let $\gamma$ be the intersection between $z=x^2+y^2$ and the plane $z=1+2x$. Calculate the work done by the field $F=(0,x,-y)$ when the curve $\gamma$ traverses on lap in positive direction seen from z-axis.

I should solve this using stokes theorem. Setting both equations above equal we get $1+2x=x^2+y^2\iff 2=(x-1)^2+y^2$. Then I set $x=s \cos t+1$ and $y=s\sin t$ with $0\leq s\leq \sqrt{2}$ and $0\leq t\leq2\pi$. Then parametrization of the plane gives $r(s,t)=(s \cos t+1,s\sin t, 2+s\cos t)$.

Further, $r_s=(\cos t, \sin t,\cos t)$ and $r_t=(-s \sin t, s \cos t, -s \sin t)$ and $r_s \times r_t=(-s,0,s)$, also $\text{curl }F=(-1,0,1) $ so $\int\int_\Gamma \text{curl } F\cdot ndS=\int^{2\pi}_0\int^{\sqrt{2}}_0 2s dsdt=4\pi $. But the answer is $6\pi$. So how do I solve this without calculating the area of the ellipse given of the intersecting line?

Does not the parametrization I wrote give the surface which is enclosed by the line of the intersection?

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    $\begingroup$ Your plane paramterization is wrong. You don't have $z = 1+2x$ $\endgroup$
    – Ninad Munshi
    Commented Apr 12 at 0:21
  • $\begingroup$ you are right I worked it out it should be $r_s\times r_t=(-2s,0,s)$ $\endgroup$
    – per persson
    Commented Apr 12 at 1:23
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    $\begingroup$ Just another lesson about ease of computation: You incorporated polar coordinates right away, and got a headache of a pair of derivatives and a cross product for your trouble. On the other hand, notice that your jacobian vector was simply $(-2,0,1)s$, which, had you used the simpler parametrization $r=(x,y,1+2x)$ and then either the cross product or the immediate shortcut $\pm(z_x,z_y,-1)$ and swapped to polar coordinates$(s,\theta)$ after doing the dot product in the integrand, would have gotten you the same result for a fraction of the work. $\endgroup$
    – Ninad Munshi
    Commented Apr 12 at 5:41

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