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Calculate the line integral directly without Stokes' theorem: \begin{gather*} \oint_\gamma \mathbf{F} \cdot d\mathbf{r} \end{gather*} \begin{gather*} \mathbf{F}(x,y,z)=(2z-3y) {\hat{\mathbf{i}}} + (3x-z){\hat{\mathbf{j}}} + (y-2x){\hat{\mathbf{k}}} \end{gather*} $\gamma$ is the intersection of the sphere \begin{gather*} x^2+y^2+z^2=1 \end{gather*} and the plane \begin{gather*} x+y+z=0 \end{gather*} Solution: First the intersection: \begin{gather*} z=\pm \sqrt{1-x^2-y^2} \\ z=-x-y \end{gather*} Solve with the positive square root: \begin{gather*} \sqrt{1-x^2-y^2} =-x-y \\ 1-x^2-y^2=(-x-y)^2 = x^2+y^2+2xy\\ \iff \\ 2x^2+2y^2+2xy=1 \end{gather*} And here I'm stuck because I can't eliminate x or y in the last equation ...

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  • $\begingroup$ The intersection is a curve. You won't be able to eliminate two out of three variables. $\endgroup$ – David Peterson Jun 26 '16 at 18:01
  • $\begingroup$ Okey, so it's not possible to parametrize the intersection? Or should I parametrize $2x^2+2y^2+2xy=1$? $\endgroup$ – JDoeDoe Jun 26 '16 at 21:20
  • $\begingroup$ Sure it is. In the $xy$ plane it's an ellipse, rotated $\pi/4$ radians. Start with the ellipse $x^2/3+y^2=1$ and hit it with a transformation matrix. See wolfram alpha $\endgroup$ – David Peterson Jun 26 '16 at 22:35
  • $\begingroup$ Thanks! Btw, how did you find the rotation $\pi/4$ radians? $\endgroup$ – JDoeDoe Jun 29 '16 at 10:15

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