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I have quoted the question. This assignment is past due and I have questions about the solution:

Computer the path integral of $\int_C{f} \, ds $ where $f(x,y,z)= x^2$ and the path C is the intersection of the sphere $x^2+y^2+z^2=1$ and the plane $x+y+z = 0$.

So the way I see it is that the intersection of the sphere and curve gives us a circle on the xy plane with radius one. So i thought the parametrization is as simple as $x = cos(t)$ and $y=sin(t)$ but it's not. Here is the correct solution.

http://imgur.com/Mcrsq

How is it that they are parametrizing using vectors. What is the reasoning/logic behind it?

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$x^2$+$y^2$+$z^2$ equals what? I am confused by the equation of the sphere. –  analysisj Nov 12 '11 at 21:08
    
sorry, it equals one. I will edit. –  Tyler Hilton Nov 12 '11 at 21:10
    
Have you considered spherical coordinates? –  analysisj Nov 12 '11 at 21:19
    
The circle is not in the x-y plane. The plane contains the point $(1,1,−2)$, e.g. So the circle contains the point ${1\over \sqrt6}(1,1,-2)$. –  David Mitra Nov 12 '11 at 21:29
    
even so, shouldnt the parametrization be the same? –  Tyler Hilton Nov 12 '11 at 21:30

1 Answer 1

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Looking at the solution provided in your link, you're asking "how do I parameterize a circle in $\Bbb R^3$".

If the circle has radius $r$, is centered at the origin, and contains the points $p_1=(x_1,y_1,z_1)$ and $p_2=(x_2,y_2,z_2)$ which, when thought of as vectors, have norm one and are orthogonal, a vector parameterization is ${\bf x}(t)= r p_1 \cos t +r p_2\sin t. $


To see that this gives a circle:

Any linear combination of $p_1$ and $p_2$ (thinking of them as vectors) lies in the plane determined by $p_1$ and $p_2$. Using the fact that $p_1$ and $p_2$ are orthogonal, it follows that $r p_1 \cos t + r p_2\sin t$ has norm r. Indeed $$ || r p_1 \cos t +r p_2\sin t||^2= r^2\cos^2 t ||p_1||^2+ r^2\sin^2 t ||p_2||^2=r^2(\cos^2t+\sin^2t)=r^2 $$

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thanks. I guess my answer was "close enough" that it was specific to xy plane and the unit circle. However now that we are not in the xy plane (but still have a unit circle in the plane) we get a different parametrization. –  Tyler Hilton Nov 12 '11 at 21:46

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