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While going through an exercise of surface integration, I got confused in this problem.The surface is the intersection of sphere $S:x^2+y^2+z^2-1=0$ and the plane $P:y-x=0$. Clearly, the curve of intersection is the circle $S=0,P=0$ whose projection on $XZ$ or $YZ$ plane is an ellipse. It seems that it can be obtained by substituting $y=x$ in $S=0$ to get $2x^2+z^2=1$ (point of confusion) but I don't think this is the projection as the calculations doesn't support the geometry.

Kindly correct me what is wrong in my approach. Thanks

Nitin Uniyal
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  • Yes, this is the correct projection. I didn't quite understand the problem though. Are you trying to do a line integral or surface integral? And on what function? – KittyL Jan 01 '16 at 14:33
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    It is a circle, but only when viewed in the direction perpendicular to the plane surface. You are looking at the circle along the y-axis and viewed from that direction it is an ellipse in the x, z plane. – Paul Jan 01 '16 at 14:35
  • @Paul Did you mean that S=0,P=0 represents an ellipse here? – Nitin Uniyal Apr 05 '16 at 15:59
  • The projection onto the XZ plane (or the projection in the direction of the Y axis) IS an ellipse - the one you wrote down. The projection of the sphere onto the plane P (or the projection in the direction normal to the plane) IS a circle. You can think of it as the circle in the XZ plane, $x^2+z^2=1$ rotated through $45^0$ about the Z axis. – Paul Apr 06 '16 at 13:29

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