How do you find the equation of the curve of intersection of the surfaces
z = x^2 and x^2 + y^2 =1?How can I show that the curve with parametric equations x = sin t, y = cos t, z = sin^2 t is the curve of intersection of these two surfaces?Hello and thank you for posting your question to Brainmass.
The solution is attached below in two files. the files are identical in content, only differ in format. The first is in MS Word format, while the other is in Adobe pdf format. Therefore you can choose the format that is most suitable to you.Feedback is always appreciated.The first surface is:
This is a cylinder aligned with the z-axis. Its cross section is a unit circle centered at the origin.When we have a cylindrical symmetry it is useful to use polar cylindrical coordinates:
Where is the distance from the origin and is the angle between the radius vector to the point and the positive x-axis measured counterclockwise.
Thus, equation (1.1) becomes:
Hence, the set of all the points the face of the cylinder (locus) is:
The second surface is:
It is a flat surface curved to form a cross section of a symmetric parabola in the plane with a vertex in the origin.
It looks like this:The set of the points on the face on this parabolic surface is:

The intersection of these two surfaces look like:All the points on the intersection curve must satisfy both coordinates of the cylinder (equation 1.4) and that of the parabola (equation 1.6)
This gives:
From the first two coordinates we can write for the curve:
And from the third coordinate:
Thus, the parametric equation of the curve can be written as:
It looks like this:This would be the curve representation if we use the standard cylindrical coordinate system.
however, no one stops us from measuring the polar angle clockwise with respect to the y-axis:In this case the transformation is:
And the coordinate of a point on the cylinder is:
The coordinate of the parabolic surface does not change (equation 1.6) and we are left with the parametric coordinates of the curve:
Or simply (as before):

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