# Intersection of an arbirary plane and sphere

• Apr 18th 2011, 09:07 PM
TaylorM0192
Intersection of an arbirary plane and sphere
Hello,

Consider the following two equations:

1. ax + by + cz = 0.
2. x^2 + y^2 + z^2 = r^2

The curve of intersection is a circle centered along the z axis. Projection onto the xy-plane is an ellipse, and the circle itself is in a rotated plane.

But...does anybody here have a solution for deriving the parametric equations which describe this curve of intersection?

Furthermore, I'm also interested in deriving parametric equations for the surface which lies on this sphere, but has a boundary which is the curve of intersection first described. From what I can so far determine, it is a rotated hemisphere about the xy-plane with center about the z-axis.

I guess a more general question would be how to find new parametric equations for surfaces and curves rotated about the xy-plane?

Thanks ~
• Apr 19th 2011, 06:22 PM
TaylorM0192
Hmm...was hoping someone might be able to answer this by now! ;}

Well, here's a simpler question that is related to the more general case given above:

Compute the line integral of the tangential component of the vector field F: R3 -> R3 = yi-xj+3yk along the curve of intersection between the sphere x^2 + y^2 + z^2 = 9 and the plane ax + by + z = 0. Evaluate the line integral by direct evaluation and by the "Curl" Theorem.

The former method requires you to parametrize the circle of intersection (a "great circle" or rotated circle lying on the sphere/plane) and the ladder method requires you to parametrize the hemisphere bisected by the plane through the center of the sphere (a rotated half-sphere).

If someone could help me obtain the correct parametrizations, I would appreciate it! I would then probably be able to determine the more general case in the OP myself.

EDIT: For anyone who is attempting the problem, the answer is [9*pi (3a - 2)] / [sqrt(1 + a^2 + b^2)]
• Apr 20th 2011, 12:13 AM
xxp9
Find two orthogonal unit vectors v and w laying on the plane, then the intersection can be parameterized as c(t)= r[cos(t)v + sin(t)w].
To find v and w, extend n=(a,b,c) to a base of R^3, then use the "Gram-Schmidt process" to get orthogonal unit vectors.
• Apr 20th 2011, 07:51 AM
TaylorM0192
I am not interested in parametrizing the curve with respect to a 2-dim basis on the intersecting plane; I want it with respect to the standard orthonormal 3-dim Euclidean space (ijk). I also need to parametrize this surface with the boundary that is the curve as well.

No one can at least do the problem in post #2? Evidently (according to the textbook I pulled it out of) it's suppose to be quite easy with the use of Stoke's Theorem (hence the need to find the parametric representation of that surface).
• Apr 20th 2011, 01:58 PM
HallsofIvy
Then what in the world do you want? You can't write a single equation in x, y, and z for any curve in three space. Three space is, of course, three dimensional. Requiring that x, y, and z satisfy a single equation only reduces the "degrees of freedom" by 1: 3- 1= 2 so any figure described by a single equation is two-dimensional- a surface. You can write a curve, in three space, by two equations. That way the "degrees of freedom" are reduced by 2: 3- 2= 1- a one dimensional figure or curve. But you already have that: ax+ by+ cz= 0 and x^2+ y^2+ z^2= r^2 are two such equations and probably the simplest.

The most common way to represent a curve, a one dimensional figure, in three space is to use parametric equation- by writing x, y, and z in terms of a fourth variable so that we have four degrees of freedom but reduced by the three equations to 4-3= 1 dimension.
• Apr 20th 2011, 03:39 PM
TaylorM0192
I need the PARAMETRIZED version of the intersection curve -- e.g., r(t) = x(t)i + y(t)j + z(t)k for some parameter domain I: a < t < b. Two equations in variables x, y and z would not be helpful; as you pointed out, they are already given. The challenging part, is determining a useful parametrization which describes this.

Similarly I need the PARAMETRIZED version of the open surface whose boundary is this curve given by r(t) -- e.g., s(u, v) = x(u, v)i + y(u, v)j + z(u, v)k for some parameter domain D: a < u < b and c < v < d.

What it seemed like xxp was suggesting was to parametrize the curve with respect to a new vector space, whose basis vectors span the intersecting plane; i.e. ignore this rotation -- but this again is not helpful because I would have to somehow rewrite (eventually) the vector fields that I integrate over this curve in terms of these new basis, which in itself might be quite challenging.
• Apr 20th 2011, 06:09 PM
xxp9
My post is the answer you're looking for. Note that v and w are vectors in the 3-dimensional space. So c(t)= r[cos(t)v + sin(t)w] has actually 3 components on x, y and z.
For the area, use the same trick: s( \rho, \theta ) = \rho ( cos\theta v + sin\theta w ), where \rho in [0, r] and \theta in [0, 2\pi].
Write the two constant vectors v and w in terms of i, j and k you'll get the result.