[calculus] question about identify boundary curve between two surface

[kougou]

Homework Statement

I have two questions.
1) generally speaking, when we are given two equations both describing surface in R3:
f1(x,y,z)=k
and f2(x,y,z)=C,

The intersection of the two will be a curve that's by solving both equations. My question is, by solving f1 and f2 to get anther equation say f3, does f3 iteself enough to describe the intersection curve, OR do we need f3 AND one of f1 or f2 to describe the intersection? I am not talking about very complicated or special case situation.

I ask this question because we need it for computing surface integral or line integral using stroke's theorem. The first step of identifying the boundary is critical

 

[LCKurtz]

Homework Statement

I have two questions.
1) generally speaking, when we are given two equations both describing surface in R3:
f1(x,y,z)=k
and f2(x,y,z)=C,

The intersection of the two will be a curve that's by solving both equations. My question is, by solving f1 and f2 to get anther equation say f3, does f3 itself enough to describe the intersection curve, OR do we need f3 AND one of f1 or f2 to describe the intersection? I am not talking about very complicated or special case situation.

I ask this question because we need it for computing surface integral or line integral using stroke's theorem. The first step of identifying the boundary is critical

That's a very general question and I don't think there is a single answer to it. I don't know what you mean by "solving $f_1$ and $f_2$ to get $f_3$". This type of problem often requires an appropriate parameterization or perhaps a piecewise description of the boundary. What method to use depends a lot on what particular equations you have, and you usually start with a good picture. Here's an example. Say your surfaces are the slanted plane $x+y+z=4$ and the cylinder $x^2+y^2 = 1$. You would normally parameterize the cylinder in cylindrical coordinates $\vec R(\theta,z) = \langle \cos\theta,\sin\theta,z\rangle$. Here $\theta$ takes you around and $z$ locates you vertically. If you want $z$ to be on the plane you could solve it for $z$: $z=4-x-y$, which, in terms of $\theta$ is $z=4-\cos\theta-\sin\theta$. Putting that together gives the intersection curve in terms of $\theta$:$$
\vec C(t) = \langle \cos\theta,\sin\theta,4-\cos\theta-\sin\theta\rangle$$where $0\le\theta\le 2\pi$. Different surfaces might require an entirely different approach.

 

[kougou]

That's a very general question and I don't think there is a single answer to it. I don't know what you mean by "solving $f_1$ and $f_2$ to get $f_3$". This type of problem often requires an appropriate parameterization or perhaps a piecewise description of the boundary. What method to use depends a lot on what particular equations you have, and you usually start with a good picture. Here's an example. Say your surfaces are the slanted plane $x+y+z=4$ and the cylinder $x^2+y^2 = 1$. You would normally parameterize the cylinder in cylindrical coordinates $\vec R(\theta,z) = \langle \cos\theta,\sin\theta,z\rangle$. Here $\theta$ takes you around and $z$ locates you vertically. If you want $z$ to be on the plane you could solve it for $z$: $z=4-x-y$, which, in terms of $\theta$ is $z=4-\cos\theta-\sin\theta$. Putting that together gives the intersection curve in terms of $\theta$:$$
\vec C(t) = \langle \cos\theta,\sin\theta,4-\cos\theta-\sin\theta\rangle$$where $0\le\theta\le 2\pi$. Different surfaces might require an entirely different approach.

Thank you
"by solving f1 and f2, to get f3"
what I meant is to get another equation f3, from f1 and f2. And f3 describes the values of the intersection

 

[Bacle2]

I assume you mean solving both f1, f2 for the same variable , and setting the equations equal to each other. Since surfaces are 2-dimensional, you should find a representation in one variable after setting the equations equal to each other. Think of a very simple case: the XZ plane and the YZ planes, intersecting in a curve. But it can get more complicated if your surfaces have volume, as in the intersection of spheres.