Jtdylktuy

If I had a region bounded by the surface $x^2 + y^2 = 9$ and a plane $x+y+z=1$, how would I determine what "boundary" I'd take the line integral over? Or would I take it over both boundaries?

The first being the circle of radius 3 with centre at the origin on the x-y plane

and the second, being the "ellipse-looking" boundary at the top (so the equation that comes out when we solve those two simultaneously)
?

Image of the region (so that small wedge above the x-y plane and below the surface plane): http://i.imgur.com/lim8gSZ.png?
Website used: https://www.geogebra.org/3d

I think there's a couple issues here. First, there is no region bounded by the cylinder $x^2+y^2=1$ and the slanted plane $x+y+z=1$. The two surfaces split $Bbb{R}^3$ into 4 unbounded regions:

• inside the cylinder and above the plane


• inside the cylinder and below the plane


• outside the cylinder and above the plane


• outside the cylinder and below the plane

Including the $xy$-plane $z=0$ creates a total of 10 regions, two of which are bounded. Adding the last condition that $zgeq 0$ finally clears up the region in question. Before we can even discuss what the boundary is, we need to have the region well-defined, so I will assume that the set we are referring to is

$$ Omega = {(x,y,z)inmathbb{R}^3 | (x^2+y^2leq 1)wedge(x+y+zleq 1)wedge(zgeq 0) } $$

From here, the components that make up the boundary are clear, and it brings up another issue. We are discussing a three-dimensional space, so we should have a two-dimensional boundary, which is consistent with what he have here. However, it is not consistent with use of the (Kelvin-)Stokes' Theorem, which is used on oriented two-dimensional surfaces with one-dimensional boundary. Without seeing the integral you intend to solve with this method, there's no way to answer the question, but this region that you seem to be defining isn't "compatible" with that form of Stokes' theorem (in the sense that the $int_Omega text{curl}Fcdot text dS = 0$ for any closed surface $Omega$).

Below, I will show how to find the boundary of the set $Omega$ independent of its use in integration.

So, the circle and ellipse curves will not be the entire boundary, since they are 1-dimensional curves. Nonetheless, they are part of the boundary. Just like the boundary of a solid cube is made up of faces (two-dimensional solid squares), edges (one-dimensional curves), and vertices (zero-dimensional points), we have the same situation for our solid here.

The faces are the truncated circular disk base, the truncated elliptical disk ceiling, and the cylindrical wall. We can get these two-dimensional boundary pieces by changing one inequalities in the definition of $Omega$ to an equality, and changing the rest to strict inequalities. So, we get the base by restricting our points to those in the $xy$-plane:

$$partialOmega_{text{base}} = {(x,y,z)inmathbb{R}^3 | (x^2+y^2< 1)wedge(x+y+z< 1)wedge(z= 0) }$$

Note that we let $z=0$, so we could have written $x+yleq 1$ explicitly for the second inequality. Similarly, we get the ceiling by restricting our points to those in the slanted plane:

$$partialOmega_{text{ceiling}} = {(x,y,z)inmathbb{R}^3 | (x^2+y^2< 1)wedge(x+y+z = 1)wedge(z> 0) }$$

and we get the wall by restricting our points to those on the cylinder:

$$partialOmega_{text{wall}} = {(x,y,z)inmathbb{R}^3 | (x^2+y^2= 1)wedge(x+y+z < 1)wedge(z> 0) }$$

Next, we can do a similar process to get the different 1-dimensional edge pieces by setting two inequalities to equalities at a time. This will give us three different edge pieces: the circular edge, the elliptical edge, and the linear edge. Explicitly,

$$partialOmega_{text{circular}} = {(x,y,z)inmathbb{R}^3 | (x^2+y^2= 1)wedge(x+y+z < 1)wedge(z= 0) } \
partialOmega_{text{elliptical}} = {(x,y,z)inmathbb{R}^3 | (x^2+y^2= 1)wedge(x+y+z = 1)wedge(z> 0) } \
partialOmega_{text{linear}} = {(x,y,z)inmathbb{R}^3 | (x^2+y^2< 1)wedge(x+y+z = 1)wedge(z= 0) }$$

Lastly, set all three inequalities to equality and you will get the two vertex points

$$partialOmega_{text{vertices}} = {(x,y,z)inmathbb{R}^3 | (x^2+y^2= 1)wedge(x+y+z =1)wedge(z=0) }$$

We can construct the full boundary by taking the union of these pieces:

$$partialOmega = bigcuplimits_{alpha} partialOmega_alpha $$

Note that this is a disjoint union; due to the use of strict inequalities mixed with equalities, it should be clear that no element of one piece could possibly be an element of another. This technique works in general for sets defined by inequalities like this, but you may find some pieces to be empty sets (such as the intersection of parallel faces of a cube).

You would take the line integral over each boundary separately then add the two results together. Be sure to orient the curves in the correct way so that the both correspond to the outward normal. In this instance the curve on the xy plane would be counterclockwise as viewed from above and the "ellipse-looking" boundary would be oriented clockwise. This is how we learned to approach it in my multivariable class but have only used it with cylinders.