Why is the path not on the xz plane? Stoke's theorem

[flyingpig]

Homework Statement

Okay I am given a vector field (which is irrelevant to my question) and a bounded surface.

S is part of the cone $$y^2 = x^2 + z^2$$ that lies between the planes y = 0 and y = 3, oriented in the direction of the positive y-axis.

The Attempt at a Solution

My book and I both evaluated it to a line integral. But my book's path was r(t) = <3sint, 3, 3cost>

while my path was r(t) = <3cost,0,3sint>

My questions

1. Why do they have sin and cos swap the other way? Isn't it a cone with the y-axis as its "major"?

2. If I want to make my life difficult and project it on the xy-plane, I will have to use hyperbolic substitutions namely zsinh²(t) + zcosh²(t) = z

3. Why is the path on y = 3? Why can't I project this on the xz-plane?

 

[tiny-tim]

hi flyingpig!

My book and I both evaluated it to a line integral. But my book's path was r(t) = <3sint, 3, 3cost>

while my path was r(t) = <3cost,0,3sint>

My questions

1. Why do they have sin and cos swap the other way? Isn't it a cone with the y-axis as its "major"?

think outside the sty …

any parameter will do … your t is their π/2 - t … either is valid

2. If I want to make my life difficult and project it on the xy-plane, I will have to use hyperbolic substitutions namely zsinh²(t) + zcosh²(t) = z

you mean y = zsinht, x = zcosht, z = t?

yes

3. Why is the path on y = 3? Why can't I project this on the xz-plane?

difficult to answer without seeing the original question , but I'll guess it asks for the flux through a surface bounded by a circle at y = 3, not y = 0

 

[Disconnected]

The cone is facing down the way, right? If you want to integrate it as a line integral with Stoke's theorem remember that this must be done on the curve that bounds the surface. This curve is at y=3, right?
You can't project it to the x/z plane because the vector field will have completely different values at y=0 then at y=3.

 

[flyingpig]

This is going to sound like a really really strange question, but why couldn't the curve enclosed be the dot at the origin?

y = zsinht, x = zcosht, z = t?

Why does z = t?

 

[tiny-tim]

Why does z = t?

oops!

should be z = constant ​

 

[HallsofIvy]

Homework Statement

Okay I am given a vector field (which is irrelevant to my question) and a bounded surface.

S is part of the cone $$y^2 = x^2 + z^2$$ that lies between the planes y = 0 and y = 3, oriented in the direction of the positive y-axis.

The Attempt at a Solution

My book and I both evaluated it to a line integral. But my book's path was r(t) = <3sint, 3, 3cost>

while my path was r(t) = <3cost,0,3sint>

You mean r(t)= <3cos t, 3, 3sin t>, right?

My questions

1. Why do they have sin and cos swap the other way? Isn't it a cone with the y-axis as its "major"?

2. If I want to make my life difficult and project it on the xy-plane, I will have to use hyperbolic substitutions namely zsinh²(t) + zcosh²(t) = z

3. Why is the path on y = 3? Why can't I project this on the xz-plane?

 

[Disconnected]

This is going to sound like a really really strange question, but why couldn't the curve enclosed be the dot at the origin?

You cannot perform a line integral at a point, Stoke's theorem can only be used for simple curves bounding a surface. This implies that the surface is closed on all but one side, and the simple curve bounds at that side. The curve is closed at y=o, but no closed at y=3, thus the point at y=0 does not bound the surface.

... I hope that made sense...

 

[flyingpig]

I don't understand a dot is still a path, but just very very small

 

[vela]

It doesn't enclose the entire area you're integrating over.

 

[tiny-tim]

… why couldn't the curve enclosed …

the curve isn't enclosed …

the curve does the enclosing …

you choose the surface you want the flux through, and its boundary is the curve …

if you choose a surface whose boundary is a point, then the surface is a closed surface minus one point, and the flux through it will be the same as through the whole closed surface

 

[Disconnected]

I don't understand a dot is still a path, but just very very small

But if you take the point as the boundary then the surface is not closed! There is a great whopping hole in it at y=3! The surface must be closed and bounded by the curve.

That is to say, the surface that is within the bounds of the curve must be continuous. It must be as a rubber sheet, stretched in any way but with the edge of the sheet bounding it.

 

[tiny-tim]

But if you take the point as the boundary then the surface is not closed! There is a great whopping hole in it at y=3! The surface must be closed and bounded by the curve.

not following you

the surface doesn't have to be closed, and usually isn't closed

a surface that isn't closed has a boundary: that boundary is the curve

 

[Disconnected]

not following you

the surface doesn't have to be closed, and usually isn't closed

a surface that isn't closed has a boundary: that boundary is the curve

Sorry, I should have said that the surface within the boundary of the curve is closed. Stoke's theorem doesn't work on closed surfaces. Thanks