What are the dimensions and boundaries for different types of integrals?

[jaus tail]

Homework Statement

[/B]

Homework Equations

Substitution.

The Attempt at a Solution

Since the circle is of unit radius and around origin,
limits are x = -1 to 1, and y = -1 to 1

I replaced x by cos t, and y by sin t.
But what to put in place of ds?
I thought about divergence theorem, but then I'd have to find curl of F.
How to break ds in terms of dx + dy or dt?

 

[jaus tail]

I wrote as 2 parts
Total integral = blue part + red part
Blue part = x = -square root (1 - y²) . limits of y are -1 to +1
For red part x = +square root (1 - y²) . limits of y are +1 to -1
But how to resolve ds in terms of x and y.

I tried with ds = dx dy but didn't get answer

I'm getting answer as 2 * 3.14 divide by 3. But that's not in options. Book answer is A.
Is my ds = dx dy right or is ds = i dx + j dy?

 

[vela]

Use $ds^2 = dx^2+dy^2$. Also, it's probably easiest to do this integral after converting to polar coordinates.

 

[jaus tail]

But doesn't ds mean surface integral? So ds = dx multiplied by dy ?

 

[Orodruin]

No, ds is the curve length.

Use $ds^2 = dx^2+dy^2$. Also, it's probably easiest to do this integral after converting to polar coordinates.

Essentially, the curve parametrisation given does this for you as it is parametrising the circle using the polar angle.

 

[jaus tail]

Okay.
and for limits:
x = cos t
x = -1 to + 1
so t = 180 to 0 degrees

y = sin t
y = -1 to + 1
so t = -90 to + 90
I'm confused which limits to take.

Is the drawing above correct of ds region?

 

[Orodruin]

No. The integral is taken around the border of the circle. Not over a two-dimensional region. The curve parameter should be chosen such that you complete one full turn.

 

[jaus tail]

theta = positive angle. So i have theta = 0 to 180 and theta = -180 to 0?
x = r cos theta. y = r sin theta. This helps. r is given to be 1 in question.

 

[Orodruin]

So i have theta = 0 to 180 and theta = -180 to 0?

Why not just -180 to 180? Also note that the angle should be given in radians in order for $ds$ to be conveniently expressed in terms of it.

 

[jaus tail]

Okay that gives the answer. Thanks. But I still didn't understand why dS = line integral and not surface.
As per book:
dl = line, ds = surface, dv = volume.

 

[Orodruin]

Do not confuse capital and small letters. Also, it is not always the case that notation is unique between different references.

 

[jaus tail]

Okay. So:
had it been like: compute integral by region bounded by circle. Then it would've been surface integral, right?
And for volume integral it's region enclosed.

 

[Mark44]

Okay that gives the answer. Thanks. But I still didn't understand why dS = line integral and not surface.
As per book:
dl = line, ds = surface, dv = volume.

Here are the commonly used symbols:
ds -- increment of arc length
dV -- increment of volume
dS or dA - increment of area of a surface (the wikipedia article on Surface Integral uses dΣ)
As Orodruin already said, don't confuse uppercase and lowercase letters -- they are often used to mean different things.

The problem could have been stated more clearly, but a clue to their intent is "compute [the integral] around the circle..."

 

[WWGD]

But doesn't ds mean surface integral? So ds = dx multiplied by dy ?

The " Infinitesimal Length" depends on the parametrization, on the region over which you are integrating. Informally, dx=1dx, dy=1dy are infinitesimal straight-line displacement. Change of variable using parametrization "corrects" this.

 

[Ray Vickson]

Okay. So:
had it been like: compute integral by region bounded by circle. Then it would've been surface integral, right?
And for volume integral it's region enclosed.

Enclosed or bounded are the same: will either be area or volume, depending on what is the dimension (2D vs. 3D).

 

[WWGD]

In general, the boundary of a manifold is of dimension one higher than the manifold they are bounding.

 

[Orodruin]

In general, the boundary of a manifold is of dimension one higher than the manifold they are bounding.

Lower.

 

[WWGD]

Lower.

FALSE Statement: In general, the boundary of a manifold is of dimension one higher than the manifold they are bounding.
(Lower-enough Orodruin ;) ?)
EDIT: The boundary of a manifold is one dimension lower than the manifold it bounds.

 

[Mark44]

FALSE Statement: In general, the boundary of a manifold is of dimension one higher than the manifold they are bounding.
EDIT: The boundary of a manifold is one dimension lower than the manifold it bounds.

And the grammar is now correct, as well, with the pronoun agreeing in number with its antecedent.

 

[jaus tail]

Meaning?
For line integral, dimension is metre. Boundary dimension is also metre.
For surface integral, dimension is dx, dy so metre square. Boundary dimension is ??
Volumte integral, dimension is metre cube.

 

[Mark44]

Meaning?
For line integral, dimension is metre. Boundary dimension is also metre.

In a line integral, you're calculating the length of the curve, which could be meters, centimeters, feet, miles, whatever.

For surface integral, dimension is dx, dy so metre square. Boundary dimension is ??

The boundary of such a surface would be a one-dimensional figure, embedded in a space of higher dimension.
Also, the integral does not have to have dx and dy -- it could be in terms of polar coordinates $dr$ and $d\theta$.

Volumte integral, dimension is metre cube.