So i get pretty far on these, but I can't integrate so i must have gone wrong soemwhere.

So a simple one

Q.1 Integral along the Line C of y dS C : x = t², y = t, between 0 and 2

I get Integral between 0 and 2 of ( t Sqrt(t^4 + t2) dt )

and then I'm stuck.

Another Question

Q.2. Integral along the Line C of x²yz dS C : x = t^3, y = t, z = t between 0 and 1

and
Q3. How do you do Integral cos²tsint?
Last edited by henza_x at Nov 12, 2007,
42.
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You're doing the first one wrong. Given your parametrization that x = t^2 and y = t, and your integral bounds 0 and 2, you can do a simple integration

Integral 0 to 2 of t^2 + t dt.

I think that should be the answer according to the information you gave me. Your notation doesn't make sense to me too much. Remember a line integral has the form of

Integral over C of f(x,y,z) and you are already given what x and y are in terms of t, and without a given velocity field, that's all you can make of it.
Quote by denizenz
I'll logic you right in the thyroid.

Art & Lutherie
Great fucking job asshole. I'm trying to help him do math and you have to get this thread closed.
Quote by denizenz
I'll logic you right in the thyroid.

Art & Lutherie
First off, thanks alot for helping,

The first one, basically you're looking for the line integral over line C of the function "y" with those parameters.. so what I did was

y(t) ds/dt dt

with ds/dt = square root of ((dx/dt)² + (dy/dt)²)
= sqrt ( 4t² + 1 )

So then it becomes Integral between 0 and 2, y(t) becomes just t

= integral t sqrt (4t² + 1) dt

I think thats how its spose to go
Damn, I haven't done line integrals in a couple of quarters and I totally forgot how they're supposed to go.

Ok, so given a function r(t) = x(t) i + y(t) j + z(t) k, let's say r(t) in our case is t^2 i + t j.

Now the line integral Integral over C of f(x,y,z) ds = Integral from a to b of f(g(t),h(t),k(t)) |v(t)|dt.

|v(t)| = |dx/dt(t^2)^2 + dx/td(t)^2| = sqrt(4t^2 + 1).

So your line integral looks like: Integral from 0 to 2 of (t * sqrt(4t^2 + 1)) dt

Explaining it, integral of y dS equates to y being t and dS being sqrt(4t^2 + 1) dt. Integrate that from 0 to 2 using substitution and viola.

EDIT: In other words, yea you were right. Just use substitution.
Quote by denizenz
I'll logic you right in the thyroid.

Art & Lutherie
Last edited by darkstar2466 at Nov 13, 2007,
ok so how do you integrate t x sqrt (4t² + 1) ?
Remember substitution from way back when. Let u = 4t^2 + 1, then du = 8t dt and dt = du/8t.

Then your integral becomes: Integral over 0 to 2 of (t * sqrt(u) * du/8t). The Ts cancel and you get: Integral over 0 to 2 of 1/8(sqrt(u) du).

Ring a bell? Yes? No?
Quote by denizenz
I'll logic you right in the thyroid.

Art & Lutherie
ohhhhh yeah!

cheers buddy
Similarly, for #2, you get |v(t)| = (dx/dt(t^3))^2 + (dy/dt(t))^2 + (dz/dt(t))^2 which gives you sqrt(9t^4 + 2).

Your line integral is then: Integral from 0 to 1 of (x^2 * y * z * sqrt(9t^4 + 2) dt

which is Integral from 0 to 1 of (t^8 * sqrt(9t^4 + 2)).
Quote by denizenz
I'll logic you right in the thyroid.

Art & Lutherie