#1

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?

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,*

#2

42.

#3

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.

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.

#4

lol i cant wait till i learn integration. looks like loads of fun

#5

Great fu

**cking job asshole. I'm trying to help him do math and you have to get this thread closed.**
#6

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

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

#7

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.

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.

*Last edited by darkstar2466 at Nov 13, 2007,*

#8

ok so how do you integrate t x sqrt (4t² + 1) ?

#9

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?

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?

#10

ohhhhh yeah!

cheers buddy

cheers buddy

#11

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)).

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)).

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