Line integral, wire problem?

Question

Find the mass of a wire wiht the shape of the curve y=x^2 between (-2,4 ) and ( 2,4) if the density is given by δ(x,y)=k|x|

Answer 1

goldenflaw (interesting name) gives you correct advice, but he ignores the first sentence of his advice, in that he did not parameterize the length of the curve.

Here's my suggested solution:
Let dt be a differential change in the length of the curve.
Then dt^2 = dx^2 + dy^2, and dt = sqrt(dx^2 + dy^2).

y = x^2
dy/dx = 2x
dy = 2x dx

So dt = sqrt (dx^2 + 4x^2 dx^2)

The mass is Int (for x = -2 to 2) of (k|x| dt)
= Int (k|x| sqrt(dx^2 + 4x^2 dx^2)
= k Int(|x| sqrt(1 + 4x^2) dx)

Since the function is symmetrical with respect to x = 0, we can just evaluate it from 0 to 2 and double the result. That way we can replace the absolute value function |x| with x, since both are positive and equal from 0 to 2.

So we now have:
2k Int(x sqrt(1 + 4x^2)dx) from 0 to 2

Integrating, we have:
2k ( (1 + 4 x^2)^(3/2) / 12) evaluated from 0 to 2
= (k / 6) (17^(3/2) - 1)
= (k / 6) (17 sqrt(17) - 1)

And that's my proposed answer.
Hope that helps.

Answer 2

First you parametrize your curve. let's say with t. So you start at (-2,4) and end at (2,4) so -2 < t < 2. You get the integral from -2 to 2 of \delta(x,y) * x^2 dx = \int_{-2}^2 k|x|*x^2. = 2 \int_0^2 kx^3 = 2 kx^4/4 |0 to 2, thus, k2^5/4=8k. EDIT: Oops, this is wrong. See next comment. Sorry.