finding the mass of a wire

Question

Let C be the line segment on the xy-plane from (−4, 3) to (1, −9).

(a) Evaluate ∫_C 2xy dy.

(b) Use a line integral to find the mass of a wire in the shape of C if the density (mass per unit length) at any point (x, y) on the wire is δ(x, y) = 3x² + 1 g/cm

Answer 1

Hey Nina!

Let me answer part a) for you and once I do, you'll see the identically procedure is used for part b) just with a different density function.

We can approach this by parameterizing our density function and our path by t. To do this, let's find a parametric form of the line segment starting from (-4,3) to (1,-9).

The general parametric form of this equation is

r(t) = p_o(1-t) + p_f(t) 0 ≤ t ≤ 1

Where p_o is our initial point and p_f is our final point. Using this equation with our points, we have

r(t) = <-4,3>(1-t) + <1,-9>t = <-4 + 4t, 3 - 3t> + <t, -9t> = <-4 + 5t, 3 - 12t>

Let's go back to our integral for a second and manipulate it slightly so it's more obvious what form it takes when we introduce the parameter.

∫2x(t)y(t)(dy/dt) dt 0 ≤ t ≤ 1

I should note, we simply expanded dy into (dy/dt)dt

Okay so now we know the form our integral takes, what we need to realize now is our parametric line segment equation actually gives us parameterizations for x as a function of t and y as a function of t, it's exactly the x and y components of the equation r(t). Because by definition

r(t) = <x(t),y(t)>

Equating this to our line segment equation where

r(t)= <-4 + 5t, 3 - 12t>

Therefore

x(t) = -4 + 5t and y(t) = 3 - 12t

We also need (dy/dt) for our equation, so simply taking the derivative with respect to t we have

dy/dt = -12

Plugging this into our integral we get

2∫(-4 + 5t)(3 - 12t)(-12) dt 0 ≤ t ≤ 1

Expanding this and moving the constant -12 out of the integral, we get

(-24)∫(-12+63t-60t²) dt 0 ≤ t ≤ 1

Now this is simply integration

(-24)(-12t+31.5t²-20t3) 0 ≤ t ≤ 1

Plugging in the bounds we have

(-24)(-12 + 31.5 - 20) = 12

Okay so to reiterate, the approach is to

1. Construct the line segment parameterization.
2. Use the parameterization to find your variables x,y,z as functions of that parameter.
3. Substitute these into your integral which will now be entirely in terms of that parameter t.
4. Integrate!