- 1). Create a line integral, using a "very small" or differential length "s or ds times the function, giving the density of the wire at different positions f(x,y).
For example, suppose the density function is f(x,y)=4xy. The integrand which will be integrated will then be 4xy ds. - 2). Write ds in terms of a parameter t that relates x and y. Suppose you are given that the vector r(t)=x(t)i+y(t)j, where r, i and j are vectors, traces out the curve of the wire. (i and j are the unit vectors in the x- and y-directions.) Note that a differential length in the x-y plane is √(dx^2+dy^2). This can be rewritten to take the differentials out from under the radical sign as follows: dt√[(dx/dt)^2+(dy/dt)^2].
- 3). Write the integrand in terms of the parameter t.
For example, suppose the wire is described by r(t)=x(t)i+y(t)j, where x(t)=t and y(t)=t^2-1, for t=1 to 3. Then continuing with our earlier example, f(x,y)= 8(y+1)/x=8t, ds becomes dt√[(dx/dt)^2+(dy/dt)^2] = dt√[(1)^2+(2t)^2] = dt√[4t^2-1]. So the full integrand is √[4t^2-1]---[8t]dt. - 4). Integrate over the length of the wire with respect to the parameter t.
For example, the line in our example is from t=1 to 3. So the mass of the wire is ∫√[4t^2-1]---[8t]dt, which is easily solved with knowledge of the chain rule to get the difference of (2/3)[4t^2-1]^1.5 evaluated at 3 and 1, or (2/3)[35^1.5-3^1.5] = 134.58. This is the mass of the wire in our example.
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