# Let f(x,y) = 2xy, and let C be the arc of the curve y = x^2 from (-1,1) to (2,4). Evaluate the given line integrals.a. ∫c f(x,y)ds b. ∫c f(x,y)dx c. ∫c f(x,y)ds

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You need to find the parametrization of the curve:

`x=(1-t)x_0 + tx_1=gt x = (1-t)(-1) + 2t=gt x = -1 + t + 2t=gtx=3t-1`

`y=(1-t)y_0 + ty_1=gty = (1-t)(1) + 4t=gty=3t+1`

The individual parametric equations are x=3t-1 and y=3t+1.

You need to determine `ds = sqrt((dx/dx)^2 + (dy/dt)^2)dt`

`dx/dt = (3t-1)' =gt dx/dt = 3`

`` `dy/dt = (3t+1)' =gt dx/dt = 3`

`ds = sqrt(3^2 + 3^2)dt =gt ds = sqrt9 dt`

Write the function f(x,y) in terms of t:

`f(x,y) = 2(3t-1)(3t+1) = 2(9t^2 - 1)`

Calculating the line integral yields:

`oint` f(x,y)ds = `2sqrt9 int_0^1 (9t^2 - 1)dt = 2sqrt9(9t^3/3 - t)(0-gt1)`

`` `oint f(x,y)ds = 2sqrt9 (3*1^3 - 3*0^3 - 1 + 0) = 4sqrt9`

**Evaluating the line integral yields `oint f(x,y)ds = 4sqrt9` .**