Question

Evaluate the integral \(\int_{C}(2 x-y) d x+(x+3 y) d y\) along the path \(C\). \(C:\) arc on \(y=1-x^{2}\) from (0,1) to (1,0)

Short answer

The evaluated line integral \(\int_{C}(2 x - y) dx + (x + 3y) dy\) over the path C is \(5/3\).

Step-by-step solution

Parametrization of the path

The path \( C \) is described by the curve \( y = 1 - x^2 \). For sake of simplicity, it can be parametrized with respect to \( t \) as follows: \( x = t \) and \( y = 1 - t^2 \). The parameter \( t \) ranges form 0 to 1 to represent the entire curve from point (0,1) to (1,0). Therefore, \( dx = dt \) and \( dy = -2t dt \).

Substitute the parameters

Substitute \( x = t \), \( y = 1 - t^2 \), \( dx = dt \), and \( dy = -2t dt \) into the integral: \[\int_{C}(2 x - y) dx + (x + 3y) dy = \int_{0}^{1}(2t - (1 - t^2)) dt + (t + 3(1 - t^2) (-2t dt)\]After simplifying, this results in:\[ \int_{0}^{1}(3t^2 + 2t - 1) dt -2t (t - 3t^2) dt = \int_{0}^{1}(5t^2) dt\]

Evaluate the integral

Now the integral can be evaluated:\[\int_{0}^{1}5t^{2}dt = [5/3 * t^{3}]_{0}^{1} = 5/3\]