Question

Evaluate the line integral along the given path. \(\int_{C} 4 x y d s\) \(C: \mathbf{r}(t)=t \mathbf{i}+(2-t) \mathbf{j}\) \(0 \leq t \leq 2\)

Short answer

The value of the line integral is -\(\frac{16\sqrt{2}}{3}\).

Step-by-step solution

Define the parametric equations

Our given curve, \(\mathbf{r}(t)=t \mathbf{i}+(2-t) \mathbf{j}\), represents a path in the xy plane. This path can be parameterized as follows: \(x = t\) and \(y = 2 - t\), with \(0 \leq t \leq 2\) as the parameter range.

Find ds

\(\mathbf{r}'(t)=\mathbf{i}-\mathbf{j}\), the magnitude of \(\mathbf{r}'(t)\) is \(ds = \sqrt{1^2 + (-1)^2} = \sqrt{2}\).

Substitute into the integral

We can now substitute \(x\), \(y\) and \(ds\) into the integral given in the problem. We obtain \(\int_{0}^{2}4ty(2-t)\sqrt{2} dt=4\sqrt{2}\int_{0}^{2}t(2-t) dt\).

Evaluate the integral

This is now a simple definite integral problem, which can be evaluated using basic calculus. \(\int_{0}^{2}t(2-t) dt\) equals to -\(\frac{4}{3}\) when calculated.

Final Calculation

Multiplying by 4\(\sqrt{2}\) and -\(\frac{4}{3}\), the line integral evaluates to -\(\frac{16\sqrt{2}}{3}\).