0
$\begingroup$

How do I integrate $f$ over the given curve?

$f(x,y)= \frac{x^3}{y};\quad C: y=\frac{x^2}{2} \quad \text{for}\; 0 \leq x \leq 2.$

I can't figure this out... can anyone show me how to solve it?

The answer is supposed to be $\displaystyle \frac{10\sqrt{5}-2}{3}$.

1 Answers 1

0

Parametrize the curve $C$ by $\gamma : [0,2] \rightarrow \mathbb{R}^2$ with $ \gamma(t) = \left( t, \frac{t^2}{2} \right).$ Then, $ \int_{\gamma} f(x,y) \mathrm{ds} = \int_0^2 f(\gamma(t)) ||\dot{\gamma}(t)|| dt = \int_0^2 \frac{t^3}{\frac{t^2}{2}} ||(1, t)|| dt = \int_0^2 2t \sqrt{1 + t^2} dt.$ Solve this integral by substitution.