Recent content by AngryStyro

Yes, whoops. So that gives: So $$ \int_{C}^{} \, \frac{x^3}{y} ds = \int_{0}^{2} \, \frac{t^3}{0.5t^2} \sqrt{1+t^2} dt $$ $$ = \int_{0}^{2} \, 2t \sqrt{1+t^2} dt $$ It is still a rather difficult integral to evaluate though... Never mind, I see that I can solve with a simple substitution now...

Hi, I just wonder if I'm doing this question correctly, as it seems like I'm stuck with a particularly difficult integral. Let C be the curve $$ y = \frac{x^2}{2} , 0 \le x \le 2 $$ , evaluate $$ \int_{C}^{} \,d \frac{x^3}{y} ds $$ So I have so far: Parameterise C as $$ r(t) = (x(t),y(t)) =...