Given $f(x,y) = x^2y$, I am computing line integral of $f$ on the curve $r(t) = (t^3,t^2)$parametrized on the interval $-2<t<2$.
I was thinking mapping $r$ means $x = t^3$ and $y = t^2$ so I directly substituted into $f$ so that
$\int x^2y ds =\int_{-2}^{2} t^6*t^2 \sqrt{(3t^2)^2+(2t)^2} dt$.
the parametrized integral looks not solvable thus I am suspecting I misunderstood the question. Am I on right track or the integral is already wrong?
You have $f(x,y)= x^2y$. But you are on the curve $r(t)= (t^3,t^2)$ for $-2 \leq t \leq 2$. But then $x= t^3$ and $y= t^2$ so that on this curve, $f(r(t))= (t^3)^2 t^2= t^6 \cdot t^2= t^8$. Now $ds= \sqrt{(x'(t))^2+(y'(t))^2}\;dt=\sqrt{(3t^2)^2+(2t)^2}\;dt= \sqrt{9t^4+4t^2}\;dt$. Then you have $$ \int f \;ds= \int_{-2}^2 t^8\sqrt{9t^4+4t^2}\;dt $$ So you are exactly correct!