Line Integrals - Part II — Question 6

Question 6 – Solution

Strategy. First calculate with $x$ as parameter, then reparametrize by $y$ to confirm invariance under a valid change of parameter.

Step 1: Use $x$ $$y=x^3,\qquad dy=3x^2\,dx,\qquad 0\le x\le 1.$$

Therefore $$\int_Cx^2\,dy =3\int_0^1x^4\,dx =3\left[\frac{x^5}{5}\right]_0^1 =\boxed{\frac 35}.$$

Step 2: Use $y$ instead On this interval, $x=y^{1/3}$, so $x^2=y^{2/3}$. Since the differential is already $dy$, $$\int_Cx^2\,dy =\int_0^1y^{2/3}\,dy =\left[\frac 35y^{5/3}\right]_0^1 =\frac 35.$$

Verification Both parametrizations preserve the motion from $(0,0)$ to $(1,1)$ and give the same value, confirming the differential conversion.