Line Integrals - Part II — Question 1

Question 1 – Solution

Strategy. For an integral with respect to $x$, use $dx=x'(t)\,dt$ rather than the arc-length element.

Step 1: Parametrize $$\mathbf r(t)=\langle t,t^2\rangle,\qquad 0\le t\le 2.$$ Thus $x=t$, $y=t^2$, and $dx=dt$.

Step 2: Evaluate $$\int_Cy\,dx =\int_0^2t^2\,dt =\left[\frac{t^3}{3}\right]_0^2 =\boxed{\frac 83}.$$

Step 3: Distinguish from $ds$ The differential is $dx=x'(t)\,dt=dt$. The speed $\sqrt{1+4t^2}$ would be used only for an integral with respect to $ds$; inserting it here would change the problem.

Verification Since $x$ increases from $0$ to $2$ and $0\le y\le 4$, the integral must lie between $0$ and $4(2)=8$. The value $8/3$ satisfies this bound.