Line Integrals - Part II — Question 2

Question 2 – Solution

Strategy. Use the graph relation to write $dy=y'(x)\,dx$.

Step 1: Convert the differential Along $y=x^2$, $$dy=2x\,dx,\qquad 0\le x\le 2.$$

Step 2: Evaluate $$\int_Cx\,dy =\int_0^2x(2x)\,dx =2\left[\frac{x^3}{3}\right]_0^2 =\boxed{\frac{16}{3}}.$$

Step 3: Compare Although both integrals use the same curve, they measure change in different coordinates: $$y\,dx=x^2\,dx,\qquad x\,dy=x(2x\,dx)=2x^2\,dx.$$ The second integrand is exactly twice the first, so its value is twice $8/3$.

Verification Here $y$ increases from $0$ to $4$ and $0\le x\le 2$, so the value must lie between $0$ and $2(4)=8$. The result $16/3$ does.