I need help with this problem:

Compute the line integral$\int_{C^+}(x^2-y)dx+(y^2+x)dy$ where $C^+$ is the parabollic arc $y=x^2+1$, $0\leq x\leq 1$ oriented from $(0,1)$ to $(1,2)$.

First I parametrized the arc $C^+$ by setting $x=t$, then $y=t^2+1$, so the parametrization would be $\alpha(t)=(t,t^2+1), t\in[0,1]$. then I tried to solve this by using the formula $\int_Cfds=\int_\alpha fds=\int_{0}^1f(\alpha(t))\Vert\alpha'(t)\Vert dt+\int_1^2 f(\alpha(t))\Vert\alpha'(t)\Vert dt$. But I don't know how to compute that, since I have an integral with respect to $x$ and one with respect to $y$.

  • $\begingroup$ You don’t have two integrals. In general, $\int(f\,dx+g\,dy)\ne(\int f\,dx)+(\int g\,dy)$. $\endgroup$ – amd Apr 21 at 20:46
  • $\begingroup$ I didn't know that, why they aren't equal? $\endgroup$ – davidllerenav Apr 21 at 20:50
  • $\begingroup$ See math.stackexchange.com/q/1971225/265466. $\endgroup$ – amd Apr 21 at 20:51

With $x=t$ and $y=t^2+1$ giving $dx=dt$ and $dy=2t \, dt$, and with $t$ going from $0$ to $1$, the integral becomes $$ \int_{C^+}(x^2-y)dx+(y^2+x)dy = \int_0^1 (t^2-(t^2+1)) \, dt + ((t^2+1)^2+t) \, 2t \, dt \\ = \int_0^1 (2t^5+4t^3+2t^2+2t-1) \, dt = \left[ \frac13 t^6 + t^4 + \frac23 t^3 + t^2 - t \right]_0^1 \\ = \frac13 + 1+ \frac23 + 1 - 1 = 2. $$

  • $\begingroup$ I see, so I also need to change the varible of $dy$ and $dx$, like in a substitution, rihgt? There's one problem, according to my book the answer is 2. $\endgroup$ – davidllerenav Apr 21 at 20:52
  • 1
    $\begingroup$ Yes, it's like a substitution. Regarding the error, I see that @michael has fixed it. Thanks! $\endgroup$ – md2perpe Apr 22 at 8:25

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