0
$\begingroup$

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 $y$.

$\endgroup$
  • $\begingroup$ 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
2
$\begingroup$

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 $$ \int_{C^+}(x^2-y)dx+(y^2+x)dy = \int_0^1 (t^2-(t^2+1)) \, dt + ((t^2+1)^2+t) \, 2t \, dt \\ = \int_0^1 (2t^5+4t^3+2t^2+2t-1) \, dt = \left[ \frac13 t^6 + t^4 + \frac23 t^3 + t^2 - t \right]_0^1 \\ = \frac13 + 1+ \frac23 + 1 - 1 = 2. $$

$\endgroup$
  • $\begingroup$ 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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.