# How to compute the line integral of $\int_{C^+}(x^2-y)dx+(y^2+x)dy$?

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 $$y$$.

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

## 1 Answer

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

• 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. – davidllerenav Apr 21 at 20:52
• Yes, it's like a substitution. Regarding the error, I see that @michael has fixed it. Thanks! – md2perpe Apr 22 at 8:25