I have to find the line integral of the following.

$$\int_Cxe^ydx+x^2ydy, C: 0 \leq x \leq 2, y = 3$$

I am trying to understand the concept of line integrals, but in this case, I am confused as to what $C$ really is. Parameterizing both $x$ and $y$, i have $x = t$ and $y = 3$. Following from this, we have:




However, upon checking with my answer key, the answer is $2e^3$. So I am rather stuck here.

  • 1
    $\begingroup$ If $y =$ constant then $dy = 0$ $\endgroup$ – jim Apr 14 '16 at 22:32

Your curve will be $r(t) = (t,3)$, for $0 \leq t \leq 2$. We can think that if $r(t)=(x(t),y(t))$, then ${\rm d}x = x'(t)\,{\rm d}t$ and ${\rm d}y = y'(t)\,{\rm d}t$, so here we'll have ${\rm d}x = {\rm d}t$ and ${\rm d}y = 0$. So: $$\int_C xe^y\,{\rm d}x + x^2y\,{\rm d}y = \int_0^2 te^3\,{\rm d}t = 2e^3.$$


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.