Evaluate the line integral $\int_C \sin(x)\sin(y)dx - \cos(x)\cos(y)dy$ where $C$ is the line segment fom $(0,-\pi)$ to $(\frac{3\pi}{2},\frac{\pi}{2})$

I parameterized and substituted into the original equation and got stuck with: $\int_0^1 \sin(\frac{3\pi}{2}t)\sin(\frac{3\pi}{2}t-\pi)(\frac{3\pi}{2}) - \cos(\frac{3\pi}{2}t)\cos(\frac{3\pi}{2}t-\pi)(\frac{3\pi}{2}) dt$

I'm not sure how to integrate this so any help is appreciated!

  • $\begingroup$ Are you required to calculate this integral directly, or are you allowed to use the Gradient Theorem? $\endgroup$ – Semiclassical Jul 20 '14 at 23:42
  • $\begingroup$ I was thinking of using the $1=sin^2(x) + cos^2(x)$ but I can't make the functions squared so I am not sure what identity you are referring to @DanielCharry $\endgroup$ – Coop Jul 20 '14 at 23:46
  • $\begingroup$ It just says to evaluate using the Fundamental Theorem of Line Integrals @Semiclassical I do not know what the Gradient Theorem is to be honest so I am assuming we aren't allowed to use it $\endgroup$ – Coop Jul 20 '14 at 23:46
  • $\begingroup$ Ahah! That's another name for the Gradient theorem. It'd help to put that into the text of your question as context. $\endgroup$ – Semiclassical Jul 20 '14 at 23:47

I used the parameterization $\alpha:(0,1)\rightarrow C\in\mathbb{R}^2,\,t\mapsto[{3\pi t\over2},{3\pi t\over2}-\pi]$. You just have to use the Sine-Cosine shift identities (when shifted by $\pi$).


Use the identities $\sin(a-\pi)=-\sin(a)$ and $\cos(a-\pi)=-\cos(a)$
$\displaystyle\left[\sin({3\pi t\over2})\sin({3\pi t\over2} - \pi), - \cos({3\pi t\over2})\cos({3\pi t\over2} - \pi)\right] ⋅ \left[{3\pi\over2}, {3\pi\over2}\right]={3\pi\cos(3\pi t)\over2}$

Future: As Semiclassical posted, you can use the FTLI to solve the problem, in fact, letting $f(x,y)=-\cos(x)\sin(y)$ gives you $\nabla f(x,y)=F(x,y)$


Recall that the fundamental integral of line integrals is $$\int_{C[\mathbf{q},\mathbf{p}]}\nabla{F}\cdot d\mathbf{r} = F(\mathbf{p})-F(\mathbf{q})$$ So if we want to compute your line integral along $C$ this way, we need to find a function $F(x,y)$ such that $$\nabla F(x,y)\cdot d\mathbf{r} = \frac{\partial F}{\partial x} dx+\frac{\partial F}{\partial y} dy = \sin x \sin y \, dx - \cos x \cos y \, dy $$ Do you see a convenient choice for $F$ that has the correct partial derivatives?

  • $\begingroup$ I read up on what the fundamental theorem of line integrals was and was able to figure it out. The answer is 0 if you were curious. Thank you for all of your help, I appreciate it! $\endgroup$ – Coop Jul 20 '14 at 23:58
  • $\begingroup$ Glad to hear it. One thing I'd suggest is plugging the integral you wrote in your post into Wolfram Alpha just so you can assure yourself that it gives the same answer. @Coop $\endgroup$ – Semiclassical Jul 21 '14 at 0:02

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