# Evaluate the integral along the stated curve

$\int_{C}{(3x+2y) \, dx + (2x-y) \, dy}$ along the curve y = sin($\pi*x\over2$) from (0,0) to (1,1). (Given that the curve is smooth).

Approach: I attempted this problem by parametrizing x = $\pi*t\over2$ and y = sin(t), but that wasn't working out since I got this: ∫3t + 2sin($\pi*t\over2$) + ($\pi$)tcos($\pi*t\over2$) - ($\pi\over2$)*cos($\pi*t\over2$)*sin($\pi*t\over2$). I then attempted x = t and y = sin($\pi*t\over2$), which didn't help.

I'm having trouble finding which parametrization works (the integration should follow easy from there). Can some help out with the setup of the parameters?

Thanks

• why is your first attempt difficult? – Vladimir Vargas Nov 25 '14 at 1:57
• I think I messed up the parameters. – 86BCP2432T Nov 27 '14 at 4:02
• It would be what Adriano suggested: \begin{align*} x = \frac{2}{\pi}t &\implies dx = \frac{2}{\pi}\, dt \\ y = \sin t &\implies dy = \cos t \, dt \\ \end{align*} I just didn't look carefully enough at the x to manipulate it to t in order to have y = sin(t). – 86BCP2432T Nov 27 '14 at 4:19

You seem to have made a typo with your first attempt. If we want $y = \sin t$, then we need to ensure that $t = \frac{\pi}{2}x$, so we have: \begin{align*} x = \frac{2}{\pi}t &\implies dx = \frac{2}{\pi}\, dt \\ y = \sin t &\implies dy = \cos t \, dt \\ \end{align*} So our line integral becomes: $$\int_0^{\pi/2} (3 \cdot \tfrac{2}{\pi}t + 2 \cdot \sin t) \cdot \tfrac{2}{\pi}\, dt + \int_0^{\pi/2} (2 \cdot \tfrac{2}{\pi}t - \sin t) \cdot \cos t \, dt$$ which is not too bad to evaluate.
$$\int_{C}{(3x+2y) \, dx + (2x-y) \, dy} = \int_{C}{3x \, dx - y \, dy + 2 \,d(xy)} =\left[\frac32 x^2 - \frac12 y^2 + 2xy \right]_{0,0}^{1,1} \\ = 3$$