Find the line integrals value $\int_{0,0}^{1,{\frac{\pi}{2}}} e^x \sin(y)dx + e^x\cos(y)dy$

Question

Here is what i have so far :

$$ \left(e^x\sin\left(y\right)\right)\vert_{0,0}^{1,{\frac{π}{2}}} + \left(e^x\sin\left(y\right)\right)\vert_{0,0}^{1,{\frac{π}{2}}} $$ $$ e^1\sin\left(\frac{π}{2}\right) - e^0\sin\left(0\right) \space + \space e^1\sin\left(\frac{π}{2}\right) - e^0\sin\left(0\right) $$ $$e +e = 2e$$

However I am fairly sure this is wrong, what is the correct way to solve this line integral ?

Answer 1

If I right to assume : $\displaystyle \int_{(0,0)}^{(1,\pi/2)}\left(e^{x}\sin(y)dx\right) + \int_{(0,0)}^{(1,\pi/2)}e^{x}\cos(y)dy = e^{x}\sin(y)|_{(0,0)}^{(1,\pi/2)}+e^{x}sin(y)|_{(0,0)}^{(1,\pi/2)}$ and you're doing fine.

Answer 2

Computing along a straight line path, you have $$ \gamma(t)=(t,t\frac{\pi}{2}) $$ and the vector field $$ F(x,y)=(e^x\sin(y),e^x\cos(y)) $$ and the integral is $$ \int_{\gamma}F(\gamma(t))\cdot \gamma'(t)\mathrm dt=\int_0^1e^t\sin\left(\frac{\pi}{2}t\right)+\frac{\pi}{2}e^t\cos\left(\frac{\pi}{2}t\right)\mathrm dt $$ both pieces of which you can integrate by parts.

Answer 3

It is path independent. The vector field $F(x, y)=(P(x,y), Q(x,y))$ where $P(x, y)=e^x\sin y$ and $Q(x,y)=e^x\cos y$ is conservative with the potential function $\phi(x,y)=e^x\sin y$. Thus the given integral is equal to the difference of initial and final points of the path:=$$\phi(1, \frac{\pi}{2})-\phi(0, 0)=1.$$