1. ## Evaluate Line Integral

Q: Evaluate $\displaystyle \int_{\gamma} y^2 e^{2\sin{x}} \cos{x} dx + ye^{2\sin{x}} dy$ where $\displaystyle \gamma$ is the arc of the curve $\displaystyle y = \cos{x}$ from the point $\displaystyle (0,1)$ to the point $\displaystyle (\frac{\pi}{2},0)$.

A: We know that the integral is exact, meaning that we can choose any path from $\displaystyle (0,1)$ to $\displaystyle (\frac{\pi}{2},0)$. In the solutions they have gone through the origin, along the coordinate axes. So the path taken is from $\displaystyle (0,1)$ to $\displaystyle (0,0)$, and then to $\displaystyle (\frac{\pi}{2},0)$

The integral is then

$\displaystyle \int^0_1 ye^{0} dy + \int^{\frac{\pi}{2}}_0 0 dx$

Not sure how they get this? It looks like they've just subbed $\displaystyle x=0$ and $\displaystyle y=\frac{\pi}{2}$ into the first and second bit of the original integral?

I know it's going to be something obvious but I'd appreciate it if someone could help me out with this.

Craig

2. Originally Posted by craig
Q: Evaluate $\displaystyle \int_{\gamma} y^2 e^{2\sin{x}} \cos{x} dx + ye^{2\sin{x}} dy$ where $\displaystyle \gamma$ is the arc of the curve $\displaystyle y = \cos{x}$ from the point $\displaystyle (0,1)$ to the point $\displaystyle (\frac{\pi}{2},0)$.

A: We know that the integral is exact, meaning that we can choose any path from $\displaystyle (0,1)$ to $\displaystyle (\frac{\pi}{2},0)$. In the solutions they have gone through the origin, along the coordinate axes. So the path taken is from $\displaystyle (0,1)$ to $\displaystyle (0,0)$, and then to $\displaystyle (\frac{\pi}{2},0)$

The integral is then

$\displaystyle \int^0_1 ye^{0} dy + \int^{\frac{\pi}{2}}_0 0 dx$

Not sure how they get this? It looks like they've just subbed $\displaystyle x=0$ and $\displaystyle y=\frac{\pi}{2}$ into the first and second bit of the original integral?
On the line from (0, 1) to (0, 0), x is identically 0. And on the line from (0, 0) to $\displaystyle (\pi/2, 0)$, y is identically 0. You could, for example, take as parametric equations for (0, 1) to (0, 0), x= 0, y= t with t from 1 to 0 and for (0, 0) to $\displaystyle (\pi/2, 0)$, x= t, y= 0 for t from 0 to $\displaystyle \pi/2$.

I know it's going to be something obvious but I'd appreciate it if someone could help me out with this.

Craig

I understand most of what you're saying, when you sub in y = 0 then that's where you get the 0 from etc.

I'm just confused how you get:

$\displaystyle \int^0_1 ye^{0} dy$ ?

What happens to the $\displaystyle y^2$ from the original integral?

Thanks again

4. Originally Posted by craig

I understand most of what you're saying, when you sub in y = 0 then that's where you get the 0 from etc.

I'm just confused how you get:

$\displaystyle \int^0_1 ye^{0} dy$ ?

What happens to the $\displaystyle y^2$ from the original integral?

Thanks again
On the first leg, from (0, 1) to (0, 0) x is identically 0 so x does not change. That means that dx is also 0.

More specifically, if we let x= 0, y= t, then dx= 0dt and dy= dt.
$\displaystyle \int y^2e^{2sin(x)}dx+ ye^{2sin(x)}dy$ becomes $\displaystyle \int t^2e^0(0dt)+ te^0(dt)= \int t dt$

5. Of course! Thankyou for that, always amazes me how simple most things are when you know the answer...