Evaluate Line Integral

**craig** · November 1st 2010, 05:15 AM

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

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

The answers then state that:

The integral is then

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

Not sure how they get this? It looks like they've just subbed $x=0$ and $y=\frac{\pi}{2}$ 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.

Thanks in advance

Craig

**HallsofIvy** · November 1st 2010, 09:13 AM

Originally Posted by craig

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

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

The answers then state that:

The integral is then

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

Not sure how they get this? It looks like they've just subbed $x=0$ and $y=\frac{\pi}{2}$ 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 $(\pi/2, 0)$ , 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 $(\pi/2, 0)$ , x= t, y= 0 for t from 0 to $\pi/2$ .

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

Thanks in advance

Craig

**craig** · November 1st 2010, 11:23 AM

Thanks for the reply.

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:

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

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

Thanks again

**HallsofIvy** · November 1st 2010, 11:54 AM

Originally Posted by craig

Thanks for the reply.

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:

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

What happens to the $y^2$ 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.
$\int y^2e^{2sin(x)}dx+ ye^{2sin(x)}dy$ becomes $\int t^2e^0(0dt)+ te^0(dt)= \int t dt$

**craig** · November 1st 2010, 11:56 AM

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

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