What is the Line Integral of xydx+4ydy along a Curve from (1,2) to (3,5)?

[qq545282501]

Homework Statement

$$\int xydx+ 4ydy$$
where C is the curve from (1,2) to (3,5) made up of the twoline segments parallel to the coordinate axes.
$$c_1:(1,2)\rightarrow(3,2)$$
$$c_2:(3,2)\rightarrow(3,5)$$

Homework Equations

The Attempt at a Solution

i got c2 correct, y=2+3t, and x = 0, for t goes from 0 to 1.
but i got c1 wrong, for c1, i see only x is changing, x=1+2t. so x'(t)= 2. if y value is not changing, it means that dy=0, my professor had y=1, by setting 2xy=2x, i guess 2xy is the first half of the initial integral by replacing dx with 2, but i don't understand what is 2x on the right side of the equation.

 

[Buzz Bloom]

Hi qq:

I don't understand why you introduce the t variable unless it is that you are used to doing that in general for arbitrary curves. I also don't understand why you use t=0 and t= 1 as the limits. For this particular problem I suggest you write down the integrals with the limits specified for both c1 and c2. Since y is constant for c1 and x is constant for c2, these two integrals should be easy to integrate.

Hope t his helps.

Regards,
Buzz

 

[BvU]

$\int xy\, dx$ with y constant is $y \int x\, dx$. For the first leg that means the same as xy = 2x and you calculate $2\int_1^3 x\, dx$.

If you want to end up with an $\int_0^1$ ( a sort of parametrization that isn't really necessary here, as BB explains), you substitute t = (x-1)/2 so dt = dx/2 to get $2 \int_0^1 2(1+2t) \, dt$

same result

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