# Line Integrals (yayyy )

1. Jan 1, 2009

Line Integrals (yayyy!!)

1. The problem statement, all variables and given/known data

Okay, so I have already done it using the surface integral; now I need to compute the 3 individual line integrals.

By definition, the integral (I will call it I since I am that creative) is given by:

$I=\oint v\cdot\, dl$

$v=<xy, 2yz, 3xz>$

$dl=<dx, dy, dz>$

$\Rightarrow v\dot dl=xydx+2yzdy+3xzdz$

So the three individual integrals are:

along the base of the triangle,

$I_1=\int_o^2 xydx+2yzdy+3xzdz$

along the height of the triangle,

$I_2=\int_o^2 xydx+2yzdy+3xzdz$

Now the hypoteneuse is what is getting me (I think),

I need to again evaluate $\int xydx+2yzdy+3xzdz$

Are my bounds just from 0--->(2-y) ?

Casey

2. Jan 1, 2009

Re: Line Integrals (yayyy!!)

I am looking at my work and now I think that I really do not know what I am doing.

I do not understand what these integrals even mean.

There are three lines i, ii, and iii that correspond to the base, height and hyp, respectively.

When I integrate along i, I am integrating v dot dl from y=0 to y=2 right?

3. Jan 1, 2009

### JaWiB

Re: Line Integrals (yayyy!!)

Don't you have to parameterize the curve?

4. Jan 1, 2009

### gabbagabbahey

Re: Line Integrals (yayyy!!)

Let's start with the base (i): What can you say about the values of x and z all the way along the base? ...that should simplify your vector function a little. Now, what variable changes along the base? Clearly that will be your integration variable for that section of the path.

The same applies to the height (ii).

The hypotenuse is a little trickier: what can you say about the values of x and y along the hypoteneuse? ....So what is the relationship between dx and dy?.... So dl=___?

5. Jan 2, 2009

Re: Line Integrals (yayyy!!)

This sounds familiar.

I think I confused myself. I forgot that the triangle is (more or less) the domain of v ...NOT the function that is being integrated. Silly me.

So... let me review my Calc notes to see how we parameterize the curves. I realize that it is probably a simple problem, but I rather do it the hard way.

6. Jan 2, 2009

### gabbagabbahey

Re: Line Integrals (yayyy!!)

I wouldn't bother parameterizing such a simple curve. The hypotenuse is just a straight line segment that can be written either as $z=m_1y+b_1$ or $y=m_2z+b_2$ with y and z both going from 0 to 2. Hence, $dz=m_1y$ and $dy=m_2z$. In this way, you can write both your vector function and your dl in terms of a single variable (y or z) and its differential and then integrate from 0 to*2.

7. Jan 2, 2009

### HallsofIvy

Re: Line Integrals (yayyy!!)

Along the base line z= 0 (and, of course, dx= dz= 0) so the integral of 2zy dy is easy.

Along the vertical x= 0 (and dx= dy= 0) so the integral of 3xz dz is easy.

Along the hypotenuse of the triangle y+ z= 2 so dy+ dz= 0 and dx=0.
y= 2- z, x= 0 and dy= -dz so 2zy dy+ 3xz dz= 2z(2-z)(-dz)+ 3(0)zdz= (4z- 2z2)(-dz)= (2z2- 4z)dz. Integrate that from z= 0 to z= 2.