# finding the line integral

• Jan 8th 2013, 12:16 AM
Kvandesterren
finding the line integral
Find the line integral:

∫C(2y2 i + x j) · dr , where C is the line segment from (3,1) to (0,0).

My answer so far (...which has gone awfully awry. the answer is supposed to be -7/2):

y = 1/3x
x = 3y
0 ≤ t ≤ 3

∫ (lower limit: t=0 upper limit: t=3) (2t^2i + tj) · (-3i-j) dt
∫ (lower limit: t=0 upper limit: t=3) (-6t^2 - t) dt
= -6(1/3)t^3 - (1/2)t^2 , evaluated from 3 to 0
= -2t^3 - (1/2)t^2 , evaluated from 3 to 0
= [-2(27) - (9/2)] - [0]
= -54 - 9/2
= -108/2 - 9/2
= -117/2 (...which is far off from the correct answer above. I am pretty sure I messed up the parameterization. anybody, please help?)

• Jan 8th 2013, 01:40 AM
BobP
Re: finding the line integral
If $\displaystyle y=\frac{1}{3}t,$ then $\displaystyle 2y^{2}=\frac{2}{9}t^{2}.$
• Jan 8th 2013, 03:12 AM
Kvandesterren
Re: finding the line integral
I still cannot arrive at the correct answer. I must be doing something fundamentally wrong. I see how 2y^2i = (2/9)t^2i.

now, when I follow the formula: ∫ F·dr = ∫ F(r(t) · r’(t) dt

r(t) = (2/9)t^2i + tj
r’(t) = (4/9)ti + (1)j
then: ∫ (lower limit: t=0 upper limit: t=3) [ (2/9)t^2i + tj ] · [ (4/9)ti + (1)j ]
= ∫ (lower limit: t=0 upper limit: t=3) (8/81)t^3 + t , which arrives at 13/2

what did I do wrong?
• Jan 8th 2013, 05:13 AM
HallsofIvy
Re: finding the line integral
Quote:

Originally Posted by Kvandesterren
Find the line integral:

∫C(2y2 i + x j) · dr , where C is the line segment from (3,1) to (0,0).

My answer so far (...which has gone awfully awry. the answer is supposed to be -7/2):

y = 1/3x
x = 3y
0 ≤ t ≤ 3

At this point you have said nothing about how x and y are related to t. I think you mean that x= 1- t (so that x goes from 3 to 0 as t goes from 0 to 3) and that y= 1- (1/3)t. That really doesn't use "x= 3y". And, of course, dx= -dt, dy= -(1/3)dt.

Quote:

∫ (lower limit: t=0 upper limit: t=3) (2t^2i + tj) · (-3i-j) dt
With t going from 0 to 3, we still have x= 1- t and y= 1- (1/3)t so that 2y^2i+ xj= 2(1- (1/3)t)^2i+ (1- t)j= (2- (4/3)t+ (2/9)t^2)i+(1- t)j and then (2y^2i+ xj)dr= [(2-(4/3)t+ (2/9)t^2)(-dt)+ (1- t)(-1/3)]dt

But that is NOT what you are doing. You are just replacing "y" with t which means your parameterization is y= t, x= 3y= 3t so that dy= dt and dx= 3dt. The integrand becomes (2y^2i+ xj)(dxi+ dyj)= (2t^2i+ 3tj)(i+ 3j)dt= (2t^2+ 9t)dt.
However, we do not now have "t" from 0 to 3. Because we are using t= y as parameter, and we are integrating from (3, 1) to (0, 0), that is, y is going from 1 to 0, the lower limit of the integral must be 1 and the upper limit 0.

Quote:

∫ (lower limit: t=0 upper limit: t=3) (-6t^2 - t) dt
= -6(1/3)t^3 - (1/2)t^2 , evaluated from 3 to 0
= -2t^3 - (1/2)t^2 , evaluated from 3 to 0
= [-2(27) - (9/2)] - [0]
= -54 - 9/2
= -108/2 - 9/2
= -117/2 (...which is far off from the correct answer above. I am pretty sure I messed up the parameterization. anybody, please help?)

• Jan 8th 2013, 07:45 AM
BobP
Re: finding the line integral
Start again from the top.

Begin by deciding what limits you are going to choose for t.

The limits on the parameter t can be whatever you want them to be, (so long as they are different), so 0 to 1, 0 to 3, 3 to 0, whatever. You seem to be choosing 0 to 3.

Choosing 0 to 3, (and check later to see that 0 to 1 works out easier and 3 to 0 is easier still), what that means is that t = 0 should generate a value of 3 for x and a value of 1 for y, and a value of t = 3 should generate zero values for both x and y. That should lead you to the equations x = 3-t and y = 1-(t/3).

(If this is a problem, let x = at + b, substitute the pairs of values (t=0, x=3), (t=3,x=0) and solve for a and b. Same routine for y.)

Substitute those for x and y in your integrand.

Also substitute for dr, dr = dx(i) + dy(j) = -dt(i) - (1/3)dt(j).

Try repeating the calculation with the other suggested limits for t, you should get -7/2 each time.