Thread: 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?)

(this is schoolwork, so only helpful hints and enlightening explanations, please, not just correct answers. thanks.)

If then

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?

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.

∫ (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.

∫ (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?)

(this is schoolwork, so only helpful hints and enlightening explanations, please, not just correct answers. thanks.)

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.