# Computing##\displaystyle\int_C f\cdot dr ## for the given vector field

• WMDhamnekar
In summary: I admit you confused me a bit here, but you seem to be thinking in terms of the integral we get after we do the parameterization.
WMDhamnekar
MHB
Homework Statement
##f(x,y) =xy^2i + xy^3 j ; ## C: the ploygonal path from (0,0) to (1,0) to (0,1) to (0,0)
Relevant Equations
None

Please refrain from posting your work as images. Even if legible (which it hardly is on my phone without zooming in) it is impossible to quote properly. Instead use the built-in LaTeX functions and type your work.

That said, you have not specified ##C_3## and your ##C_2## does not correspond to any part of the desired curve.

WMDhamnekar
The integral for ##C_3## should be zero, still you should 've said that explicitly.

I think your parameterization ##x=t,y=t## for ##C_2## is wrong (according to that parameterization the point (1,1) for t=1 belongs to the path, but we know it does NOT), I think the correct one is ##x=t,y=1-t##.

WMDhamnekar
Delta2 said:
I think the correct one is x=t,y=1−t.
This is not correct. Signs matter. OP should be able to deduce the correct path.

Orodruin said:
This is not correct. Signs matter. OP should be able to deduce the correct path.
Sorry I don't see a sign error, its the straight line path from (1,0) to (0,1) (t has to vary from 1 to 0 though not from 0 to 1).

WMDhamnekar
Delta2 said:
Sorry I don't see a sign error, its the straight line path from (1,0) to (0,1) (t has to vary from 1 to 0 though not from 0 to 1).
OP has all integrals from 0 to 1. Unless explicitly stated that would be the typical assumption.

Delta2
Write ##C = C_1 \cup C_2 \cup C_3## where C1 is the curve given by x = t, y =0, 0≤t≤1 , C2 is the curve given by x= (1-t), y =t, 0≤ t ≤1, and C3 is the curve given by x=0, y = 1-t, 0≤ t≤ 1. Then,

## \displaystyle\int_C xy^2 dx +xy^3 dy= \displaystyle\int_{C_1} xy^2 dx + xy^3 dy + \displaystyle\int_{C_2}xy^2 dx +xy^3 dy + \displaystyle\int_{C_3}xy^2 dx + xy^3 dy ##

## \displaystyle\int_C xy^2 dx +xy^3 dy = \displaystyle\int_0^1 t(0^2)(1) + t(0^3)(0) dt + \displaystyle\int_0^1(1-t)(t^2)(-1) + (1-t)(t^3)(1) dt + \displaystyle\int_0^1 0 (0^2) 0 + 0(0^3)0dt##

## \displaystyle\int_C xy^2 dx +xy^3 dy = 0 +\displaystyle\int_0^1 -t^2 +2t^3 -t^4 dt +0##

## \displaystyle\int_C xy^2 dx +xy^3 dy = -\frac{t^3}{3} + 2\frac{t^4}{4} - \frac{t^5}{5} \big|_0^1##
## \displaystyle\int_C xy^2 dx +xy^3 dy = -\frac13 +\frac12 -\frac15 =-\frac{1}{30}##

If correct, why is the answer negative? What does this negative answer indicate?

Yes it seems correct to me
You should be able to infer what the negative sign means. When does the dot product ##\mathbf{f}\cdot\mathbf{dr}## becomes negative?

WMDhamnekar said:
Yes.
WMDhamnekar said:
If correct, why is the answer negative? What does this negative answer indicate?
That you went backwards, in a sense.

WMDhamnekar
PS In general ##\int_a^b = -\int_b^a##. And, if the integrand is positive and ##a < b## then ##\int_a^b## is positive.

@PeroK your interpretation though correct is a bit shallow , there is something a bit more deep relating on how the vector field's direction relates to the direction we walk the path.

Delta2 said:
@PeroK your interpretation though correct is a bit shallow , there is something a bit more deep relating on how the vector field's direction relates to the direction we transverse the path.
In general, it won't be easy to see whether the answer is positive or negative. But, if you have a positive integrand in the first quadrant, then left to right and upwards is positive; right to left and upwards is negative (what we have here); left to right and downwards is negative; and, right to left and downwards is positive.

WMDhamnekar
PeroK said:
In general, it won't be easy to see whether the answer is positive or negative. But, if you have a positive integrand in the first quadrant, then left to right and upwards is positive; right to left and upwards is negative (what we have here); left to right and downwards is negative; and, right to left and downwards is positive.
I admit you confused me a bit here, but you seem to be thinking in terms of the integral we get after we do the parameterization.

I would say to think in terms of the line integral representing the work of a force.

WMDhamnekar
It is true that a work integral takes the form of a line integral, but the concept is of course much more general than that. Mathematically, the circulation integral being negative can be roughly interpreted as the field being integrated pointing more against the curve direction than in the curve direction on average. The circulation integral being non-zero also let's us conclude that the field is not conservative (and hence cannot be written as the gradient of a scalar field etc).

Delta2 and PeroK

## 1. What is the meaning of "Computing##\displaystyle\int_C f\cdot dr## for a vector field?

The notation ##\displaystyle\int_C f\cdot dr## represents the line integral of a vector field, where f is the vector field and C is the curve over which the integral is being calculated. This integral measures the total effect of the vector field along the curve C.

## 2. How do you calculate the line integral of a vector field?

To calculate ##\displaystyle\int_C f\cdot dr##, you first need to parameterize the curve C with a function r(t). Then, substitute this function into the integral and evaluate it using the appropriate integration techniques.

## 3. What is the significance of the line integral of a vector field?

The line integral of a vector field measures the work done by the vector field along the curve C. It can also be used to calculate other physical quantities, such as flux and circulation.

## 4. Can the line integral of a vector field be negative?

Yes, the line integral of a vector field can be negative. This indicates that the vector field is doing work in the opposite direction of the curve C.

## 5. What are some real-world applications of computing line integrals of vector fields?

Line integrals of vector fields have various applications in physics, engineering, and economics. They can be used to calculate the work done by a force field, the flow of a fluid, or the circulation of a current. They are also used in optimization problems and in calculating the total cost of a project.

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