- #1

- 371

- 28

- 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

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- Thread starter WMDhamnekar
- Start date

- #1

- 371

- 28

- 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

- #2

- 20,004

- 10,653

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

- #3

- 5,695

- 2,475

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##.

- #4

- 20,004

- 10,653

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

- #5

- 5,695

- 2,475

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).This is not correct. Signs matter. OP should be able to deduce the correct path.

- #6

- 20,004

- 10,653

OP has all integrals from 0 to 1. Unless explicitly stated that would be the typical assumption.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).

- #7

- 371

- 28

## \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}##

Is this answer correct?

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

- #8

- 5,695

- 2,475

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

- #9

- 24,232

- 15,968

Yes.Is this answer correct?

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

- #10

- 24,232

- 15,968

- #11

- #12

- 24,232

- 15,968

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.@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.

- #13

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- 2,475

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

- #14

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- 10,653

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