- #1
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- 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
Computing##\displaystyle\int_C f\cdot dr ## for the given vector field
- Thread starter WMDhamnekar
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Answers and Replies
- #2
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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.
That said, you have not specified ##C_3## and your ##C_2## does not correspond to any part of the desired curve.
- #3
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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##.
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
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This is not correct. Signs matter. OP should be able to deduce the correct path.
- #5
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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).
- #6
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OP has all integrals from 0 to 1. Unless explicitly stated that would be the typical assumption.
- #7
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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}##
Is this answer correct?
If correct, why is the answer negative? What does this negative answer indicate?
## \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
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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?
You should be able to infer what the negative sign means. When does the dot product ##\mathbf{f}\cdot\mathbf{dr}## becomes negative?
- #9
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- #10
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PS In general ##\int_a^b = -\int_b^a##. And, if the integrand is positive and ##a < b## then ##\int_a^b## is positive.
- #11
- #12
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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.
- #13
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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.
- #14
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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).