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anonymous

  • one year ago

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  1. rainbow_rocks03
    • one year ago
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    What is your question?

  2. anonymous
    • one year ago
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    How to solve an integral of the type \[\int\limits (f(x,y)dx+g(x,y)dy)\]

  3. hartnn
    • one year ago
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    distribute the integral \(\int f + \int g\) integrate f w.r.t x treating y as a constant! integrate g w.r.t y, treating x as constant!

  4. anonymous
    • one year ago
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    I tried that but it turned out to be wrong answer

  5. hartnn
    • one year ago
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    initial conditions are given? or what other conditions are given? when you integrate f w.r.t x, you get some function in x and y + some constant the thing to note here is., "the constant" here is actually the function of y !

  6. anonymous
    • one year ago
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    \[f(x,y)=y^2\]\[g(x,y)=2xy\]\[\int\limits f(x,y)dx+\int\limits g(x,y)dy=\int\limits y^2dx+\int\limits 2xydy=xy^2+xy^2=2xy^2+C\] But answer is \[xy^2+C\]

  7. hartnn
    • one year ago
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    that goes like this: \(\int fdx = xy^2 +f(y)\) \(\int gdy = xy^2 +f(x)\) comparing those 2, we can conclude there are no extra functions(terms) which only have y or x so f(y) = 0, f(x) = 0

  8. anonymous
    • one year ago
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    \[\vec F=(x^2-y^2+x)\hat i-(2xy+y)\hat j\] \[d \vec r=dx \hat i+dy \hat j+dz \hat k\] \[\phi=\int\limits \vec F.d \vec r=\int\limits [(x^2-y^2+x)dx-(2xy+y)dy]+C\] Instead of -xy^2 I get -2xy^2

  9. ikram002p
    • one year ago
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    the answer u have from ur book or some where else ??

  10. ikram002p
    • one year ago
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    i still think in both questions ur answer is correct hmmm how it could be something else :-\

  11. anonymous
    • one year ago
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    Answer is \[\frac{x^3}{3}+\frac{x^2}{2}-\frac{y^2}{2}-xy^2+C\] and I'm getting \[\frac{x^3}{3}+\frac{x^2}{2}-\frac{y^2}{2}-2xy^2+C\]

  12. hartnn
    • one year ago
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    x^3/3 -xy^2 +x^2/2 + f(y) ----(A) -xy^2 +y^2/2 +f(x) ----(B) comparing the above 2, we get f(y) = y^2/2 and f(x) = x^3/3 +x^2/2 now substitute that in (A) or (B) to get your answer

  13. hartnn
    • one year ago
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    -y^2/2 *

  14. anonymous
    • one year ago
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    Why are we comparing them?Is it necessary that they are equal?

  15. hartnn
    • one year ago
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    thats how I was taught, didn't raise this question then... maybe ikram can give some more insight into this...

  16. anonymous
    • one year ago
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    From what I understand they've done something like Let \[h \equiv h(x,y)\] be defined such that \[f(x,y)=\frac{\partial h}{\partial x}\] and \[g(x,y)=\frac{\partial h}{\partial y}\] so that \[f(x,y)dx+g(x,y)dy=\frac{\partial h}{\partial x}dx+\frac{\partial h}{\partial y}dy=dh\] \[\implies \int\limits [f(x,y)dx+g(x,y)dy]=\int\limits dh=h+C\] But why and even then how would we find our h, it's not always obvious in all cases

  17. anonymous
    • one year ago
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    One can see at a glance that \[-\int\limits (y^2dx+2xy dy)=-\int\limits d(xy^2)=-xy^2+C\]

  18. anonymous
    • one year ago
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    I'll leave this question open, maybe @IrishBoy123 would be able to help out later, the real deal is how we will find our h(x,y) and why can't we just simply integrate while keeping one variable constant

  19. anonymous
    • one year ago
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    One way I thought if since x and y are independent of each other we have \[\frac{dx}{dy}=\frac{dy}{dx}=0\] Dividing dh by dx we have \[\therefore \frac{dh}{dx}=\frac{\partial h}{\partial x} \implies dh=\frac{\partial h}{\partial x}dx=f(x,y)dx\] Similarly for dy we have \[dh=g(x,y)dy\] Adding these two we get \[2.dh=f(x,y)dx+g(x,y)dy\] \[2h=\int\limits f(x,y) dx + \int\limits g(x,y) dy\] Here we mean integration while keeping other variable constant so we will in fact write \[2h=\int\limits f(x,y) \partial x+\int\limits g(x,y) \partial y\] \[\implies h=\frac{\int\limits f(x,y) \partial x+\int\limits g(x,y) \partial y}{2}+C\] Where C is some constant independent of x or y, I'm not sure if it's correct but that's a way to think about it

  20. anonymous
    • one year ago
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    calculus is weird sometimes lol

  21. IrishBoy123
    • one year ago
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    these are *line integrals* so you can either solve them as DE's a shown above or you integrate *over a path*. ie if you are to integrate, be careful. as you are travelling over a path, there is actually only one independent variable in line integrals, hence why people often parameterise line integrals into an expression in t only. that's often OTT but it at least is a mechanism that gets answers. doing that here for the first one; \(\int y^2 dx+\int 2xy \ dy\), we choose the following path in 2 steps [because we can]. |dw:1440446154961:dw| for Part 1 of path y = dy = 0 so the integral is \(\int_{0}^{x} 0^2 \ dx = 0\) for Part 2 dx = 0, x = const so the integral is \(\int_{0}^{y} 2xy \ dy = xy^2\) adding gives \(\Phi = xy^2 = const. \) for the second one: \(\int x^2−y^2+x) \ dx + \int −(2xy+y) \ dy\) we chose the same (x,y) destination but go a different way just to illustrate: |dw:1440446941344:dw| this time part 1, dx = x = 0, integral is \(\int_{0}^{y} -y \ dy = -\frac{y^2}{2}\) part 2, dy = 0, y = const, integral is \(\int_{0}^{x} x^2 - y^2 + x \ dx = \frac{x^3}{3} - xy^2 + \frac{x^2}{2} \) adding gives \(\Phi = \frac{x^3}{3} - xy^2 + \frac{x^2}{2} -\frac{y^2}{2} = const.\) you can choose your path because these are conservative vector fields, which is why the also give rise to exact differential equations. furthermore, and this is useful, you can always check you answer by looking at \(\nabla \Phi = \vec F\). ie the gradient of the supposed solution should correspond to the integrands of line integral you started with. if this is news to you, then honestly, forget about it and solve the way @hartnn was doing above.

  22. anonymous
    • one year ago
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    thanks I understand it now first we will check if \[\vec \nabla \times \vec F=\vec 0\] If it is true then F is a conservative vector field, so to find the scalar potential phi, such that \[\vec F=\vec \nabla\phi\] We can find \[\phi=\int\limits \vec F.d\vec r\] along any path in the vector field. Taking a path along the axes or parallel to the axes is most suitable as a variable becomes 0 or constant.

  23. IrishBoy123
    • one year ago
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    @Nishant_Garg yeah, good summary!

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