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TomLikesPhysics Group Title

I need help with an integral.

  • 2 years ago
  • 2 years ago

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  1. ajit.shirsat Group Title
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    which

    • 2 years ago
  2. amistre64 Group Title
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    \[ \begin{array}l\color{red}{\text{I}}\color{orange}{\text{n}}\color{#e6e600}{\text{t}}\color{green}{\text{e}}\color{blue}{\text{g}}\color{purple}{\text{r}}\color{purple}{\text{a}}\color{red}{\text{t}}\color{orange}{\text{i}}\color{#e6e600}{\text{o}}\color{green}{\text{n}}\color{blue}{\text{ }}\color{purple}{\text{n}}\color{purple}{\text{e}}\color{red}{\text{e}}\color{orange}{\text{d}}\color{#e6e600}{\text{s}}\color{green}{\text{ }}\color{blue}{\text{n}}\color{purple}{\text{o}}\color{purple}{\text{ }}\color{red}{\text{h}}\color{orange}{\text{e}}\color{#e6e600}{\text{l}}\color{green}{\text{p}}\color{blue}{\text{ }}\color{purple}{\text{.}}\color{purple}{\text{.}}\color{red}{\text{.}}\color{orange}{\text{.}}\color{#e6e600}{\text{ }}\color{green}{\text{✌}}\color{blue}{\text{}}\end{array} \]

    • 2 years ago
  3. TomLikesPhysics Group Title
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    \[V=\int\limits_{?}^{?}f(x,y,z)dV\] \[f(x,y,z)=xy , x,y \ge , x^2+y^2+z^2\le1\]

    • 2 years ago
  4. TomLikesPhysics Group Title
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    Somehow I felt sphere and used the jacobi-determinant to rewrite the integral and to get V=4/3 * pi but to rewrite it I used a z component which f does not have so I am kind of clueless.

    • 2 years ago
  5. TomLikesPhysics Group Title
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    \[f(x,y,z)=xy\] \[x,y \ge0\] \[x^2+y^2+z^2\]

    • 2 years ago
  6. TomLikesPhysics Group Title
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    \[x^2+y^2+z^2 \le 1\]

    • 2 years ago
  7. TomLikesPhysics Group Title
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    Those are the conditions I have.

    • 2 years ago
  8. amistre64 Group Title
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    well, it does appear to be an 4th a sphere then

    • 2 years ago
  9. amistre64 Group Title
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    |dw:1353510305015:dw|

    • 2 years ago
  10. TomLikesPhysics Group Title
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    Here is my calculation.

    • 2 years ago
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  11. TomLikesPhysics Group Title
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    f(x,y,z)=xy so it is a surface I thought and if I integrate that I get a 4-dim sphere?

    • 2 years ago
  12. amistre64 Group Title
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    yeah, ill have to review this later, work is calling nice work on that jacobian tho

    • 2 years ago
  13. TomLikesPhysics Group Title
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    yeah but I think nobody ask for it.^^

    • 2 years ago
  14. TomLikesPhysics Group Title
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    I mean I guess I don´t need that jacobi-det. here, right?

    • 2 years ago
  15. TuringTest Group Title
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    I'm confused as to what your final integral is....

    • 2 years ago
  16. TomLikesPhysics Group Title
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    4/3*pi

    • 2 years ago
  17. TomLikesPhysics Group Title
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    After I got that I checked the Jacobi-Det. to make sure I did not misread something on Wikipedia.

    • 2 years ago
  18. TuringTest Group Title
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    well, let me try it, but I just don't quite see the set up

    • 2 years ago
  19. TomLikesPhysics Group Title
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    The set up is everything on the left side of the picture above the line with the 1 in a circle underneath it.

    • 2 years ago
  20. TuringTest Group Title
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    I seem to get zero

    • 2 years ago
  21. TuringTest Group Title
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    \[x=\rho\sin\phi\cos\theta\]\[y=\rho\sin\phi\sin\theta\]\[z=\cos\phi\]\[dV=\rho^2\sin\phi d\theta d\rho d\phi\]\[f(x,y,z)=xy=\rho^2\sin^2\phi\sin\theta\cos\theta\]\[\iiint\limits_Ef(x,y,z)dV=\int_0^\pi\int_0^1\int_0^{2\pi}\rho^4\sin^3\phi\sin\theta\cos\theta d\theta d\rho d\phi\]\[=\int_0^\pi\int_0^1\rho^4\sin^3\phi\left[\left.\frac{\sin^2\theta}2\right|_0^{2\pi}\right] d\rho d\phi=0\] did I make a mistake? I hate it when I get zero...

    • 2 years ago
  22. TuringTest Group Title
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    that should be \(z=\rho\cos\theta\) of course

    • 2 years ago
  23. TomLikesPhysics Group Title
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    I think the theta integral goes from 0 to pi. The phi integral goes from 0 to 2pi.

    • 2 years ago
  24. TomLikesPhysics Group Title
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    But that does not change a thing I think...

    • 2 years ago
  25. TomLikesPhysics Group Title
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    Ups, my mistake, you change the angles.

    • 2 years ago
  26. TuringTest Group Title
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    yeah I switched your theta for my hi just cuz that's how I usually do it

    • 2 years ago
  27. TuringTest Group Title
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    my phi*

    • 2 years ago
  28. TomLikesPhysics Group Title
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    So perhaps sphericalcoordinates do not help with this problem?

    • 2 years ago
  29. TomLikesPhysics Group Title
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    but I ended up with 4/3*pi when I used the Jacobi-Det.

    • 2 years ago
  30. TomLikesPhysics Group Title
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    but I also threw the function f=xy right out of the window and did not include it in the calculation...

    • 2 years ago
  31. TomLikesPhysics Group Title
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    Perhaps my calculation on the picture which comes right after the set up is right? The block with the 1, I mean.

    • 2 years ago
  32. TuringTest Group Title
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    you can't just throw out the thing you are integrating, and by the 'jacobi-det' I can only presume you mean the jacobian dV in spherical coordinates \(\rho^2\sin\phi\) which I used as well the change in coordinates should not affect the answer if we do it right, which I think we did, so I think the answer is zero 4/3pi is just the volume of the shpere over which we are integrating, but the integral itself is zero

    • 2 years ago
  33. TomLikesPhysics Group Title
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    We are neglecting the part where it is mandatory that x,y are greater or equal to zero. The x in terms of sphericalcoordinates can be negative but it shouldn´t be. I think we must somehow work that and the condition that the sum of the squares is 1 into the integral.

    • 2 years ago
  34. TuringTest Group Title
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    oh, I missed that :P if x and y are greater than zero, we are in the first quadrant of the xy-plane, so theta varies from 0 to pi/2\[\iiint\limits_Ef(x,y,z)dV=\int_0^\pi\int_0^1\int_0^{\frac\pi2}\rho^4\sin^3\phi\sin\theta\cos\theta d\theta d\rho d\phi\]

    • 2 years ago
  35. TomLikesPhysics Group Title
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    So the volume or whatever we just calculated is just 4/30? Did I do that right? Thx a lot so far Turing Test.

    • 2 years ago
  36. TuringTest Group Title
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    \[\iiint\limits_Ef(x,y,z)dV=\int_0^\pi\int_0^1\int_0^{\frac\pi2}\rho^4\sin^3\phi\sin\theta\cos\theta d\theta d\rho d\phi\]\[=\frac12\int_0^\pi\int_0^1\rho^4\sin^3\phi d\rho d\phi=\frac1{10}\int_0^\pi \sin^3\phi d\phi=\]\[\frac1{10}\int_0^\pi\sin\phi-\cos^2\phi\sin\phi d\phi=\frac1{10}\left(\left.-\cos\phi+\frac{\cos^3\phi}3\right|_0^\pi\right)\]\[=\frac1{10}(\frac23-(-\frac23))=\frac2{15}\]so yes, I guess that's right. btw the thing that we found is the integral of the function f(x,y,z)=xy over the given volume of a quarter-sphere, E.

    • 2 years ago
  37. TomLikesPhysics Group Title
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    Since is has something to do with a sphere isn´t it weird that we got 2/15 and not something with pi?

    • 2 years ago
  38. TuringTest Group Title
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    No, not really. Just because the region you are integrating in is a sphere doesn't meant you should expect pi to show up. If you think about any integral in either polar, cylindrical, or spherical coordinates, if theta shows up in the integrand as the argument of a trig function, then evaluation ion the d(theta) integration will cause he angle (which is where the pi would come from, of course) converted into whatever the resulting trig function results in. example in polar coordinates for simplicity:\[\int_0^1\int_0^{\pi}rdrd\theta=\frac\pi2\](with no integrand, this integral just represents the area of the region of integration; a circle with radius 1) However if we wanted to integrate \(y=r\sin\theta\) *over* that region of a half-circle we get\[\int_0^1\int_0^\pi r^2\sin\theta drd\theta=\frac23\]Notice the \(\pi\) got swallowed up by the fact that it was evaluated as the argument of a trig function, which is very common an in no way unusual in these situations.

    • 2 years ago
  39. TomLikesPhysics Group Title
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    Ok. Thank you Turing Test for your durability and stamina. You made the whole thing a lot clearer.

    • one year ago
  40. TuringTest Group Title
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    Happy I could help :)

    • one year ago
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