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anonymous

  • 5 years ago

Let sigma be the upper hemisphere with radius R. Use cartesian coordinates to write a triple integral that represents the volume V of sigma. Indicate both terminals on each integral?

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  1. anonymous
    • 5 years ago
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    \[\int\limits_{-R}^{R}\int\limits_{-\sqrt{R^2-x^2}}^{\sqrt{R^2-x^2}}\int\limits_{0}^{\sqrt{R^2-x^2-y^2}}1dzdydx\] is this correct? because I'm getting 0 out of this.

  2. anonymous
    • 5 years ago
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    hii nikola !!!

  3. anonymous
    • 5 years ago
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    hey dan, whenever u integrate a constant, in this case 1, u get the variable here wrt x, u get an x similarly with respect to y & z.. are u familiar with basic integrals?? :)

  4. anonymous
    • 5 years ago
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    u der??

  5. anonymous
    • 5 years ago
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    The integrals looks right. You sure you're getting zero?

  6. anonymous
    • 5 years ago
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    yes, I do familiar with the basic integral, but when I get 0 that means something wrong with the terminals

  7. anonymous
    • 5 years ago
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    yeah I got 0 after I integrated it wrt to y

  8. anonymous
    • 5 years ago
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    I have a question on how to do this as well. I'm learning this right now too. How are you integrating with respect to y? You have y^2 in the sqrt and no y outside.

  9. anonymous
    • 5 years ago
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    you got: \[\int\limits_{-R}^{R}\int\limits_{-\sqrt{R^2-x^2}}^{\sqrt{R^2-x^2}} z dy dx\] and you put in sqrt(R^2-x^2-y^2) So: \[\int\limits_{-R}^{R}\int\limits_{-\sqrt{R^2-x^2}}^{\sqrt{R^2-x^2}} \sqrt{R^2-x^2-y^2} dy dx\] then u integrate again wrt to y and when u put the terminals in, it becomes 0 :(

  10. anonymous
    • 5 years ago
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    I don't know if this would help...but doing this in spherical coordinates would give you the answer of (R^2 * pi^2)/4 I'm pretty sure you can convert that to cartesian by changing the R^2, which, in spherical to Cartesian would be x^2 + y^2 + z^2 (sorry, i don't know how to do the formula thing to make it look pretty)

  11. anonymous
    • 5 years ago
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    yeah, i know we can solve it via spherical coordinates, but the question ask me to do in cartesian and it's obvious the answer wouldn't be a nice and simple one

  12. anonymous
    • 5 years ago
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    the general formula of spherical would be, \[x^{2}+y^{2}+z^{2} = r^{2}\] use the equation box below the chat box :)

  13. anonymous
    • 5 years ago
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    I'm going to try this. I'm pretty sure you're doing the integration wrt y wrong. I'm going to substitute (R^2 - x^2 - y^2)^.5 as u and integrate using substitution, it shouldn't be zero. and lol, thanks, but I clicked it and I can't seem to make heads or tails out of the interface. so I'll stick to plain and simple text :D

  14. anonymous
    • 5 years ago
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    correction: just R^2 - x^2 - y^2 as u

  15. anonymous
    • 5 years ago
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    yup, i got something that's not 0... now let's see if i can't figure this equation thing out...

  16. anonymous
    • 5 years ago
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    hahaha, ok i'll wait. let's see.

  17. anonymous
    • 5 years ago
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    \[1/3 \int\limits\limits_{-R}^{R} (R ^{2} - x ^{2} - y ^{2})^{3/2} dz\] where y = plus or minus (R^2 - x^2) ^-- i didn't plug it in yet and I dont know how to use the equations.... sighs

  18. anonymous
    • 5 years ago
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    ahh..i see what you mean....

  19. anonymous
    • 5 years ago
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    yeah? u got 0 after u plug the terminals in? that must be something wrong with the terminals, but I don't know which part :(

  20. anonymous
    • 5 years ago
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    i think it's -1/3y instead of 1/3 because we use the chain rule of integral?

  21. anonymous
    • 5 years ago
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    i don't believe so...but i'm not sure

  22. anonymous
    • 5 years ago
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    what r u studying btw? vector calculus?

  23. anonymous
    • 5 years ago
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    because if you substitute, you would just get \[\int\limits_{long equation}^{longequation} 1/2 \sqrt{u}du\] which would be 1/2 * 2/3 * u ^ 3/2 which is 1/3 * u ^ 3/2 take the 1/3 out caz it's a constant and replace u with original equation. I'm doing line integrals right now. and no one has helped me with my question T-T

  24. anonymous
    • 5 years ago
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    ok, i have a solution or rather, i found somewhere there's a solution the question here is the same but it's a whole sphere, so the z value is different. but the part where we are stuck, you need trig identities, which has always been one of my weakest areas here's the link http://www.physicsforums.com/showthread.php?t=386226

  25. anonymous
    • 5 years ago
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    i still don't quite understand it....but I believe that the solution makes sense

  26. anonymous
    • 5 years ago
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    yup but if you let \[u = R^{2}-x{2}-y{2}\] and \[du/dy = -2y\] so, \[dy= du/-2y\] then, \[-1/2y\int\limits\limits_{}^{} u^{1/2} du\]

  27. anonymous
    • 5 years ago
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    ok thanks i'll check it out

  28. anonymous
    • 5 years ago
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    Yeah I think you're "more" right than I am. I think i just broke every single rule of integration that I've ever learned. But I don't think you can take out a variable like that... because the y doesn't necessarily need to be the same inside and outside of the (u)^/5. I looked that up as well and it says that you need trig identities again. you have to substitute x, or y in our case, with 2 *cos(u) and onwards! I'm completely lost now. :( I have to go now, but I hope that site will help. And also, if you have time, can you take a look at my problem? It's two questions below yours. The one that starts with a intimidating integral...and relating to line integrals Good luck!

  29. anonymous
    • 5 years ago
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    ok i'll try, that physics guy made me lost too haha.. thanks :)

  30. anonymous
    • 5 years ago
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    got it, it's a very helpful link up there. :)

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