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anonymous

  • 5 years ago

triple integral from -3 to 2, 0 to sqrt(9-x^2), 0 to 9-x^2-y^2 of sqrt(x^2+y^)2 dzdydx

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  1. anonymous
    • 5 years ago
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    use cylindrical coordinates. right?

  2. anonymous
    • 5 years ago
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    yes(:

  3. anonymous
    • 5 years ago
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    i get 93.8

  4. anonymous
    • 5 years ago
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    i get to the integral from 0 to pi/2 of (2/35)(cos^3theta + costheta sin^2theta) dtheta did you get that too? after the dz and dr are integrated. thank youu by the way! (:

  5. anonymous
    • 5 years ago
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    give me a sec, let me make sure i'm right...let me do it again...

  6. anonymous
    • 5 years ago
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    oh wait i'm looking at the wrong problem sorry...one second

  7. anonymous
    • 5 years ago
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    can you show me the first steps of how to find the bounds and what the problem turns into after converting to polar?

  8. anonymous
    • 5 years ago
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    (x^2+y^2)^(1/2) = r (9-x^2)^(1/2) is the top half of a circle w/radius 3... it's easier if you sketch it. can you sketch it?

  9. anonymous
    • 5 years ago
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    yes. so is it 0 to pi for dtheta, 0 to three for dz, and 0 to 3 for dr?

  10. anonymous
    • 5 years ago
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    yes.

  11. anonymous
    • 5 years ago
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    are you positive about the dz and dr? i kind of just guessed lol(:

  12. anonymous
    • 5 years ago
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    no, that shoud be right.

  13. anonymous
    • 5 years ago
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    don't forget the extra r when you evaluate it though....

  14. anonymous
    • 5 years ago
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    i got 84.8

  15. anonymous
    • 5 years ago
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    integral from zero to pi, from zero to three, and from zero to three of r^2 dzdrdtheta then, integral from zero to pi and from zero to three of r^2z evaluated from zero to three of drdtheta then, integral from zero to pi and from zero to three of 3r^2 drdtheta then, integral from zero to pi of r^3 evaluated from zero to three of dtheta then, integral from zero to pi of 27 dtheta then, 27theta evaluated from zero to pi which is 27 pi i think?

  16. anonymous
    • 5 years ago
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    sorry, i just re-read your question earlier, r should be from 0 to 9 - r^2...

  17. anonymous
    • 5 years ago
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    oh okay that makes much more sense(: thank you so much!

  18. anonymous
    • 5 years ago
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    omg, i just did this problem, and now i keep messing up...sorry...look...z=0 to z-9-x^2-y^2 <--- is 9-(x^2+y^2)... so, z is actually from 0 to 9-r^2...

  19. anonymous
    • 5 years ago
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    and r is from zero to three?

  20. anonymous
    • 5 years ago
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    yes, sorry, haha, i'll just leave you alone before i mess you up anymore...tired...i think you get it...

  21. anonymous
    • 5 years ago
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    do you have software to graph these problems with? just in case you run into trouble graphing them? it really helps, you can just pick off the limits from the graph, so the key is being able to graph it.

  22. anonymous
    • 5 years ago
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    no i don't :( i'm frustrated because i keep getting different answers than you got originally... integral from zero to pi, from zero to three, from zero to 9-r^2 of r^2 dzdrdtheta then it should become the integral from zero to pi and zero to three of r^2(9-r^2)drdtheta then, integral from zero to pi of 32.4dtheta which gives you 32.4pi where did i go wrong..?

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