anonymous
  • anonymous
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
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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chestercat
  • chestercat
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anonymous
  • anonymous
use cylindrical coordinates. right?
anonymous
  • anonymous
yes(:
anonymous
  • anonymous
i get 93.8

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anonymous
  • anonymous
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! (:
anonymous
  • anonymous
give me a sec, let me make sure i'm right...let me do it again...
anonymous
  • anonymous
oh wait i'm looking at the wrong problem sorry...one second
anonymous
  • anonymous
can you show me the first steps of how to find the bounds and what the problem turns into after converting to polar?
anonymous
  • anonymous
(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?
anonymous
  • anonymous
yes. so is it 0 to pi for dtheta, 0 to three for dz, and 0 to 3 for dr?
anonymous
  • anonymous
yes.
anonymous
  • anonymous
are you positive about the dz and dr? i kind of just guessed lol(:
anonymous
  • anonymous
no, that shoud be right.
anonymous
  • anonymous
don't forget the extra r when you evaluate it though....
anonymous
  • anonymous
i got 84.8
anonymous
  • anonymous
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?
anonymous
  • anonymous
sorry, i just re-read your question earlier, r should be from 0 to 9 - r^2...
anonymous
  • anonymous
oh okay that makes much more sense(: thank you so much!
anonymous
  • anonymous
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...
anonymous
  • anonymous
and r is from zero to three?
anonymous
  • anonymous
yes, sorry, haha, i'll just leave you alone before i mess you up anymore...tired...i think you get it...
anonymous
  • anonymous
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.
anonymous
  • anonymous
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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