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anonymous
 5 years ago
triple integral from 3 to 2, 0 to sqrt(9x^2), 0 to 9x^2y^2 of sqrt(x^2+y^)2 dzdydx
anonymous
 5 years ago
triple integral from 3 to 2, 0 to sqrt(9x^2), 0 to 9x^2y^2 of sqrt(x^2+y^)2 dzdydx

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anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0use cylindrical coordinates. right?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0i 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
 5 years ago
Best ResponseYou've already chosen the best response.0give me a sec, let me make sure i'm right...let me do it again...

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0oh wait i'm looking at the wrong problem sorry...one second

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0can you show me the first steps of how to find the bounds and what the problem turns into after converting to polar?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0(x^2+y^2)^(1/2) = r (9x^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
 5 years ago
Best ResponseYou've already chosen the best response.0yes. so is it 0 to pi for dtheta, 0 to three for dz, and 0 to 3 for dr?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0are you positive about the dz and dr? i kind of just guessed lol(:

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0no, that shoud be right.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0don't forget the extra r when you evaluate it though....

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0integral 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
 5 years ago
Best ResponseYou've already chosen the best response.0sorry, i just reread your question earlier, r should be from 0 to 9  r^2...

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0oh okay that makes much more sense(: thank you so much!

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0omg, i just did this problem, and now i keep messing up...sorry...look...z=0 to z9x^2y^2 < is 9(x^2+y^2)... so, z is actually from 0 to 9r^2...

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0and r is from zero to three?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0yes, sorry, haha, i'll just leave you alone before i mess you up anymore...tired...i think you get it...

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0do 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
 5 years ago
Best ResponseYou've already chosen the best response.0no 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 9r^2 of r^2 dzdrdtheta then it should become the integral from zero to pi and zero to three of r^2(9r^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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