Question

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

Answer 1

use cylindrical coordinates. right?

Answer 2

yes(:

Answer 3

i get 93.8

Answer 4

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! (:

Answer 5

give me a sec, let me make sure i'm right...let me do it again...

Answer 6

oh wait i'm looking at the wrong problem sorry...one second

Answer 7

can you show me the first steps of how to find the bounds and what the problem turns into after converting to polar?

Answer 8

(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?

Answer 9

yes. so is it 0 to pi for dtheta, 0 to three for dz, and 0 to 3 for dr?

Answer 10

yes.

Answer 11

are you positive about the dz and dr? i kind of just guessed lol(:

Answer 12

no, that shoud be right.

Answer 13

don't forget the extra r when you evaluate it though....

Answer 14

i got 84.8

Answer 15

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?

Answer 16

sorry, i just re-read your question earlier, r should be from 0 to 9 - r^2...

Answer 17

oh okay that makes much more sense(:

thank you so much!

Answer 18

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...

Answer 19

and r is from zero to three?

Answer 20

yes, sorry, haha, i'll just leave you alone before i mess you up anymore...tired...i think you get it...

Answer 21

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.

Answer 22

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..?