Question

please help me!! Triple integrals.

Answer 1

are you asking someone named Tripple Integrals to help you?

Answer 2

Find the volume enclosed by the parabaloids z=x^2+y^2 and z = 8-x^2-y^2

Answer 3

no lol I need help with tripple and double integrals

Answer 4

looks ripe for cylindrical coordinates

Answer 5

what will the projection of this shape on the xy-plane be?

Answer 6

i think it makes kind of an egg shape

Answer 7

a circular region

Answer 8

but the projection. i.e its shadow...
yes, a circle of radius what?

Answer 9

I'm not sure how do I figure this out?

Answer 10

wait is it radius 8

Answer 11

find when the two equations intersect
z=z
8-x^2-y^2=x^2+y^2

Answer 12

ah, ok

Answer 13

i got x^2+y^2=4 so the radius is 2

Answer 14

yes, now change it to cylindrical coordinates
what are the bounds on z, r, and theta?

Answer 15

\[\int\limits_{?}^{?}\int\limits_{?}^{?}\int\limits_{z=r^2}^{z=8-r^2} rdrd\]

Answer 16

what is the maximum r can be (think of the circle)

Answer 17

2

Answer 18

and what is the minimum r can be?

Answer 19

0

Answer 20

and theta should be 2pi and 0

Answer 21

yes, so what is your integral?

Answer 22

\[\int\limits_{0}^{2PI}\int\limits_{0}^{2}\int\limits_{r^2}^{8-r^2} rdrd \theta \]

Answer 23

where is your z differential?

Answer 24

triple integral=three differentials

Answer 25

thats right dz should be on the end of that

Answer 26

the end? what are we going to integrate wrt first?

Answer 27

r^2 and 8-r^2 are the bounds on what?

Answer 28

ok im confused by what your saying

Answer 29

\[\int_0^{2\pi}\int_0^2\int_{r^2}^{8-r^2}rdrd\theta dz\]would suggest that \[r^2\le r\le8-r^2\]but that's not right, what should we have between r^2 and 8-r^2 ??

Answer 30

oh I made a typo to, but the inner integral is the important one here

Answer 31

too*

Answer 32

oh i see \[\int\limits_{0}^{2\pi}\int\limits_{0}^{2} \int\limits_{0}^{8} (8-r^2)-r^2 rdrd \theta dz\]

Answer 33

doeas that look better?

Answer 34

no, not quite

Answer 35

crap

Answer 36

didn't you earlier make it clear that\[r^2\le z\le8-r^2\]?
you wrote that for an earlier integral...

Answer 37

right so then that is my limit and the integral is rdr dtheta dz

Answer 38

no, you need to work from the inside out
since your internal integral has the bounds r^2 and 8-r^2, which are the bounds for z, you innermost differential should be dz
make sense?

Answer 39

ya it does so the integral is rdz dr dtheta

Answer 40

yes :)

Answer 41

ok finally lol