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well, what part are you having an issue with?

can you guide the first couple steps, im really confused please

trying to see if i can see the boundary
|dw:1436633020633:dw|

is it the inner part of cylinder we are looking for?

are your limits valid? other than that it really is just a mechanical process.

there valid but im having trouble with the integration

yes

where did the sin^2t and cos come from

did you do squareroot (9-y^2) ==> squareroot (9-((9-x^2)^(1/2))?

where did you plug in the limits for y?

right i see that...so my limits for dz are wrong?

uh, you originally asked for the volume of the intersection ...

|dw:1436634359956:dw|
so you want the volume of something like this?

it looks like this

right, and you want the volume of .. what now?

|dw:1436634627932:dw|

yes ignore the red cylinder

exactly like that

from that part, how do you find the limits?

interesting part

good to hear its not completey wrong. y limits how to find them?

still working on that ....

its oki ! im still here

thank you so much

would you happen to remember greens thm?

i never really covered greens thrm in my courses. so no.

its oki..would you be staying online today? if its oki to ask you other questions

oh so the set up is the same?

\[\int_{y=0}^{y=3}\int_{x=0}^{x=\sqrt{9-y^2}}\int_{z=0}^{z=\sqrt{9-y^2}}~dz~dx~dy\]

oki can you show me the frst integrration please

well, using your fundamental thrm of calculus ...
\[\int_{0}^{b}dz=F(b)-F(0)\]

what is the integration of 1?

well, z in this case, since we are integrating with respect to z but yeah
so
z(b) - z(0) = z(b)

so with z...then we plug in the limits of x?

limits of z ...

\[\int_{0}^{\sqrt{9-y^2}}dz=z_b-z_a=\sqrt{9-y^2}-0\]

now, what is the middle integral?
\[\int_{0}^{\sqrt{9-y^2}}\sqrt{9-y^2}~dx=??\]

\[\int\limits_{0}^{3} (1/2)(\sqrt{9-y^2}y+9\sin^{-1} (y/3)\]

x^2/2

no, try again.

what was the integration of 1?

the respective variable yes ...
in this case we are doing dx, so its an x

for some constant k
\[\int_{0}^{b}dx=b(k)-0(k)\]

in this case, k=sqrt(9-y^2), and b=sqrt(9-y^2)
leaving us:
\[\int_{0}^{3}9-y^2~dy\]

oh ok so this is our middle integral

yes, we did the first, then the middle, and we are now left with the end

is the final answer 19?

dunno, i havent gotten to the end of it yet

i dont think 19 is it

18

what is:
\[\int_{0}^{3}9-y^2~dy?\]
work the steps for me

\[9y-y\]

9y-y^3/3

144...can you remind me again why we work with 1/8

the limits for x and z why are they the same?

those are better explained in the youtube video i posted.

oki i understand...so much more clear. thank you

good luck :)