MEDAL!! Please help...! use triple integration to find the volume of the solid bounded by the cylinders x^2+y^2=9 and y^2+z^2=9

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MEDAL!! Please help...! use triple integration to find the volume of the solid bounded by the cylinders x^2+y^2=9 and y^2+z^2=9

Mathematics
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well, what part are you having an issue with?
the limit points im using if -3 to 3, -squareroot (9-x^2), squareroot (9-x^2), -squareroot (9-y^2), -squareroot (9-y^2)
can you guide the first couple steps, im really confused please

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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
well lets see what we can work then dz from -sqrt(9-y^2) to sqrt(9-y^2) dy from -sqrt(9-x^2) to sqrt(9-x^2) dx from -3 to 3 you agree that this is how we want to move?
yes
then we simply work it from the inside out \[\int \left(\int\left[\int dz\right]~dy\right)~dx\] \[\int \left(\int z_b-z_a~dy\right)~dx\] to start with
\[\int \left(2\int \sqrt{9-y^2}~dy\right)~dx\] let y = 3sin(t) , dy = 3cos(t) dt \[\int \left(2\int \sqrt{9-(3sin(t))^2}~3cos(t)~dt\right)~dx\] \[\int \left(2\int \sqrt{9(1-sin^2(t))}~3cos(t)~dt\right)~dx\] \[\int \left(2\int 3cos^2(t)*3cos(t)~dt\right)~dx\] \[\int \left(18\int cos^3(t)~dt\right)~dx\] etc ..
where did the sin^2t and cos come from
i have a small error ... but its just using a trig substitution we let y = 3sin(t) and substitute all the y parts. sqrt(1-sin^2) = cos ... not cos^2 tho
did you do squareroot (9-y^2) ==> squareroot (9-((9-x^2)^(1/2))?
where did you plug in the limits for y?
no sqrt(9 - y^2) dy if we let y = 3sin(t), the derivative allows us to replace dy by dt parts dy = 3cos(t) dt by replacing the appropriate parts ... sqrt(9 - (3sin(t))^2) * 3cos(t) dt
my only concern for this is your limits ... im just not sure that the intersection of 2 cylindars is a sphere ... and that is what your limits are suggesting.
right i see that...so my limits for dz are wrong?
i believe your limits are bad ... im just not sure at the moment how we would correct them off hand. the draw tool is not that good at 3d imaging for me
is there a formula... in mathetica it demonstrated 2 cylinders crossing eachother like a cross and what i was told to look for the volume of the non-intersecting upper part
uh, you originally asked for the volume of the intersection ...
|dw:1436634359956:dw| so you want the volume of something like this?
1 Attachment
it looks like this
right, and you want the volume of .. what now?
if its just 2 cylndars intersecting, i thing the profile of the intersection is a "pringle" chip .. or a saddle back
|dw:1436634627932:dw|
yes ignore the red cylinder
exactly like that
from that part, how do you find the limits?
now, you want to find the volume of the top part? or of the intersecting part? if its the top part, then we need to have a plane of something that cuts it off from going to infinity
interesting part
ok, interSECTing part :) yeah z limits are fine, -sqrt(9-y^2) to sqrt(9-y^2) x limits are fine, -3 to 3 its your y limits that need refined if anything
good to hear its not completey wrong. y limits how to find them?
still working on that ....
its oki ! im still here
https://www.youtube.com/watch?v=rASKzEbMXIQ this might be a bit more useful then me trying to remember things long dead past :) at least for a review
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
im not going to be online much today .. got family stuff to do. but after the youtube video i do see where the issue was :) it was in the limits
oh so the set up is the same?
the setup is the same, but we need to better define the limits ... and using symmetry helps x^2+y^2=9 y^2+z^2=9 they both have a y^2 in common so lets define z and x in terms of y x = 0 to sqrt(9-y^2) z = 0 to sqrt(9-y^2) y = 0 to 3 and this defines 1/8 of our volume dz dx dy is a good order of integration
\[\int_{y=0}^{y=3}\int_{x=0}^{x=\sqrt{9-y^2}}\int_{z=0}^{z=\sqrt{9-y^2}}~dz~dx~dy\]
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?
x
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? \[\int_{0}^{\sqrt{9-y^2}}\sqrt{9-y^2}~dx=??\]
\[\int\limits_{0}^{3} (1/2)(\sqrt{9-y^2}y+9\sin^{-1} (y/3)\]
not quite sqrt(9-y^2) , with respect to x is just some constant ... so, what is the integration of a constant again?
x^2/2
no, try again.
what was the integration of 1?
z
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
9y - y^3/3 , given that y=3 yes 27 - 27/3 = 27(2/3) = 18 but recall that we only worked 1/8 of it, so we need to mulitply this by 8 to get the total volume.
144...can you remind me again why we work with 1/8
the limits for x and z why are they the same?
we dont have to, but due to the symmetry of the intersection, it makes life a little simpler for some.
those are better explained in the youtube video i posted.
oki i understand...so much more clear. thank you
good luck :)

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