anonymous
  • anonymous
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
  • Stacey Warren - Expert brainly.com
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jamiebookeater
  • jamiebookeater
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amistre64
  • amistre64
well, what part are you having an issue with?
anonymous
  • anonymous
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)
anonymous
  • anonymous
can you guide the first couple steps, im really confused please

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amistre64
  • amistre64
trying to see if i can see the boundary |dw:1436633020633:dw|
anonymous
  • anonymous
is it the inner part of cylinder we are looking for?
amistre64
  • amistre64
are your limits valid? other than that it really is just a mechanical process.
anonymous
  • anonymous
there valid but im having trouble with the integration
amistre64
  • amistre64
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?
anonymous
  • anonymous
yes
amistre64
  • amistre64
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
amistre64
  • amistre64
\[\int \left(2\int \sqrt{9-y^2}~dy\right)~dx\] 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 ..
anonymous
  • anonymous
where did the sin^2t and cos come from
amistre64
  • amistre64
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
anonymous
  • anonymous
did you do squareroot (9-y^2) ==> squareroot (9-((9-x^2)^(1/2))?
anonymous
  • anonymous
where did you plug in the limits for y?
amistre64
  • amistre64
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
amistre64
  • amistre64
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.
anonymous
  • anonymous
right i see that...so my limits for dz are wrong?
amistre64
  • amistre64
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
anonymous
  • anonymous
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
amistre64
  • amistre64
uh, you originally asked for the volume of the intersection ...
amistre64
  • amistre64
|dw:1436634359956:dw| so you want the volume of something like this?
anonymous
  • anonymous
1 Attachment
anonymous
  • anonymous
it looks like this
amistre64
  • amistre64
right, and you want the volume of .. what now?
amistre64
  • amistre64
if its just 2 cylndars intersecting, i thing the profile of the intersection is a "pringle" chip .. or a saddle back
amistre64
  • amistre64
|dw:1436634627932:dw|
anonymous
  • anonymous
yes ignore the red cylinder
anonymous
  • anonymous
exactly like that
anonymous
  • anonymous
from that part, how do you find the limits?
amistre64
  • amistre64
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
anonymous
  • anonymous
interesting part
amistre64
  • amistre64
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
anonymous
  • anonymous
good to hear its not completey wrong. y limits how to find them?
amistre64
  • amistre64
still working on that ....
anonymous
  • anonymous
its oki ! im still here
amistre64
  • amistre64
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
anonymous
  • anonymous
thank you so much
anonymous
  • anonymous
would you happen to remember greens thm?
amistre64
  • amistre64
i never really covered greens thrm in my courses. so no.
anonymous
  • anonymous
its oki..would you be staying online today? if its oki to ask you other questions
amistre64
  • amistre64
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
anonymous
  • anonymous
oh so the set up is the same?
amistre64
  • amistre64
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
amistre64
  • amistre64
\[\int_{y=0}^{y=3}\int_{x=0}^{x=\sqrt{9-y^2}}\int_{z=0}^{z=\sqrt{9-y^2}}~dz~dx~dy\]
anonymous
  • anonymous
oki can you show me the frst integrration please
amistre64
  • amistre64
well, using your fundamental thrm of calculus ... \[\int_{0}^{b}dz=F(b)-F(0)\]
amistre64
  • amistre64
what is the integration of 1?
anonymous
  • anonymous
x
amistre64
  • amistre64
well, z in this case, since we are integrating with respect to z but yeah so z(b) - z(0) = z(b)
anonymous
  • anonymous
so with z...then we plug in the limits of x?
amistre64
  • amistre64
limits of z ...
amistre64
  • amistre64
\[\int_{0}^{\sqrt{9-y^2}}dz=z_b-z_a=\sqrt{9-y^2}-0\]
amistre64
  • amistre64
now, what is the middle integral? \[\int_{0}^{\sqrt{9-y^2}}\sqrt{9-y^2}~dx=??\]
anonymous
  • anonymous
\[\int\limits_{0}^{3} (1/2)(\sqrt{9-y^2}y+9\sin^{-1} (y/3)\]
amistre64
  • amistre64
not quite sqrt(9-y^2) , with respect to x is just some constant ... so, what is the integration of a constant again?
anonymous
  • anonymous
x^2/2
amistre64
  • amistre64
no, try again.
amistre64
  • amistre64
what was the integration of 1?
anonymous
  • anonymous
z
amistre64
  • amistre64
the respective variable yes ... in this case we are doing dx, so its an x
amistre64
  • amistre64
for some constant k \[\int_{0}^{b}dx=b(k)-0(k)\]
amistre64
  • amistre64
in this case, k=sqrt(9-y^2), and b=sqrt(9-y^2) leaving us: \[\int_{0}^{3}9-y^2~dy\]
anonymous
  • anonymous
oh ok so this is our middle integral
amistre64
  • amistre64
yes, we did the first, then the middle, and we are now left with the end
anonymous
  • anonymous
is the final answer 19?
amistre64
  • amistre64
dunno, i havent gotten to the end of it yet
amistre64
  • amistre64
i dont think 19 is it
anonymous
  • anonymous
18
amistre64
  • amistre64
what is: \[\int_{0}^{3}9-y^2~dy?\] work the steps for me
anonymous
  • anonymous
\[9y-y\]
anonymous
  • anonymous
9y-y^3/3
amistre64
  • amistre64
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.
anonymous
  • anonymous
144...can you remind me again why we work with 1/8
anonymous
  • anonymous
the limits for x and z why are they the same?
amistre64
  • amistre64
we dont have to, but due to the symmetry of the intersection, it makes life a little simpler for some.
amistre64
  • amistre64
those are better explained in the youtube video i posted.
anonymous
  • anonymous
oki i understand...so much more clear. thank you
amistre64
  • amistre64
good luck :)

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