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anonymous
 one year ago
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
anonymous
 one year ago
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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amistre64
 one year ago
Best ResponseYou've already chosen the best response.1well, what part are you having an issue with?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0the limit points im using if 3 to 3, squareroot (9x^2), squareroot (9x^2), squareroot (9y^2), squareroot (9y^2)

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0can you guide the first couple steps, im really confused please

amistre64
 one year ago
Best ResponseYou've already chosen the best response.1trying to see if i can see the boundary dw:1436633020633:dw

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0is it the inner part of cylinder we are looking for?

amistre64
 one year ago
Best ResponseYou've already chosen the best response.1are your limits valid? other than that it really is just a mechanical process.

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0there valid but im having trouble with the integration

amistre64
 one year ago
Best ResponseYou've already chosen the best response.1well lets see what we can work then dz from sqrt(9y^2) to sqrt(9y^2) dy from sqrt(9x^2) to sqrt(9x^2) dx from 3 to 3 you agree that this is how we want to move?

amistre64
 one year ago
Best ResponseYou've already chosen the best response.1then we simply work it from the inside out \[\int \left(\int\left[\int dz\right]~dy\right)~dx\] \[\int \left(\int z_bz_a~dy\right)~dx\] to start with

amistre64
 one year ago
Best ResponseYou've already chosen the best response.1\[\int \left(2\int \sqrt{9y^2}~dy\right)~dx\] let y = 3sin(t) , dy = 3cos(t) dt \[\int \left(2\int \sqrt{9(3sin(t))^2}~3cos(t)~dt\right)~dx\] \[\int \left(2\int \sqrt{9(1sin^2(t))}~3cos(t)~dt\right)~dx\] \[\int \left(2\int 3cos^2(t)*3cos(t)~dt\right)~dx\] \[\int \left(18\int cos^3(t)~dt\right)~dx\] etc ..

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0where did the sin^2t and cos come from

amistre64
 one year ago
Best ResponseYou've already chosen the best response.1i have a small error ... but its just using a trig substitution we let y = 3sin(t) and substitute all the y parts. sqrt(1sin^2) = cos ... not cos^2 tho

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0did you do squareroot (9y^2) ==> squareroot (9((9x^2)^(1/2))?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0where did you plug in the limits for y?

amistre64
 one year ago
Best ResponseYou've already chosen the best response.1no 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
 one year ago
Best ResponseYou've already chosen the best response.1my 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
 one year ago
Best ResponseYou've already chosen the best response.0right i see that...so my limits for dz are wrong?

amistre64
 one year ago
Best ResponseYou've already chosen the best response.1i 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
 one year ago
Best ResponseYou've already chosen the best response.0is 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 nonintersecting upper part

amistre64
 one year ago
Best ResponseYou've already chosen the best response.1uh, you originally asked for the volume of the intersection ...

amistre64
 one year ago
Best ResponseYou've already chosen the best response.1dw:1436634359956:dw so you want the volume of something like this?

amistre64
 one year ago
Best ResponseYou've already chosen the best response.1right, and you want the volume of .. what now?

amistre64
 one year ago
Best ResponseYou've already chosen the best response.1if its just 2 cylndars intersecting, i thing the profile of the intersection is a "pringle" chip .. or a saddle back

amistre64
 one year ago
Best ResponseYou've already chosen the best response.1dw:1436634627932:dw

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0yes ignore the red cylinder

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0from that part, how do you find the limits?

amistre64
 one year ago
Best ResponseYou've already chosen the best response.1now, 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

amistre64
 one year ago
Best ResponseYou've already chosen the best response.1ok, interSECTing part :) yeah z limits are fine, sqrt(9y^2) to sqrt(9y^2) x limits are fine, 3 to 3 its your y limits that need refined if anything

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0good to hear its not completey wrong. y limits how to find them?

amistre64
 one year ago
Best ResponseYou've already chosen the best response.1still working on that ....

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0its oki ! im still here

amistre64
 one year ago
Best ResponseYou've already chosen the best response.1https://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
 one year ago
Best ResponseYou've already chosen the best response.0would you happen to remember greens thm?

amistre64
 one year ago
Best ResponseYou've already chosen the best response.1i never really covered greens thrm in my courses. so no.

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0its oki..would you be staying online today? if its oki to ask you other questions

amistre64
 one year ago
Best ResponseYou've already chosen the best response.1im 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
 one year ago
Best ResponseYou've already chosen the best response.0oh so the set up is the same?

amistre64
 one year ago
Best ResponseYou've already chosen the best response.1the 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(9y^2) z = 0 to sqrt(9y^2) y = 0 to 3 and this defines 1/8 of our volume dz dx dy is a good order of integration

amistre64
 one year ago
Best ResponseYou've already chosen the best response.1\[\int_{y=0}^{y=3}\int_{x=0}^{x=\sqrt{9y^2}}\int_{z=0}^{z=\sqrt{9y^2}}~dz~dx~dy\]

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0oki can you show me the frst integrration please

amistre64
 one year ago
Best ResponseYou've already chosen the best response.1well, using your fundamental thrm of calculus ... \[\int_{0}^{b}dz=F(b)F(0)\]

amistre64
 one year ago
Best ResponseYou've already chosen the best response.1what is the integration of 1?

amistre64
 one year ago
Best ResponseYou've already chosen the best response.1well, z in this case, since we are integrating with respect to z but yeah so z(b)  z(0) = z(b)

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0so with z...then we plug in the limits of x?

amistre64
 one year ago
Best ResponseYou've already chosen the best response.1\[\int_{0}^{\sqrt{9y^2}}dz=z_bz_a=\sqrt{9y^2}0\]

amistre64
 one year ago
Best ResponseYou've already chosen the best response.1now, what is the middle integral? \[\int_{0}^{\sqrt{9y^2}}\sqrt{9y^2}~dx=??\]

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0\[\int\limits_{0}^{3} (1/2)(\sqrt{9y^2}y+9\sin^{1} (y/3)\]

amistre64
 one year ago
Best ResponseYou've already chosen the best response.1not quite sqrt(9y^2) , with respect to x is just some constant ... so, what is the integration of a constant again?

amistre64
 one year ago
Best ResponseYou've already chosen the best response.1what was the integration of 1?

amistre64
 one year ago
Best ResponseYou've already chosen the best response.1the respective variable yes ... in this case we are doing dx, so its an x

amistre64
 one year ago
Best ResponseYou've already chosen the best response.1for some constant k \[\int_{0}^{b}dx=b(k)0(k)\]

amistre64
 one year ago
Best ResponseYou've already chosen the best response.1in this case, k=sqrt(9y^2), and b=sqrt(9y^2) leaving us: \[\int_{0}^{3}9y^2~dy\]

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0oh ok so this is our middle integral

amistre64
 one year ago
Best ResponseYou've already chosen the best response.1yes, we did the first, then the middle, and we are now left with the end

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0is the final answer 19?

amistre64
 one year ago
Best ResponseYou've already chosen the best response.1dunno, i havent gotten to the end of it yet

amistre64
 one year ago
Best ResponseYou've already chosen the best response.1i dont think 19 is it

amistre64
 one year ago
Best ResponseYou've already chosen the best response.1what is: \[\int_{0}^{3}9y^2~dy?\] work the steps for me

amistre64
 one year ago
Best ResponseYou've already chosen the best response.19y  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
 one year ago
Best ResponseYou've already chosen the best response.0144...can you remind me again why we work with 1/8

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0the limits for x and z why are they the same?

amistre64
 one year ago
Best ResponseYou've already chosen the best response.1we dont have to, but due to the symmetry of the intersection, it makes life a little simpler for some.

amistre64
 one year ago
Best ResponseYou've already chosen the best response.1those are better explained in the youtube video i posted.

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0oki i understand...so much more clear. thank you
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