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

- Stacey Warren - Expert brainly.com

Hey! We 've verified this expert answer for you, click below to unlock the details :)

- jamiebookeater

I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!

- amistre64

well, what part are you having an issue with?

- 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

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

Looking for something else?

Not the answer you are looking for? Search for more explanations.

## More answers

- amistre64

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

- anonymous

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

- amistre64

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

- anonymous

there valid but im having trouble with the integration

- 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

yes

- 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

\[\int \left(2\int \sqrt{9-y^2}~dy\right)~dx\]
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

where did the sin^2t and cos come from

- 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

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

- anonymous

where did you plug in the limits for y?

- 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

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

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

- 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

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

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

- amistre64

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

- anonymous

##### 1 Attachment

- anonymous

it looks like this

- amistre64

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

- amistre64

if its just 2 cylndars intersecting, i thing the profile of the intersection is a "pringle" chip .. or a saddle back

- amistre64

|dw:1436634627932:dw|

- anonymous

yes ignore the red cylinder

- anonymous

exactly like that

- anonymous

from that part, how do you find the limits?

- 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

interesting part

- 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

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

- amistre64

still working on that ....

- anonymous

its oki ! im still here

- 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

thank you so much

- anonymous

would you happen to remember greens thm?

- amistre64

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

- anonymous

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

- 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

oh so the set up is the same?

- 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

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

- anonymous

oki can you show me the frst integrration please

- amistre64

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

- amistre64

what is the integration of 1?

- anonymous

x

- amistre64

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

- anonymous

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

- amistre64

limits of z ...

- amistre64

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

- amistre64

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

- anonymous

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

- 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

x^2/2

- amistre64

no, try again.

- amistre64

what was the integration of 1?

- anonymous

z

- amistre64

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

- amistre64

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

- 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

oh ok so this is our middle integral

- amistre64

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

- anonymous

is the final answer 19?

- amistre64

dunno, i havent gotten to the end of it yet

- amistre64

i dont think 19 is it

- anonymous

18

- amistre64

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

- anonymous

\[9y-y\]

- anonymous

9y-y^3/3

- 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

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

- anonymous

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

- amistre64

we dont have to, but due to the symmetry of the intersection, it makes life a little simpler for some.

- amistre64

those are better explained in the youtube video i posted.

- anonymous

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

- amistre64

good luck :)

Looking for something else?

Not the answer you are looking for? Search for more explanations.