Question

Suppose W is the region outside the cylinder
x^2+y^2 = 1 and inside the sphere
x^2+y^2+z^2 = 2.
Calculate W

There are a couple ways I'm thinking of doing this - I want to take the triple integral, in spherical coordinates, of (x^2+y^2)drd(theta)d(phi) but I can't figure out what the limits for dr would be. I could also do cylindrical coordinates but again I'm having difficulty finding the limits for dr. I could also take the volume of the sphere in spherical coordinates and subtract from that the volume of the cylinder in cylindrical coordinates, which is easier I think.

Answer 1

I'm thinking just doing it in rectangular coordinates should be best. we can get the z value where the shapes intercept with a quick substitution, from which we can get the radius of the area of integration for the cylinder.

Answer 2

actually I think spherical coordinates are best. Have you found the z value at which the shpaes intersect?

Answer 3

@flynnali

Answer 4

I gave rectangular coordinates a shot - what I did was take the volume of the cylinder and subtract it from the volume of the sphere. I think the z value where the shapes intersect should be at 1? Since if you draw a right triangle with the hypotenuse as the radius going to the point where the cylinder first touches the surface of the sphere you can take that with the radius of the cylinder and use the pythagorean theorem to find sqrt(2)^2-1^2=1

I have no idea if my integral for the sphere is correct or not, though, wolfram-alpha won't give me an answer so probably it isn't

Answer 5

Thank you for your help, by the way!

Answer 6

my pleasure, it's been a minute since I've done these... the intersection won't be when z=1. Best first to make a drawing...

Answer 7

to find the z value we can just sub in the value for the circle to the sphere\[x^2+y^2=1\\x^2+y^2+z^2=2\]intersects when\[1+z^2=2\implies z=\pm\sqrt3\]

Answer 8

Shouldn't that be 1, though? since 2-1 = 1, and sqrt 1 =1?

Answer 9

also how do you make such tidy pictures, that's awesome

Answer 10

Also, I thought that the z value of the cylinder just extended out past the sphere - like a bead on a string, I guess? I thought this since there was no z value really defined in the equation, but it did confuse me

Answer 11

Oh yes of course, thank you
I've used this graphic for a long time :)
we can elaborate on our picture now a bit (which I messed up)

Answer 12

Would this be right?

Answer 13

yes the cylinder goes through it, but I just wanted to draw the area of integration D over which wecan get the volume of the cylinder

Answer 14

so we can get the limits of\(\phi\) from this, and double the integration on the upper half to find the volume of the shape the cylinder cuts out

Answer 15

Okay, got'cha!
the limit is just from 0 to pi/4, right?
so
\[\int\limits_{0}^{?}\int\limits_{0}^{2 \pi }\int\limits_{0}^{1}rdrd \theta dh\]

Answer 16

I'm not quite sure I understand how to get the limits of h, since that seems to be the difficult part; the cylinder has a rounded top since it extends past the sphere, right? So I could either just find the volume of the cylinder and then use spherical coordinates to find the volume of the section of sphere that's cut out, or I could try and come up with the limits for that weirdo round cylinder?

Answer 17

Also, do you know if I'm integrating the right thing? I don't know how to choose what goes in the place of "r" since it asked us to integrate (x^2+y^2); should I be integrating r^3?

Answer 18

yeah, that is an issue... i think the way I was going to do it was say that 0<r<sqrt2 and 0<phi<pi/4, but that will be a cone-shape... gotta think a bit more, sorry

Answer 19

No problem - don't worry about it :)

Answer 20

ok new plan; let's hope it works

Answer 21

I'm ready

Answer 22

let me rephrase that whole thing lol

Answer 23

I was trying to think about it but then I got confused xD thanks, haha

Answer 24

the caps are bounded like so

Answer 25

The picture makes it a lot clearer, but I'm still a bit confused about sin phi

Answer 26

yeah ignore that part :P
actually you could just say \[1\le z\le\sqrt2-r\]