Question

Evaluate the triple integral of f(x,y,z)=cos(x^2+y^2) over the solid cylinder with height 3 and with base of radius 3 centered on the z- axis at z=−1.

I need help starting this problem. Do I need to convert to cylindrical coordinates? (How do I know when I need to do this?) What do I do with the information about the z? Will the bounds for my z be -1 to 3 if I stay in Cartesian?

Answer 1

The choice of coordinates is completely arbitrary, but it's usually helpful to change coordinates in cases where you see something like \(x^2+y^2\) or when there's something round like a cylinder or sphere. The coordinates don't change the problem, they just kinda like change the wire frame perspective in which you're looking at the problem and if you use a coordinate system that matches the symmetry of your problem better, the easier it usually is!

Figuring out how to do that is kind of an art form and it's sorta something you just gotta practice to get used to. One of the main things to keep yourself on track is to draw out the region of integration and it usually gives you some kind of grounding on how to make sure your bounds of integration are right when you change... That being said, I say go for it, draw the region of integration as best as you can and try to change to cylindrical coordinates as best as you can and I'll try to help you figure it out.

Answer 2

I gave it a shot!

Answer 3

if it is centred on z = -1 and has height 3 then i think you are looking at this

\(\Large \int\limits_{\theta = 0}^{2 \pi} ~ \int\limits_{z = -2.5}^{0.5} ~ \int\limits_{r = 0}^{3}\)

plus it's \(\large \cos r ^ {\color{red}{2} }\) which makes the integration easier :-)

Answer 4

Thanks, that seems like a correct interpretation of z! But are you saying that cos(r)r=cos(r^2)? I'm not so sure that's right. I integrated both ways and still didn't get the right answer :/

Answer 5

hi

i'm going by how you framed the question in the OP: \(f(x,y,z)= \cos(x^2+y^2) = \cos r^\color{red}{2}\)

i make it \(2 \pi \times 3 \times \dfrac{\sin 9}{2} = 3.8814......\), no need to integrate...

9 in rads?

or the long way round:
https://www.wolframalpha.com/input/?i=%5Cint_%7Bt+%3D+0%7D%5E%7B2+pi%7D+%5Cint_%7Bz%3D-2.5%7D%5E%7B0.5%7D+%5Cint_%7Br%3D0%7D%5E%7B3%7D+r+cos+(r%5E2)+dr+dz+dt

Answer 6

sorry, that was a really dumb thing to say, of course you need to integrate. what i mean was that i think the 2 outer integrals you can just multiply as you get a constant from the inner one...hope that makes more sense.

Answer 7

ah, ok I see. I was so convinced this whole time that r=x^2+y^2. My mistake, haha. I got the right answer now. Thanks a bunch!!