H Howard Lifer Oct 14, 1999 47,989 10 81 Oct 20, 2006 #1 S(C(R^2 - r^2) 2pi*r dr) so that becomes 2*pi*C*S(r(R^2 - r^2)dr) R is a constant. Any ideas?
B BrownTown Diamond Member Dec 1, 2005 5,314 1 0 Oct 20, 2006 #2 http://integrals.wolfram.com/index.jsp
Random Variable Lifer Aug 10, 2001 10,424 2 0 Oct 20, 2006 #3 I would help you if I understood your notation. Is S supposed to be an integral sign? And is C another constant?
I would help you if I understood your notation. Is S supposed to be an integral sign? And is C another constant?
H Howard Lifer Oct 14, 1999 47,989 10 81 Oct 20, 2006 #5 OK, now I feel like an idiot. I hate forgetting elementary stuff, but more than that, I hate having to remember it. Damn fluid mechanics.
OK, now I feel like an idiot. I hate forgetting elementary stuff, but more than that, I hate having to remember it. Damn fluid mechanics.
H Howard Lifer Oct 14, 1999 47,989 10 81 Oct 20, 2006 #7 Couldn't you just expand the brackets and integrate each term?
Random Variable Lifer Aug 10, 2001 10,424 2 0 Oct 20, 2006 #8 Originally posted by: Howard Couldn't you just expand the brackets and integrate each term? Click to expand... yes
Originally posted by: Howard Couldn't you just expand the brackets and integrate each term? Click to expand... yes
H Howard Lifer Oct 14, 1999 47,989 10 81 Oct 20, 2006 #9 http://psu.thetruck.net/Chapter%203%20solutions.pdf 1st question. I can't get the same answer. ugh, i'm fvcked for this chapter.
http://psu.thetruck.net/Chapter%203%20solutions.pdf 1st question. I can't get the same answer. ugh, i'm fvcked for this chapter.
Random Variable Lifer Aug 10, 2001 10,424 2 0 Oct 20, 2006 #10 let u = R^2-r^2 then du = -2rdr -Pi*C?udu = -Pi*C*u^2/2 = -Pi/2*C*(R^2-r^2)^2 now evaluate it at the limits -Pi/2*C*[0-R^2]^2 = Pi/2*C*R^4
let u = R^2-r^2 then du = -2rdr -Pi*C?udu = -Pi*C*u^2/2 = -Pi/2*C*(R^2-r^2)^2 now evaluate it at the limits -Pi/2*C*[0-R^2]^2 = Pi/2*C*R^4
Random Variable Lifer Aug 10, 2001 10,424 2 0 Oct 20, 2006 #12 Originally posted by: Howard oh, right I'm screwed Click to expand... You can't follow what I did? It's really quite a simple integral.
Originally posted by: Howard oh, right I'm screwed Click to expand... You can't follow what I did? It's really quite a simple integral.
H Howard Lifer Oct 14, 1999 47,989 10 81 Oct 20, 2006 #13 No, I can follow it, but I can't do integrals like I used to.
Random Variable Lifer Aug 10, 2001 10,424 2 0 Oct 20, 2006 #14 I'm fairly certain that in a fluid dynamics course you're going to come across all kinds of integrals. particularly surface integrals
I'm fairly certain that in a fluid dynamics course you're going to come across all kinds of integrals. particularly surface integrals