Question

Why do "Odd integrands" = 0?

I'm trying to derive the formula for the volume of a torus. I'm using this explanation:

http://answers.yahoo.com/question/index?qid=20101104151437AAcr752

and at one point the equation is equal to
4π [∫(z = -r to r) z√(r² - z²) dz + ∫(z = -r to r) R√(r² - z²) dz]

apparently the integral z√(r² - z²) dz is equal to zero "since the first integrand is odd on [-r, r]". I don't know what this means. Can anyone please explain? Thanks a lot!!

Answer 1

consider \[\int\limits_{-1}^{1}x dx=(x ^{2}/2)_{-1}^{1}=0\]

Answer 2

another way to make it zero ,x is an odd function as if f(x)=x
f(-x)=-f(x)=-x=-f(x)
so,f(x) is x is an odd function
so,the integrand is zero

Answer 3

fine ???

Answer 4

ahh thanks so much! i think i understand so far...

Answer 5

how come the second integrand is not equal to zero though, even though it's practically the same thing but with a different constant?

Answer 6

which is the second integrand

Answer 7

see if the function between limits -r to r is odd ,,,,,,then only the answer is zero

Answer 8

odd function.....i guess you know

Answer 9

hmm..

Answer 10

i'm confused..but thank you for the help

Answer 11

confused at which step

Answer 12

it is because of the definition of the integral.
the area u define either through Riemman's or Darboux's is "algebraic", i.e. it has a sign

Answer 13

hmm, well I appreciate it guys

Answer 14

not sure exactly where im lost. seems like i need to brush up on some concepts. Don't see why the sign makes it = 0