Question

Show that the integral of x/(x^2+y^2) from 0 to 2pi is zero, where r = 1+ sin(θ).

Answer 1

there is no r

Answer 2

integrate w.r.t which variable?

Answer 3

\[\int\limits_{0}^{2\pi} x/(x^2+y^2)\]

Answer 4

\[x ^{2}+y ^{2}=r ^{2}\]

Answer 5

Convert to polar first where x = rcos θ and y = sin θ

Answer 6

is the eqwuation r=1+sin(theta) right?

Answer 7

ok ,but is it dx or dy

Answer 8

and y=sin theta????

Answer 9

In order to show that the integral is zero, I think i need to show that the integrand function is an odd function.

Answer 10

i think it is d theea

Answer 11

yea its dθ

Answer 12

indeed it is a odd function....as only x is there...

Answer 13

the denominator is x^2+y^2...which is an even func...

Answer 14

but i have to show that x is an oddd function.

Answer 15

see there is some problem which u r saying....u say r=1+sin(theta)
again u say x=rcos(theta) y=sin(thetea) how is it possible?

Answer 16

ans is 0

Answer 17

rony did u considered r=1+sin(theta)?

Answer 18

otherwise it is zero as u only have to integrate cos theta over a full cycle....

Answer 19

hello.........rony r u there?

Answer 20

\[\int\limits_{0}^{2 \pi}\frac{rcos \theta }{r^2}d \theta=\frac1r \int\limits_{0}^{2 \pi }\cos \theta d \theta=0\]

Answer 21

Well Im actually trying to integrate the function as a line integral over the cardiod curve r = 1+ sinθ.

Answer 22

See, x = rcosθ = (1+sinθ) cosθ
But I cant prove that x is an odd function.

Answer 23

Of course not, because it is not. What you need is \[\int_0^π (1+\sin θ)\cos θ = - \int_π^{2π}(1+\sinθ)\cos θ\] So that \((1+\sin (θ+π))\cos (θ+π)\) is odd.

Answer 24

\[x=\cos \theta + \sin \theta *\cos \theta =\cos \theta +\frac{\sin 2 \theta}{2}\]

Answer 25

nowhereman is close to what im trying

Answer 26

Yeah, but it won't work. dipankar did a good step there. The reason the integral is zero is that you are integrating over integral multiples of the period.