Is the Integral of f(x, y) Over the Unit Circle Zero?

[iloveannaw]

Homework Statement

Express

f(x,y) = 1/sqrt(x^2 + y^2) . (y/sqrt(x^2 + y^2)) .exp(-2sqrt(x^2 + y^2))

in terms of polar coordinates \rho and \varphi then evaluate the integral over a circle of radius 1, centered at the origin.

Homework Equations

x = \rhocos\varphi
y = \rhosin\varphi

sin^2\varphi + cos^2\varphi = 1

The Attempt at a Solution

ok so here's my effort

after rearranging and substituting: f(\rho,\varphi) = sin\varphiexp(-2\rho)

now let's integrate!
limits are 0 \leq \rho \leq1
and 0 \leq \varphi \leq 2\pi

\int\int sin\varphiexp(-2\rho) d(fi) d(rho)

the problem is sin becomes -cos so, -cos(2pi) - -cos(0) = 0

giving a final answer of zero doesn't make much sense, does it? so what arent i getting?

 

[Altabeh]

Homework Statement

Express

f(x,y) = 1/sqrt(x^2 + y^2) . (y/sqrt(x^2 + y^2)) .exp(-2sqrt(x^2 + y^2))

in terms of polar coordinates \rho and \varphi then evaluate the integral over a circle of radius 1, centered at the origin.

Homework Equations

x = \rhocos\varphi
y = \rhosin\varphi

sin^2\varphi + cos^2\varphi = 1

The Attempt at a Solution

ok so here's my effort

after rearranging and substituting: f(\rho,\varphi) = sin\varphiexp(-2\rho)

now let's integrate!
limits are 0 \leq \rho \leq1
and 0 \leq \varphi \leq 2\pi

\int\int sin\varphiexp(-2\rho) d(fi) d(rho)

the problem is sin becomes -cos so, -cos(2pi) - -cos(0) = 0

giving a final answer of zero doesn't make much sense, does it? so what arent i getting?

Didn't you drop a \rho in the integrand?

AB

 

[vela]

You made a mistake when converting f to polar coordinates, and you made another one in writing down the integral. It turns out they cancel each other, so you got the right answer, which is 0.

Note that the original integrand is an odd function of y. Since the unit circle is symmetric about the y-axis, the integral turns out to be 0.