How Do Double Integrals and Polar Coordinates Relate to Symmetry Arguments?

[jenc305]

Please help. Thank you.

 

[Data]

Alright. The unit disk is

S = \{(x, \ y) | x^2 + y^2 \leq 1 \}

Changing to polar coordinates,

S = \{ (\rho \cos{\theta}, \ \rho \sin{\theta}) | \rho \leq 1, \ 0 \leq \theta \leq 2\pi \}

ie. x = \rho \cos{\theta}, \ y = \rho \sin{\theta}.

Thus

\int \int_S xy\sqrt{x^2 + y^2} dA = \int_0^{2\pi} \int_0^1 \rho \sin{\theta}\rho \cos{\theta} \sqrt{\rho^2 \sin^2{\theta} + \rho^2 \cos^2{\theta}} \ \left|\frac{\partial (x, \ y)}{\partial(\rho, \ \theta)}\right| \ d\rho \ d\theta = \int_0^{2\pi} \int_0^1 \rho^4\cos{\theta}\sin{\theta} \ d\rho \ d\theta

I trust you can work out the rest.

I will note that it is quite easy to evaluate this integral without actually doing any calculations, by appealing to symmetry.

 

[jenc305]

What if D is a closed disk with radius 1 and center (1,0).
That would make the polar coord. x=(r cos(theta)-1) and y=(r sin (theta)).

 

[Data]

It's actually still the same answer. You can make the same symmetry argument.

Here's a hint:

Obviously, if f is 2\pi-periodic, then \int_0^{2\pi} f(x)\sin{x} \ dx = \int_{-\pi}^{\pi} f(x)\sin{x} \ dx. Thus if f is an even function, since \sin is odd, then f(x)\sin{x} is odd and

\int_0^{2\pi} f(x)\sin{x} \ dx = 0