# Double integral and polar coord

1. Mar 15, 2005

### jenc305

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2. Mar 15, 2005

### 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.

Last edited: Mar 15, 2005
3. Mar 15, 2005

### 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)).

Last edited: Mar 15, 2005
4. Mar 16, 2005

### 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

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

Last edited: Mar 16, 2005