Polar Coordinate Double Integral Area Problem

**1005** · July 18th 2009, 08:28 AM

The answer in the book is 1/4. I googled, and secant cubed is integration by parts. Before I went through all that, suspecting I set something up incorrectly, I used my ti-89 to calculate a numeric estimate. The number is negative and is not of the magnitude 1/4. What did I do wrong? Edit: My picture says tan when it should say arctan.

**galactus** · July 18th 2009, 09:03 AM

Try it this way:

$\frac{1}{((rcos{\theta})^{2}+(rsin{\theta})^{2})^{ 2}}=\frac{1}{r^{4}}$

$\int_{\frac{-\pi}{4}}^{\frac{\pi}{4}}\int_{sec(\theta)}^{\sqrt{ 2}}\frac{1}{r^{3}}drd{\theta}$

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