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anonymous

  • 5 years ago

Evaluate the double integral D4x−4ydA , D= is the quarter circle in the first quadrant with center at the origin and radius 5

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  1. anonymous
    • 5 years ago
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    You can do this a couple of ways. You can stay in the rectilinear system, or shift into polar coordinates. In the polar system, you can make the following transformations,\[x=r \cos \theta, y = r \sin \theta\]and\[dA=r dr d \theta\]Given the description of your domain, you're integrating over a radial distance 0 to 5, and your angle swept is from 0 to pi/2. Your integral is then,\[\int\limits_{0}^{r}\int\limits_{0}^{\pi/2}(4r \cos \theta - 4 r \sin \theta )r dr d \theta \]\[=4\int\limits_{0}^{r=5}\int\limits_{0}^{\theta = \frac{\pi}{2}}r^2(\cos \theta - \sin \theta) d \theta dr\]\[=4\int\limits_{0}^{5}r^2dr \int\limits_{0}^{\pi/2}\cos \theta - \sin \theta d \theta\]\[=4\int\limits_{0}^{5}r^2dr \left[ \sin \theta + \cos \theta \right]_0^{\pi/2}=4\int\limits_{0}^{5}r^2dr (0)=0\]

  2. anonymous
    • 5 years ago
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    You get the same answer sticking to the rectilinear system. If you need help with that, let me know.

  3. anonymous
    • 5 years ago
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    cheers

  4. anonymous
    • 5 years ago
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    np

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