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

1. anonymous

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

You get the same answer sticking to the rectilinear system. If you need help with that, let me know.

3. anonymous

cheers

4. anonymous

np