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Integrate by changing to polar coordinates.

  1. Mar 20, 2012 #1

    ElijahRockers

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    Gold Member

    EDIT: I found my mistake. Theta goes from 90 to 270, not -90 to 90. Wrong side of the y axis. That changes last integration to -2 instead of 2, making final answer -126.

    Evaluate the given integral by changing to polar coordinates.

    [itex]\int\int_R 3(x+y) dA[/itex]

    where R is the region that lies to the left of the y-axis between the circles x2 + y2 = 1 and x2 + y2 = 16.

    3. The attempt at a solution

    First I converted the region to integrate over in terms of theta and r.

    It would be [itex]\frac{-\pi}{2}\leq \theta \frac{\pi}{2}[/itex]

    and [itex]1 \leq r \leq 4[/itex]

    right? Usually this is the part I have problems with, so this may be wrong.

    Next I convert dA which will turn into [itex]rdrd\theta[/itex].

    Finally, the integrand 3(x+y) turns into 3r(cos(theta) + sin(theta)), and adding the extra r from dA, altogether getting

    [itex]\int_\frac{-\pi}{2}^\frac{\pi}{2}\int_{1}^{4} 3r^{2}(\cos\theta + \sin\theta) drd\theta[/itex].

    I got 126, but apparently that is wrong... did I set it up right? here's how I solved

    The 3(cos + sin) would be a constant with respect to r. integrating r2 from 1 to 4 gives me 21. 3*21=63, which is constant, so i am integrating (cos + sin) from -90° to 90°. I get 2 for this, and 63 times 2 is 126.
     
    Last edited: Mar 20, 2012
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