Calculating Area of Plane Regions | r2sin(2theta) and r^2 < 4

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Homework Statement


Find the area of the region in the plane where, r2sin(2theta) >2(sq.root3) , r^2 < 4

Homework Equations

The Attempt at a Solution


To try to visualize the problem a little better I converted from r2sin(2theta) to 2xy. However I'm confused after this, since I don't know what the upper limit to integrate is. Also, in the context of the question what does r^2 < 4 mean? Thanks very much. :)
 
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nanostudy said:

Homework Statement


Find the area of the region in the plane where, r2sin(2theta) >2(sq.root3) , r^2 < 4

Homework Equations

The Attempt at a Solution


To try to visualize the problem a little better I converted from r2sin(2theta) to 2xy. However I'm confused after this, since I don't know what the upper limit to integrate is. Also, in the context of the question what does r^2 < 4 mean? Thanks very much. :)

Recall that r = \sqrt{x^2 + y^2}, so the inequality r^2 &lt; 4 covers all points within a circle of radius 2 about the origin in the xy-plane. Sketch the graph of this bounding circle, and the graph of the bounding curve 2xy = 2\sqrt{3} (a rectangular hyperbola). Then shade in the regions that satisfy the inequality. Use the boundaries of those regions to define your integrals.
 
r^2&lt; 4, in polar coordinates, is the same as "r< 2" since r is not negative. That is the interior of a circle, centered at the origin, with radius 2
 
There are two things I don't understand about this problem. First, when finding the nth root of a number, there should in theory be n solutions. However, the formula produces n+1 roots. Here is how. The first root is simply ##\left(r\right)^{\left(\frac{1}{n}\right)}##. Then you multiply this first root by n additional expressions given by the formula, as you go through k=0,1,...n-1. So you end up with n+1 roots, which cannot be correct. Let me illustrate what I mean. For this...
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