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Unformatted text preview: he graphs of the polar equations involved
to get a sense of the geometric situation. Since all of the equations in this example are found in
either Example 11.5.2 or Example 11.5.3, most of the work is done for us.
1. We know from Example 11.5.2 number 3 that the graph of r = 5 sin(2θ) is a rose. Moreover,
we know from our work there that as 0 ≤ θ ≤ π , we are tracing out the ‘leaf’ of the rose
which lies in the ﬁrst quadrant. The inequality 0 ≤ r ≤ 5 sin(2θ) means we want to capture
all of the points between the origin (r = 0) and the curve r = 5 sin(2θ) as θ runs through
0, π . Hence, the region we seek is the leaf itself.
4 π θ
x −5 (r, θ) : 0 ≤ r ≤ 5 sin(2θ), 0 ≤ θ ≤ π
2 2. We know from Example 11.5.3 number 3 that r = 3 and r = 6 cos(2θ) intersect at θ = π , so
the region that is being described here is the set of points whose directed distance r from the
origin is at least 3 but no more than 6 cos(2θ) as θ runs from 0 to π . In other words, we are
looking at the points outside or on the circle (since r ≥ 3) but inside or on the rose (since
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