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Mathematics
Mathematics
Problem
Briefly explain why, when computing the exact area enclosed by the curve
r
=
3
sin
θ
r = 3\sin \theta
r
=
3
sin
θ
for
0
≤
θ
≤
π
0\le \theta \le \pi
0
≤
θ
≤
π
, one may integrate over
0
≤
θ
≤
π
2
0 \le \theta \le \tfrac{\pi}{2}
0
≤
θ
≤
2
π
and then double the result.
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