To determine an area in polar coordinates, it first helps to look at the graph that is depicted by the given parameters:
r=−θsin(−16θ2+7π12) with θ∈[0,π4]
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Not every bit of the graph in within [0,π4] is an enclosed area or loop. Once we find the r and θ values corresponding to the three loops, we can use them as limits of integration. Each area formed by a loop can be determined by the formula:
A=∫ba12f[θ]2dθ
Three loops, means three integrals. The roots of radius, r, will give us the limits of integration.
−θsin(−16θ2+7π12)=0
The trivial solution is when θ=0. Because the sine function is zero at πn, we can say sin(−16θ2+7π12)=0 whenever
−16θ2+7π12=πn
−16θ2=πn−7π12
θ2=−πn16+7π192
θ=±√−πn16+7π192
We know θ is restricted to the interval [0,π/4]. Because our interval involves only positive values, then we can write:
0≤√−πn16+7π192≤π4
0≤−πn16+7π192≤π216
0≤−πn+7π12≤π2
−7π12≤−πn≤π2−7π12
712≥n≥−π+712
−2.55826≤n≤0.583333
The integers, n, that lie on this interval are −2,−1,0. The respective angles, θ, with those values of n are θ=0 and:
θ≈0.3384,0.5576,0.7122
Now we have our limits of integration for three integrals.
A1=∫0.3384012[−θsin(−16θ2+7π12)]2dθ
A2=∫0.55760.338412[−θsin(−16θ2+7π12)]2dθ
A3=∫0.71220.557612[−θsin(−16θ2+7π12)]2dθ
Using a computer algebra system, we get
A1=0.0057972
A2=0.0225589
A3=0.0313656
ATotal=0.0597217
