Next you will have to take some care in nding the

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Unformatted text preview: β. E 5 S In polar coordinates, the curve can be rewritten as 4 =( 3 cos3 θ − 3( cos θ )( sin θ )2 ) Or, = (cos3 θ − 3 cos θ sin2 θ ) Note that, we have cos(3θ ) = cos(θ + 2θ ) = cos θ cos(2θ ) − sin θ sin(2θ ) = cos θ (cos2 θ − sin2 θ ) − 2 sin2 θ cos θ = cos3 θ − 3 cos θ sin2 θ Thus, the curve is given by = cos 3θ in polar coordinates. Further, since ≥ 0, we must have cos 3θ ≥ 0, that is, − π + 2π ≤ 3θ ≤ π + 2π or − π + 2π ≤ θ ≤ π + 2π . 2 2 6 3 6 3 Further, we consider values of θ on a particular interval of length 2π . Here it’s convenient to let θ ∈ − π 11π and then we see that solutions of cos 3θ ≥ 0 for − π ≤ 6 6 6 π π π π θ ≤ 11π are intervals − π π , 36 56 , 76 96 . Obviously, all three integrals are equal (it 6 66 means that the curve has a rotational symmetry — it doesn’t change under rotation by 120◦ ). Apply the result in exercise 4, we conclude that the area is 3 2 π 6 2 −π 6 π 6 3 (θ ) θ = 2 2 −π 6 Evaluate R2 − − (The improper integral R2 − 2π Cart...
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