Find the parametrization of the curve x^(2/3) + y^(2/3) = 1 and compute the area of the interior.
I've had no trouble with other Green's Theorem problems but can't get this one. Thanks.
I've had no trouble with other Green's Theorem problems but can't get this one. Thanks.
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Parameterization: x = cos^3(t), y = sin^3(t).
So, the area (by Green's theorem) equals
(1/2) ∫c (x dy - y dx)
= ∫(t = 0 to 2π) [cos^3(t) * 3 sin^2(t) cos t - sin^3(t) * -3 cos^2(t) sin t] dt
= ∫(t = 0 to 2π) 3 cos^2(t) sin^2(t) [cos^2(t) + sin^2(t)] dt
= ∫(t = 0 to 2π) 3 cos^2(t) sin^2(t) dt
= ∫(t = 0 to 2π) (3/4) sin^2(2t) dt
= ∫(t = 0 to 2π) (3/4) * (1/2)(1 - cos(4t)) dt
= (3/8) (t - sin(4t)/4) {for t = 0 to 2π}
= 3π/4.
I hope this helps!
So, the area (by Green's theorem) equals
(1/2) ∫c (x dy - y dx)
= ∫(t = 0 to 2π) [cos^3(t) * 3 sin^2(t) cos t - sin^3(t) * -3 cos^2(t) sin t] dt
= ∫(t = 0 to 2π) 3 cos^2(t) sin^2(t) [cos^2(t) + sin^2(t)] dt
= ∫(t = 0 to 2π) 3 cos^2(t) sin^2(t) dt
= ∫(t = 0 to 2π) (3/4) sin^2(2t) dt
= ∫(t = 0 to 2π) (3/4) * (1/2)(1 - cos(4t)) dt
= (3/8) (t - sin(4t)/4) {for t = 0 to 2π}
= 3π/4.
I hope this helps!
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That took me so long but I finally got it. My answer was also 3π/4.
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