Double Integrals in Polar Coordinates
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Double Integrals in Polar Coordinates

[From: ] [author: ] [Date: 12-10-31] [Hit: ]
Clearly, y = 0 ==> θ = 0.= (4/3)(√3 + 1).I hope this helps!......
A book would be useful if it gave you detailed steps to solving problems, unfortunately my book doesn't. Anyone care to help? Thanks.

Express the integral as an interated integral in polar coordinates, and then evaluate it.

∫ ∫ (x+y) dA, where R is the region in the first quadrant bounded by the lines y=0 and y=sqrt(3)x and the circle r=2.

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Note that when y = mx, we have tan θ = m.

So, y = x√3, we have tan θ = √3 ==> θ = π/3.
Clearly, y = 0 ==> θ = 0.

Converting to polar coordinates:
∫∫ (x + y) dA
= ∫(θ = 0 to π/3) ∫(r = 0 to 2) (r cos θ + r sin θ) * (r dr dθ)
= ∫(θ = 0 to π/3) (cos θ + sin θ) dθ * ∫(r = 0 to 2) r^2 dr
= [(sin θ - cos θ) {for θ = 0 to π/3}] * [(1/3)r^3 {for r = 0 to 2}]
= [(√3/2 - 1/2) - (0 - 1)] * (8/3)
= (4/3)(√3 + 1).

I hope this helps!
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