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6

Mathematica makes this fairly simple. We can define a region region = RegionDifference[Disk[{a, 0}, a], Disk[{0, 0}, b]] (* BooleanRegion[#1 && ! #2 &, {Disk[{a, 0}, a], Disk[{0, 0}, b]}] *) You can visualise this quite simply, for specific values of a and b with the following line Block[{a = 3, b = 1}, Region[region]] and an integrand ...


3

val = Values@Solve[y^2 == 3 x && y == 2 x - 6, {x, y}]; p1 = Plot[{Sqrt[3 x], -Sqrt[3 x], 2 x - 6}, {x, 0, 6}, AspectRatio -> 1, PlotStyle -> Black]; p2 = Plot[{Sqrt[3 x], -Sqrt[3 x]}, {x, 0, val[[1, 1]]}, AspectRatio -> 1, Filling -> {1 -> {2}}, FillingStyle -> LightBlue, PlotStyle -> Black]; p3 = Plot[{Sqrt[3 x], 2 ...


2

Area@ImplicitRegion[{y^2 < 3*x && y > 2*x - 6}, {{x, 0, 5}, {y, -3, 10}}] -(3/16) (3 Sqrt[17] - Sqrt[2] (9 - Sqrt[17])^(3/2) - Sqrt[2] (9 + Sqrt[17])^(3/2)) and BoundaryDiscretizeRegion[ ImplicitRegion[{y^2 < 3*x && y > 2*x - 6}, {{x, 0, 5}, {y, -3, 10}}], MaxCellMeasure -> (1 -> 0.1) ] Alternatively, you can ...


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