# Thread: Archimedes' Parabolic Area Formula

1. ## Archimedes' Parabolic Area Formula

The area under a parabolic curve = (2/3)bh. Sketch the graph of the parabolic arch y = h- (4h/b^2)x^2, (-b/2)=< x =< b/2 , assuming that h & b are positive. Then use calculus to find the area of the region enclosed between the arch and the x-axis.

I have no idea how to go about even starting to graph that, can anyone please explain how this would be achieved? Thanks

2. I don't know how to draw here but I can describe the graph.
Because of the -x^2, the parabola opens downward. Its axis of symmetry is x = 0 por the y-axis. Its vertex is at (0,h). It intersects the x-axis at (-b/2,0) and (b/2,0)

The area in question:
dA = y*dx
Boundaries for dx are from x = -b/2 to x = b/2
So,
A = INT.(-b/2 --> b/2)[h -(4h/(b^2))(x^2)]dx
A = [h*x -(1/3)(4h /(b^2))(x^3)] | (-b/2 --> b/2)
A = [h(b/2) -(1/3)(4h /b^2)(b^3 / 8)] -[h(-b/2) -(1/3)(4h /b^2)(-b^3 / 8)]
A = bh/2 -(1/3)(bh/2) +bh/2 -(1/3)(bh/2)
A = bh -(1/3)bh

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