Parabolic Segment

The Arc Length of the parabolic segment shown above is given by

$$s=\sqrt{4x^2+y^2}+{y^2\over 2x}\ln\left({2x+\sqrt{4x^2+y^2}\over y}\right).$$ (1)

The Area contained between the curves

$$y = x^2$$ (2)

$$y = ax+b$$ (3)

can be found by eliminating $y$,

$$x^2-ax-b=0,$$ (4)

so the points of intersection are

$$x_\pm ={\textstyle{1\over 2}}(a\pm\sqrt{a^2+4b}\,).$$ (5)

Therefore, for the Area to be Nonnegative, $a^2+4b>0$, and

$$x_\pm = {\textstyle{1\over 4}}(a^2\pm2a\sqrt{a^2+b^2}+a^2+4b)$$

$$= {\textstyle{1\over 4}}(2a^2+4b\pm 2a\sqrt{a^2+4b}\,)$$

$$= {\textstyle{1\over 2}}(a^2+2b\pm a\sqrt{a^2+4b}\,),$$ (6)

so the Area is

$$A = \int_{x_-}^{x_+} [(ax+b)-x^2]\,dx$$

$$= \left[{{\textstyle{1\over 2}}ax^2+bx-{\textstyle{1\over 3}}x^3}\right]^{(a+\sqrt{a^2+4b}\,)/2}_{(a-\sqrt{a^2+4b}\,)/2}.$$ (7)

Now,

$${x_+}^2-{x_-}^2 = {\textstyle{1\over 4}}\left[{(a^2+2a\sqrt{a^2+4b}+a^2+4b)-(a^2-2a\sqrt{a^2+4b}+a^2+4b)}\right]$$

$$= {\textstyle{1\over 4}}\left[{4a\sqrt{a^2+4b}\,}\right]= a\sqrt{a^2+4b}$$ (8)

$${x_+}^3-{x_-}^3 = (x_+-x_-)({x_+}^2+x_-x_++{x_-}^2)$$

$$= \sqrt{a^2+4b}\left\{{{\textstyle{1\over 4}}(a^2+2a\sqrt{a^2+4b}+a... ...r 4}}[a^2-(a^2+4b)]+{\textstyle{1\over 4}}(a^2-2a\sqrt{a^2+4b}+a^2+4b)}\right\}$$

$$= {\textstyle{1\over 4}}\sqrt{a^2+4b}\, (4a^2+4b) = \sqrt{a^2+4b}\,(a^2+b).$$ (9)

So

$$A = {\textstyle{1\over 2}}a^2\sqrt{a^2+4b}+b\sqrt{a^2+4b}={\textstyle{1\over 3}}(a^2+b)\sqrt{a^2+4b}$$

$$= \sqrt{a^2+4b}\left[{({\textstyle{1\over 2}}-{\textstyle{1\over 3}}) a^2+b(1-{\textstyle{1\over 3}})}\right]$$

$$= ({\textstyle{1\over 6}}a^2+{\textstyle{2\over 3}}b)\sqrt{a^2+4b}$$

$$= {\textstyle{1\over 6}}(a^2+4b)\sqrt{a^2+4b}={\textstyle{1\over 6}} (a^2+4b)^{3/2}.$$ (10)

We now wish to find the maximum Area of an inscribed Triangle. This Triangle will have two of its Vertices at the intersections, and Area

$$A_\Delta = {\textstyle{1\over 2}}(x_-y_*-x_*y_--x_+y_*+x_*y_++x_+y_--x_-y_+).$$ (11)

But $y_*={x_*}^2$, so

$$A_\Delta = {\textstyle{1\over 2}}(x_-{x_*}^2-x_*y_--x_+{x_*}^2+x_*y_*+x_+y_--x_-y_+)$$

$$= {\textstyle{1\over 2}}[-{x_*}^2(x_+-x_-)+x_*(y_+-y_-)+(x_+y_--x_-y_+)].$$ (12)

The maximum Area will occur when

$${\partial A_\Delta\over \partial x_*} = {\textstyle{1\over 2}}[-2(x_+-x_-)x_*+(y_+-y_-)]=0.$$ (13)

But

$$x_+-x_- = \sqrt{a^2+4b}$$ (14)

$$y_+-y_- = a\sqrt{a^2+4b},$$ (15)

so

$$x_*={1\over 2} {y_+-y_-\over x_+-x_-} = {\textstyle{1\over 2}}a$$ (16)

and

$$A_\Delta = {\textstyle{1\over 2}}[-({\textstyle{1\over 2}}a)^2(x_+-x_-)+({\textstyle{1\over 2}}a)(y_+-y_-)+(x_+y_--x_-y_+)].$$ (17)

Working on the third term

$$x_+y_- = {\textstyle{1\over 4}}(a+\sqrt{a^2+4b}\,)(a^2+2b-a\sqrt{a^2+4b}\,)$$

$$= {\textstyle{1\over 4}}\left[{a^3+2ab-a^2\sqrt{a^2+4b}+a^2\sqrt{a^2+4b}+2b\sqrt{a^2+4b}-a(a^2+4b)}\right]$$

$$= {\textstyle{1\over 4}}[-2ab+2b\sqrt{a^2+4b}\,]$$ (18)

$$x_-y_+ = {\textstyle{1\over 4}}(a-\sqrt{a^2+4b}\,)(a^2+2b+a\sqrt{a^2+4b}\,)$$

$$= {\textstyle{1\over 4}}\left[{a^3+2ab+a^2\sqrt{a^2+4b}-a^2\sqrt{a^2+4b}-2b\sqrt{a^2+4b}-a(a^2+4b)}\right]$$

$$= {\textstyle{1\over 4}}[-2ab-2b\sqrt{a^2+4b}\,],$$ (19)

so

$$x_+y_--x_-y_+={\textstyle{1\over 4}}(4b\sqrt{a^2+4b}) = b\sqrt{a^2+4b}$$ (20)

and

$$A_\Delta = {\textstyle{1\over 2}}(-{\textstyle{1\over 4}}a^2\sqrt{a^2+4b}+{\textstyle{1\over 2}}a^2\sqrt{a^2+4b}+b\sqrt{a^2+b^2}\,)$$

$$= {\textstyle{1\over 2}}\sqrt{a^2+4b} \left[{({\textstyle{1\over 2}... ...2+b}\right]= {\textstyle{1\over 2}}\sqrt{a^2+4b}\,({\textstyle{1\over 4}}a^2+b)$$

$$= {\textstyle{1\over 8}} \sqrt{a^2+4b}\,(a^2+4b) = {\textstyle{1\over 8}} (a^2+4b)^{3/2},$$ (21)

which gives the result known to Archimedes in the third century BC that

$${A\over A_\Delta} = {{\textstyle{1\over 6}}\over {\textstyle{1\over 8}}} = {4\over 3}\,.$$ (22)

The Area of the parabolic segment of height $h$ opening upward along the y-Axis is

$$A=2\int_0^h \sqrt{y}\,dy={\textstyle{1\over 3}} h^{3/2}.$$ (23)

The weighted mean of $y$ is

$$\left\langle{y}\right\rangle{}=2\int_0^h y\sqrt{y}\,dy=2\int_0^h y^{3/2}\,dy={\textstyle{4\over 5}}h^{5/2}.$$ (24)

The Centroid is then given by

$$\bar y={\left\langle{y}\right\rangle{}\over A}={\textstyle{3\over 5}}h.$$ (25)

See also Centroid (Geometric), Parabola, Segment

References

Beyer, W. H. (Ed.) CRC Standard Mathematical Tables, 28th ed. Boca Raton, FL: CRC Press, p. 125, 1987.