Area, moment of inertia, and radius of gyration of parabolic section

$3x^2 + 40y - 4800 = 0$

$3x^2 = -40y + 4800$

$x^2 = -\frac{40}{3}y + 1600$

$x^2 = -\frac{40}{3}(y - 120)$   →   Downward parabola with vertex at (0, 120)

$x^2 = -4ay$   →   Standard equation of downward parabola with vertex at the origin
 

At (40, -120)
$40^2 = -4a(-120)$

$a = 10/3$
 

Thus,
$x^2 = -4(10/3)y$

$x^2 = -\frac{40}{3}y$

$x = \sqrt{-\frac{40}{3}y}$
 

 

$\displaystyle I_x = 2\int_{-120}^0 (120 - y)^2 (x~dy)$

$\displaystyle I_x = 2\int_{-120}^0 (120 - y)^2 \sqrt{-\frac{40}{3}y}~dy$

$\displaystyle I_x = 2\sqrt{-\frac{40}{3}}\int_{-120}^0 (14\,400 - 240y + y^2)(y^{1/2})~dy$

$\displaystyle I_x = 2\sqrt{-\frac{40}{3}}\int_{-120}^0 (14\,400y^{1/2} - 240y^{3/2} + y^{5/2})~dy$

$I_x = 2\sqrt{-\dfrac{40}{3}}\left[ \dfrac{14\,400y^{3/2}}{3/2} - \dfrac{240y^{5/2}}{5/2} + \dfrac{y^{7/2}}{7/2} \right]_{-120}^0$

$I_x = 2\sqrt{-\dfrac{40}{3}}\left[ 9\,600y^{3/2} - 96y^{5/2} + \dfrac{2y^{7/2}}{7} \right]_{-120}^0$

$I_x = 0 - 2\sqrt{-\dfrac{40}{3}}\left[ 9\,600(-120)^{3/2} - 96(-120)^{5/2} + \dfrac{2(-120)^{7/2}}{7} \right]$

$I_x = 21,065,142.86 \, \text{ unit}^4$       (okay!)

Recall the following moment of inertia for a rectangle of height h and width b.
Centroidal moment of inertia = bd³/12
Moment of inertia at the base = bd³/3
 

For the vertical strip shown below, b = dx and d = y
 

 

$3x^2 + 40y - 4800 = 0$

$40y = 4800 - 3x^2$

$y = 120 - \dfrac{3}{40}x^2$
 

$\displaystyle I_x = 2\int_0^{40} \dfrac{dx~y^3}{3}$

$\displaystyle I_x = \dfrac{2}{3}\int_0^{40} y^3~dx$

$\displaystyle I_x = \dfrac{2}{3}\int_0^{40} (120 - \dfrac{3}{40}x^2)^3~dx$

$\displaystyle I_x = \dfrac{2}{3}\int_0^{40} [ \, 120^3 - 3(120^2)(\dfrac{3}{40}x^2) + 3(120)(\dfrac{3}{40}x^2)^2 - (\dfrac{3}{40}x^2)^3~dx$

$\displaystyle I_x = \dfrac{2}{3}\int_0^{40} (1728000 - 3240x^2 + \dfrac{81}{40}x^4 - (\dfrac{27}{64\,000}x^6)~dx$

$I_x = \dfrac{2}{3} \left[1728000x - 1080x^3 + \dfrac{81x^5}{200} - (\dfrac{27x^7}{448\,000} \right]_0^{40}$

$I_x = \dfrac{2}{3}\left[1728000(40) - 1080(40^3) + \dfrac{81(40^5)}{200} - (\dfrac{27(40^7)}{448\,000} \right] - 0$

$I_x = 21\,065\,142.86 \, \text{ unit}^4$       (okay!)