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

Situation
Given the parabola 3x² + 40y – 4800 = 0.

Part 1: What is the area bounded by the parabola and the X-axis?
A. 6 200 unit²
B. 8 300 unit²
C. 5 600 unit²
D. 6 400 unit²

Part 2: What is the moment of inertia, about the X-axis, of the area bounded by the parabola and the X-axis?
A. 15 045 000 unit⁴
B. 18 362 000 unit⁴
C. 11 100 000 unit⁴
D. 21 065 000 unit⁴

Part 3: What is the radius of gyration, about the X-axis, of the area bounded by the parabola and the X-axis?
A. 57.4 units
B. 63.5 units
C. 47.5 units
D. 75.6 units

Solution

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

$y = \frac{1}{40}(4800 - 3x^2)$

$x = \sqrt{\frac{1}{3}(4800 - 40y)}$

When x = 0, y = 120 → Vertex (0, 120)
When y = 0, x = ±40 → (40, 0) and (-40, 0)

Another way to locate the vertex is by reducing the given equation into standard form

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

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

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

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

Area of parabolic section above the x-axis
$A = \frac{2}{3}bh = \frac{2}{3}(80)(120)$

$A = 6\,400 \, \text{ unit}^2$ Part 1: [ D ]

Using horizontal strip to find the moment of inertia about the x-axis
$\displaystyle I_x = \int_{a}^{b} y^2 ~ dA$

$\displaystyle I_x = \int_{y_1}^{y_2} y^2(2x~dy)$

$\displaystyle I_x = 2\int_{0}^{120} y^2\sqrt{\frac{1}{3}(4800 - 40y)} ~ dy$

Using the calculator,
$I_x = 21\,065\,142.86 \, \text{ unit}^4$ Part 2: [ D ]

For easier manual integration, we can translate the x-axis to the vertex of the parabola.

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$x^2 = -4ay$ → 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!)

We can also use vertical strip in finding the moment of inertia about the x-axis. The solution also leads to even more easier manual integration.

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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!)

Radius of gyration
$r = \sqrt{\dfrac{I}{A}}$

$r = \sqrt{\dfrac{21\,065\,142.86}{6\,400}}$

$r = 57.37 \, \text{ units}$ Part 3: [ A ]

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