Answers
9514 1404 393
Answer:
(x, y) = (5.76, 1 5/7)
Explanation:
The location of the centroid in the x-direction is the ratio of the first moment of area about the y-axis to the total area. Similarly, the y-coordinate of the centroid is the first moment of area about the x-axis, divided by the area.
For the moment about the y-axis, we can define a differential of area as ...
dA = (y2 -y1)dx
where y2 = √(x/k2) and y1 = k1·x^3
The distance of that area from the y-axis is simply x.
So, the x-coordinate of the centroid is ...
[tex]\displaystyle c_x=\frac{a_x}{a}=\frac{\int{x\cdot dA}}{\int{dA}}\\\\a_x=\int_0^{12}{x(k_2^{-1/2}\cdot x^{1/2}-k_1x^3)}\,dx=\frac{2}{5k_2^{1/2}}\cdot12^{5/2}-\frac{k_1}{5}12^5\\\\a=\int_0^{12}{(k_2^{-1/2}\cdot x^{1/2}-k_1x^3)}\,dx=\frac{2}{3k_2^{1/2}}\cdot12^{3/2}-\frac{k_1}{4}12^4\\[/tex]
For k1 = 4/12^3 and k2=12/4^2, these evaluate to ...
[tex]a_x=115.2\\a=20\\c_x=5.76[/tex]
The y-coordinate of the centroid requires we find the distance of the differential of area from the x-axis. We can use (y2 +y1)/2 for that purpose. Then the y-coordinate is ...
[tex]\displaystyle c_y=\frac{a_y}{a}\\\\a_y=\int_0^{12}{(\frac{y_2+y_1}{2}(y_2-y_1))}\,dx=\frac{1}{2}\int_0^{12}{(\frac{x}{k_2}-(k_1x^3)^2)}\,dx\\\\a_y=\frac{12^2}{4k_2}-\frac{k_1^212^7}{14}=\frac{240}{7}\\\\c_y=\frac{12}{7}\approx1.7143[/tex]
The centroid of the shaded area is ...
(x, y) = (5.76, 1 5/7)
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