Question

Locate the centroid of the shaded area between the two curves​

Answer 1

9514 1404 393

Answer:

(x, y) = (5.76, 1 5/7)

Step-by-step 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 ...

`\displaystyle c_x=(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=(2)/(5k_2^(1/2))\cdot12^(5/2)-(k_1)/(5)12^5\\\\a=\int_0^(12){(k_2^(-1/2)\cdot x^(1/2)-k_1x^3)}\,dx=(2)/(3k_2^(1/2))\cdot12^(3/2)-(k_1)/(4)12^4\\`

For k1 = 4/12^3 and k2=12/4^2, these evaluate to ...

`a_x=115.2\\a=20\\c_x=5.76`

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 ...

`\displaystyle c_y=(a_y)/(a)\\\\a_y=\int_0^(12){((y_2+y_1)/(2)(y_2-y_1))}\,dx=(1)/(2)\int_0^(12){((x)/(k_2)-(k_1x^3)^2)}\,dx\\\\a_y=(12^2)/(4k_2)-(k_1^212^7)/(14)=(240)/(7)\\\\c_y=(12)/(7)\approx1.7143`

The centroid of the shaded area is ...

(x, y) = (5.76, 1 5/7)