Question

Find the centroid of the region bounded by the given curves. Give your answers correct to two decimal places. y = x^3, x + y = 10, y = 0

Answer 1

The centroid is the point (
`(584)/(135),(496)/(189)`)=(4.32,2.62)

Explanation:

We graph the functions given in the attached picture. As you can see the region R is surrounded by the functions:

y=x³

y=10-x

y=0

This region can be divided by 2, where the first is delimited by y=x³ (0≤x≤2) and the second is y=10-x (2≤x≤10).

The centroid of region is given by:

`x_c=\frac{\int\limits_R {x} \, dydx }{\int\limits_R  \, dydx}`

`y_c=\frac{\int\limits_R {y} \, dydx }{\int\limits_R  \, dydx}`

First, we have to find the value of
`\int\limits_R  \, dydx`

`\int\limits_R  \, dydx=\int\limits^2_0 {\int\limits^{x^(3)}_0} \, dydx +\int\limits^(10)_2 {\int\limits^(10-x)_0} \, dydx=\int\limits^2_0 {y|^{x^(3)}_(0)} \, dx +\int\limits^(10)_2 y \, dx`

=
`\int\limits^2_0 {x^3} \, dx +\int\limits^(10)_2 {(10-x)} \, dx=(x^4)/(4) |^(2)_(0)+(10x-(x^2)/(2))|^(10)_(2)`

=(16/4-0)+(100-50-20+2)=4+32=36

Now we find
`\int\limits_R {x} \, dydx`:

`\int\limits_R {x} \, dydx=\int\limits^2_0 {\int\limits^{x^(3)}_0} {x}\, dydx +\int\limits^(10)_2 {\int\limits^(10-x)_0}{x} \, dydx=\int\limits^2_0 ^(x^3)_0\, dx +\int\limits^(10)_2 x(y) \, dx`

=
`\int\limits^2_0 {x^4}\, dx +\int\limits^(10)_2 {(10x-x^2)} \, dx=(x^5)/(5) |^(2)_(0) +(5x^2-(x^3)/(3) )|^(10)_2`

=
`((32)/(5)-0)+(500-(1000)/(3)-20+(8)/(3))=(2336)/(15)`

Then, we find
`\int\limits_R {y} \, dydx`:

`\int\limits_R {x} \, dydx=\int\limits^2_0 {\int\limits^{x^(3)}_0} {y}\, dydx +\int\limits^(10)_2 {\int\limits^(10-x)_0}{y} \, dydx=\int\limits^2_0 ^(x^3)_0\, dx +\int\limits^(10)_2 ((y^2)/(2)) \, dx`

=
`\int\limits^2_0 {(x^6)/(6)}\, dx +\int\limits^(10)_2 {((10-x)^2)/(2)} \, dx=\int\limits^2_0 {(x^6)/(2)}\, dx +\int\limits^(10)_2 {(50-10x+(x^2)/(2))} \, dx\\=(x^7)/(14) |^(2)_(0) +(50x-5x^2+(x^3)/(6) )|^(10)_2`

=
`(x^7)/(14) |^(2)_(0) +(50x-5x^2+(x^3)/(6) )|^(10)_2=((64)/(7)-0)+(500-500+(500)/(3))-(100-20+(4)/(3))=(1984)/(21)`

Hence,

`x_c=((2336)/(15))/(36)=(584)/(135)`

`y_c=((1984)/(21))/(36)=(496)/(189)`

The centroid is the point (
`(584)/(135),(496)/(189)`)=(4.32,2.62)