Question

Determine the centroid or the area bounded by the lines: y= 4-x^2 y= x + 2

Answer 1

`\left(\overline{x},\overline{y}\right)=\left(-(1)/(2),(12)/(5)\right)`

Explanation:

The centroid of an area between two curves refers to the point that represents the "center of mass" or "average position" of that region.

To determine the centroid of the region bounded by the curve y = 4 - x² and line y = x + 2, we first need to find the points of intersection of the two functions.

To find the points of intersection, set the two equations equal to each other and solve for x:

`\begin{aligned}4-x^2&=x+2\\4-x^2-x-2&=0\\-x^2-x+2&=0\\x^2+x-2&=0\\x^2+2x-x-2&=0\\x(x+2)-1(x+2)&=0\\(x+2)(x-1)&=0\\\\x+2&=0\implies x=-2\\x-1&=0 \implies x=1\\\end{aligned}`

Now that we have the x-coordinates of the points of intersection, we can find the corresponding y-coordinates by plugging these values into one of the equations:

`\begin{aligned}x=-2 \implies y&=-2+2\\y&=0\end{aligned}`

`\begin{aligned}x=1 \implies y&=1+2\\y&=3\end{aligned}`

So, the two points of intersection are (-2, 0) and (1, 3).

The formula for finding the area between two curves, where y = f(x) is the upper function and y = g(x) is the lower function over an interval [a, b] on the x-axis, is given by:

`\displaystyle A = \int_(a)^(b) \left[f(x)-g(x)\right] \text{d}x`

In this case, f(x) = 4 - x² and g(x) = x + 2. The values of a and b are the x-values of the points of intersection of the two functions. Therefore:

`\displaystyle A = \int_(-2)^(1) \left[(4 - x^2) - (x + 2)\right] \text{d}x`

Simplify:

`\displaystyle A = \int_(-2)^(1) (2 - x^2-x) \;\text{d}x`

Now, integrate:

`\displaystyle A =\left[2x - (x^3)/(3)-(x^2)/(2)\right]_(-2)^(1)`

`\displaystyle A =\left(2(1) - ((1)^3)/(3)-((1)^2)/(2)\right)-\left(2(-2) - ((-2)^3)/(3)-((-2)^2)/(2)\right)`

`\displaystyle A =\left(2-(1)/(3)-(1)/(2)\right)-\left(-4+(8)/(3)-2\right)`

`\displaystyle A =(7)/(6)+(10)/(3)`

`\displaystyle A =(9)/(2)`

So, the area bounded by the curve and line is 9/2 square units.

The x-coordinate of the centroid (
`\overline{x}`) can be calculated using the following formula:

`\displaystyle \overline{x}=(1)/(A)\int_(a)^(b)x[f(x)-g(x)]\;\text{d}x`

Therefore:

`\displaystyle \overline{x}=(1)/((9)/(2))\int_(-2)^(1)x((4-x^2)-(x+2))\;\text{d}x`

`\displaystyle \overline{x}=(2)/(9)\int_(-2)^(1)(2x-x^3-x^2)\;\text{d}x`

`\overline{x}=(2)/(9)\left[x^2-(x^4)/(4)-(x^3)/(3)\right]_(-2)^(1)`

`\overline{x}=(2)/(9)\left[\left((1)^2-((1)^4)/(4)-((1)^3)/(3)\right)-\left((-2)^2-((-2)^4)/(4)-((-2)^3)/(3)\right)\right]`

`\overline{x}=(2)/(9)\left[(5)/(12)-(8)/(3)\right]`

`\overline{x}=-(1)/(2)`

The y-coordinate of the centroid (
`\overline{y}`) can be calculated using the following formula:

`\displaystyle \overline{y}=(1)/(2A)\int_(a)^(b)[f(x)]^2-[g(x)]^2\;\text{d}x`

Therefore:

`\displaystyle \overline{y}=(1)/(2\cdot (9)/(2))\int_(-2)^(1)(4-x^2)^2-(x+2)^2\;\text{d}x`

`\displaystyle \overline{y}=(1)/(9)\int_(-2)^(1)(x^4-9x^2-4x+12)\;\text{d}x`

`\overline{y}=(1)/(9)\left[(x^5)/(5)-(9x^3)/(3)-(4x^2)/(2)+12x\right]_(-2)^(1)`

`\overline{y}=(1)/(9)\left[(x^5)/(5)-3x^3-2x^2+12x\right]_(-2)^(1)`

`\overline{y}=(1)/(9)\left[\left(((1)^5)/(5)-3(1)^3-2(1)^2+12(1)\right)-\left(((-2)^5)/(5)-3(-2)^3-2(-2)^2+12(-2)\right)\right]`

`\overline{y}=(1)/(9)\left[(36)/(5)+(72)/(5)\right]`

`\overline{y}=(12)/(5)`

Therefore, the centroid of the area bounded by the curve y = 4 - x² and line y = x + 2 is:

`\Large\boxed{\boxed{\left(\overline{x},\overline{y}\right)=\left(-(1)/(2),(12)/(5)\right)}}`