# Find the centroid of the area bounded by:x^2=4y ; y^2=4x. (Area)Please show a graph or illustration and explain thoroughly.Thank you enotes... "NEED BADLY"

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### 1 Answer

You need to use the following formula to find the centroid of area bounded by two curves such that:

`bar x = (1/A)int_a^b x*f(x) dx`

You need to write the given curves in terms of x such that:

`y^2 = 4x => y = 2sqrt x`

`x^2 = 4y => y = x^2/4`

You need to find the limits of integration, hence, you need to solve the following equation, such that:

`2sqrt x = x^2/4 => sqrt x = x^2/8`

You need to raise to square both sides such that:

`x = x^4/64`

Moving all terms to one side yields:

`x - x^4/64 = 0`

Factoring out x yields:

`x(1 - x^3/64) = 0 => {(x = 0),(1 - x^3/64 = 0):} => {(x = 0),(x^3/64 = 1):} => {(x = 0),(x^3= 64):} => {(x = 0),(x^3 - 64 = 0):} => {(x = 0),((x - 4)(x^2 + 4x + 16) = 0):}`

Since `x^2 + 4x + 16 > 0 => x - 4 = 0 => x = 4`

Hence the limits of integration are `x = 0` and `x = 4` .

You need to evaluate the area of the region bounded by the curves `y = 2sqrt x ` and `y = x^2/4` , over the interval [0,4] such that:

`A = int_0^4 (2sqrt x -x^2/4) dx`

Using the property of linearity of integral yields:

`A = int_0^4 (2sqrt x) dx -int_0^4 (x^2/4) dx`

`A = (4/3)xsqrt x|_0^4 - (1/12)x^3|_0^4`

Using the fundamental theorem of calculus yields:

`A = (4/3)(4sqrt 4 - 0sqrt 0) - (1/12)(4^3 - 0^3)`

`A = 32/3 - 16/3 => A = 16/3`

You may evaluate the x coordinate of centroid such that:

`bar x = 1/(16/3) int_0^4 x*(2sqrt x -x^2/4) dx`

`bar x = 3/16 int_0^4 2x^(3/2)dx - 3/64 int_0^4 x^3 dx`

`bar x = (6/16)(2/5)*x^2sqrtx|_0^4 - 3/256(x^4)|_0^4`

`bar x = 6/40(32 - 0) - 3 => bar x = 24/5 - 3`

`bar x = 9/5`

You may find y coordinate of centroid such that:

`bar y = (1/A)int_c^b y(x_2 - x_1)`

You need to write the equations of the given curves in terms of y such that:

`x^2 = 4y => x = 2sqrt y`

`x = y^2/4`

`bar y = (3/16)int_0^4 y(2sqrt y - y^2/4) = 9/5`

**Hence, evaluating the coordinates of the given centroid yields **`(9/5,9/5).`