Final answer:
The centroid of the given region is approximately at (1.44, 8.55).
Step-by-step explanation:
To find the centroid (x, y) of the region in the right-half plane (x, y) | x ≥ 0, bounded by the curves y = 3x^2 + 3x, y = 0, x = 0, and x = 7, we can use the formula for finding the centroid of a region with respect to the x-axis and y-axis. The centroid (x, y) is given by:
x = (1/A) ∫[a, b] x*f(x) dx
y = (1/(2A)) ∫[a, b] f(x)^2 dx
Where A is the area of the region and f(x) represents the equation of the curve bounding the region.
Finding the Area of the Region:
The area A of the region can be found by integrating the difference between the upper and lower curves with respect to x over the interval [0, 7].
The upper curve is y = 3x^2 + 3x and the lower curve is y = 0.
A = ∫[0, 7] (3x^2 + 3x - 0) dx
A = ∫[0, 7] (3x^2 + 3x) dx
A = [x^3 + (3/2)x^2] | from 0 to 7
A = [7^3 + (3/2)7^2] - [0^3 + (3/2)0^2]
A = [343 + (3/2)*49] - [0 + 0]
A = 343 + (147/2)
A = 343 + 73.5
A = 416.5 square units
Step 2: Finding X:
To find x, we need to evaluate the integral ∫[0, 7] x*(3x^2 + 3x) dx and divide it by A.
x = (1/A) ∫[0, 7] x(3x^2 + 3x) dx
= (1/416.5) ∫[0, 7] (3x^3 + 3x^2) dx
= (1/416.5) [(3/4)x^4 + (3/3)x^3] | from 0 to 7
= (1/416.5) [(3/4)7^4 + (3/3)7^3] - [(3/4)0^4 + (3/3)0^3]
= (1/416.5) [(1029/4) + (1029/3)]
= (1/416.5) [(257.25 + 343)]
= (1/416.5) 600.25
≈ 1.44 units
Step 3: Finding y:
To find y, we need to evaluate the integral ∫[0, 7] ((3x^2 + 3x)^2) dx and divide it by twice A.
y = (1/(2A)) ∫[0, 7] ((3x^2 + 3x)^2) dx
= (1/(833)) ∫[0, 7] ((9x^4 +18x^3+9x^2)) dx
= (1/(833)) [(9/5)x^5 +(18/4)x^4+(9/3)x^3]| from 0 to 7
≈ 8.55 units
Therefore, the centroid of the given region is approximately at (1.44, 8.55).
Complete quesstion:
Find the centroid ( x , y ) of the region that is contained in the right-half plane (x,y)∣x≥0, and is bounded by the curves: y=3x^2 +3x,y=0,x=0, and x=7.