Calculus 1: Find the centroid of the region bounded by the given curves.

Question

Find the centroid of the region bounded by the given curves.

y=x^3, x+y=30, y=0

(x,y)=

I've been stuck on this problem for 2 hours now and all I can think of is whether a and b are (30,0) or (3,0), also I don't know how to find the area from this to find x and y.

Please help.

Answer 1

We know where y = 0 meets both x+y = 30 (30, 0) and y=x^3 (0, 0), so we only need to find where x+y = 30 meets y = x^3

If you plot these two, it is clear that they meet when x = 3, where y = 27 with both equations. (At the bottom, I derive this algebraically).

From here, you can use the standard centroid equations.

The regions is now defined as being between y = 0 and y = x^3 for x in [0, 3] and y = 0 and y = 30 - x on [3, 30]

In the denominator of the equation for the x and y values of the centroid, we have the area of the region.

I[0, 30] f(x) dx

I[0, 3] x^3 dx = x^4/4 E[0, 3] = 81/4

I[3, 30] 30 - x dx = (noting that this is a triangle with l=w=27), 1/2*27^2 = 729/2

729/2 + 81/4 = 1539/4

Then, we must find the integral of x over the region and the integral of y over the region.

For x,

I[0,3] x*x^3 dx + I[3,30](30-x)x dx =

x^5/5 E[0,3} + 15x^2 - x^3/3 E[3, 30] =

243/5 + 15*30^2 - 30^3/3 -(15*3^2 - 3^3/3) =

22113/5

Then, the x-value at the centroid is 22113/5/(1539/4) = 1092/95 = 11.4947

For the integral of y, as we have y ranging from 0 to f(x)

I[0,30]f²(x)/2 dx=

I[0,3] x^6/2 dx + I[3,30] 1/2(30-x)^2 dx =

x^7/14 E[0, 3} - 1/6 (30-x) ^3 E[3, 30] =

2187/14 + 6561/2 =24057/7

24057/7/(1539/4) = 1188/133 = 8.9323

The centroid is

(1092/95, 1188/133) = (11.4947, 8.9323)

(NOTE: you could solve for the x solution of where x+y=30 and y=x^3 by subtracting the second from the first; If we tried to factor

x^3 + x - 30 = 0, this would seem difficult.

However, as we know x = 3 is a solution, we get

(x - 3)(x^2 + 3x + 10) = 0, with the second term having only imaginary solutions.)