Find the centroid of the region in the first quadrant bounded by the given curves.

Question

Find the centroid of the region in the first quadrant bounded by the given curves.

y=x³, x+y=10, y=0

Find (xbar, ybar).

Please show detailed steps, I am stuck on this homework problem. Thank you!

Answer 1

We will use I[a, b] for the integral from a to b and E[a, b] for evaluation from a to b

We are in the first quadrant (x, y >= 0) The bottom of the region is y = 0. The top of the region is x^3 when x^3 < 10 - x and 10 - x when x^3 > 10 - x

x^3 = 10 - x when x^3 + x - 10 = 0

x = 2 jumps out as a solution.

The top of the region is x^3 when x <= 2

For x >= 2, 10 - x >= 0 means x <= 10

You can see that x = 2 by plugging 2 into x^3 + x - 10 and seeing that it is 0.

If you feel that you must, you can use synthetic division

2 | 1 0 1 -10

2 4 10

1 2 5 0

(x - 2) (x^2 + 2x + 5) = 0

(Note that x^2 + 2x + 5 = (x+1)^2 + 4 has no real roots)

Then, the area is I[0,2] x^3 dx + I[2, 10] 10 - x dx = x^4/4 E[0, 2] + 10x - x^2/2 E[2, 10] =

2^4/4 - 0^4/4 + 10 * 10 - 10^2/2 - (2 * 10 - 2^2/2) = 16/4 + 100 - 50 - (20 - 2) = 36

y-bar = I[A] y dA/ A

x-bar = I[A] x dA/ A

Thus, we now must calculate the numerators for these equations.

I[A] x dA = I[0,2]I[0,x^3] x dy dx + I[2,10]I[0,10 -x] x dy dx =

I[0,2] xy E[0, x^3] dx + I[2,10] xy E[0, 10 - x] dy dx =

I[0, 2] (x^4 - 0) dx + I[2, 10] x(10 - x - 0) dx =

I[0, 2] x^4 dx + I[2, 10] 10x - x^2 dx =

x^5/5 E[0, 2] + (5 x^2 - x^3/3) E[2, 10] =

32/5 + (5 * 10^2 - 10^3/3) - (5 * 2^2 - 2^3/3) =

32/5 + 500 - 1000/3 - (20 - 8/3) =

6 2/5 + 480 - 330 2/3 = 155 11/15

Then, x-bar = (155 11/15) / 36 = 584/135 = 4 44/135 or 4.32592592592593

I[A] y dA/ A =

I[0,2]I[0,x^3] y dy dx + I[2,10]I[0,10 -x] y dy dx =

I[0,2]y^2/2 E[0, x^3] dx + I[2,10] y^2/2 E[0,10 -x] dx =

I[0, 2] (x^6/2 - 0) dx + I[2, 10] (10-x)^2/2 dx =

I[0, 2] x^6/2 dx + I[2, 10] (10-x)^2/2 dx =

x^7/14 E[0, 2] - (10-x)^3/6 E[2, 10] =

128/14 - (0 - 8^3/6) =

64/7 + 256/3 = 94 10/21

y-bar = (94 10/21)/36 = 496/189 = 2 118/189 = 2.62433862674552

(x-bar, y-bar) = (4 44/135, 2 118/189) or (4.32592592592593, 2.62433862674552)