Question

Find the centroid of the region bounded by the given curves.
y = x^3, x + y = 10, y = 0

Answer 1

I have found the area to be 36.

Answer 2

Yet, my x,y values for the centroid are incorrect.

Answer 3

How did you calculate your x and y values? This can possibly help us determine where you went wrong.

Answer 4

For x:
\[1/A \int\limits_{0}^{2} x(10-x-x^3)dx\]

For y:
\[1/A \int\limits_{0}^{2} 1/2 (10-x-x^3)^2dx\]

Answer 5

What happened to the region from 2 to 10? This was only from 0 to 2. :)
I think we will have to add up two integrals, one for the section [0,2] and one for the section [2,10]. y = x^3 makes up the upper boundary of the region over [0,2], and y= -x + 10 makes the upper bound of the region over [2, 10].

Adding them together ensures that you are getting contribution from each section in the final answer.

Answer 6

Oh, gosh. So for x:
\[1/36 [\int\limits_{0}^{2}(x^3-(-x+10))dx + \int\limits_{2}^{10}(-x+10-x^3)dx ]\]
?

Answer 7

The piece of the region on the interval [0,2] is bounded from above by x^3 and below by y=0. The piece of the region on the interval [2, 10] is bounded from above by y = -x + 10 and below by y=0.
So f(x) = x^3 and g(x) = 0 for [0,2] and f(x) = -x+10 and g(x) = 0 for the other.

Answer 8

For x:
\[1/36[\int\limits_{0}^{2}(x^3)dx+\int\limits_{2}^{10}(-x+10)dx]\]

Answer 9

Almost there. I think you just missed a small detail. That setup looks like the area for the region, but I believe we need a little extra multiplied on the integrand for these coordinates.

Answer 10

Multiply by x?

Answer 11

Yes. These would be called moments of the region. They are already set up that way if you are using the formulas.
http://tutorial.math.lamar.edu/Classes/CalcII/CenterOfMass.aspx
For the x-coordinate, we are multiplying both integrands by x before we integrate. The y-coordinate is somewhat different but we apply the same strategy suggested by the formula in both integrals to find the y-coordinate.

Answer 12

if you're familiar wid double integrals,

\(\large \overline{x} = \frac{1}{A}\iint x dA\)

Answer 13

Thank you!

Answer 14

using that, gives u \(\overline{x} = 4.3\)

http://www.wolframalpha.com/input/?i=1%2F%28%5Cint_0%5E8+10-y+-+y%5E%281%2F3%29dy%29%28%5Cint_0%5E8%5Cint_%28y%5E%281%2F3%29%29%5E%2810-y%29+x+dxdy%29

Answer 15

wat value u getting for x ?

Answer 16

Can you stick around so im sure my setup for y is right?

Answer 17

ahh never saw those formulas before, they look cool :)

Answer 18

Hmm.. I haven't seen it approached like that. :o
Nice!

And yes, I can stick around. :)

Answer 19

Ok, umm for y:
\[1/72[\int\limits_{0}^{2}(x^3)^2dx+\int\limits_{2}^{10}(-x+10)^2dx]\]

Answer 20

I think that looks good as well.

Answer 21

Thank you so much!

Answer 22

glad to help!

plotting all the graph information together, this is what turned out:
http://www.wolframalpha.com/input/?i=Plot+y+%3D+x%5E3%2C+y+%3D+-x%2B10%2C++y+%3D+0%2C+x%3D+4.3%2C+y+%3D+2.6%3B+from+y%3D0+to+y%3D10
i think the point looks fairly accurate to where i'd place the center of mass.

Answer 23

My online hw says the answer is incorrect

Answer 24

what was the input?

Answer 25

584/135, 332/135

Answer 26

hm.. i think there was an error evaluating the y-value's integral?
wolfram computes the integral to be 496/189. http://www.wolframalpha.com/input/?i=%28integral+from+0+to+2+of+%28x%5E3%29%5E2+dx+%2B+integral+from+2+to+10+of+%2810-x%29%5E2+dx%29%2F72

Answer 27

if you have the work written down, you could look it over step-by-step to make sure there wasn't any fishy business in the middle.

Answer 28

Thanks, I have the work. I made an error.

Answer 29

I have the correct answer now, thank you so much again!

Answer 30

Yup, you're welcome! :)