Two curves, centroid, integration

**26-4-15**

Find centroid of region bounded by following curves:

y = 4x^2

y = 2x^3

I found two y intercepts: x = 2 and x = -2

obvious x intercept is zero

Attempt:

A = (int) 2x^3 - 2x^2

boundary 0 and 2

1/2x^4 - 2/3x^3 =

1/2(2^4) - 2/3(2^3) = 8 - 2.666 = 5 1/3

X centroid:

1/2 * 1/(5.333) (int) [(2x^3)^2 - (4x^2)^2]

boundaries 2 and -2

= .09375 (int) 4/7x^7 - 16/5x^5

plugging in 2

= 4/7(2^7) - 16/5(2^5) = -29.26

completely wrong... so where'd I go wrong? Thanks

Re: Two curves, centroid, integration

Quote:

Originally Posted by

**togo** **26-4-15**

Find centroid of region bounded by following curves:

y = 4x^2

y = 2x^3

I found two y intercepts: x = 2 and x = -2

obvious x intercept is zero

Those really aren't relevant. What is relevant is where the two curves intersect- where y= 4x^2= 2x^3. Obviously, one root is x= 0. If x is not 0 we can divide by 2x^2 to get x= 2.

Quote:

Attempt:

A = (int) 2x^3 - 2x^2

boundary 0 and 2

Yes, the boundary is 0 and 2. But where did you get 2x^3- 2x^2? Your two functions are y= 4x^2 and 2x^3. Further, for all x between 0 and 2, 4x^2> 2x^3 so you want to integrate 4x^2- 2x^3, not 2x^3- 2x^2.

Be careful about your notation. Many people would interpret "1/2x^4" as 1/(2x^4) not (1/2)x^4.

In any case, the integral of 4x^2- 2x^3 is (4/3)x^3- (1/2)x^4 which, evaluated at 2 and 4 is 32/3- 8= 32/3- 24/3= 8/3.

Quote:

1/2(2^4) - 2/3(2^3) = 8 - 2.666 = 5 1/3

X centroid:

1/2 * 1/(5.333) (int) [(2x^3)^2 - (4x^2)^2]

boundaries 2 and -2

What? The integral should be just x times the area integrand: x(2x^3- 4x^2). I have no idea where those squares came from

[tex]= .09375 (int) 4/7x^7 - 16/5x^5

plugging in 2

= 4/7(2^7) - 16/5(2^5) = -29.26

completely wrong... so where'd I go wrong? Thanks[/QUOTE]

Pretty much everything is wrong! You need to review "centroid".

Re: Two curves, centroid, integration

thanks I will be reworking the question this morning. I often copy numbers and signs wrong and forget to check for simple mistakes like that. I really appreciate getting an answer on this post, have made many posts in the past and they were ignored by the community.

Re: Two curves, centroid, integration

I get -2.666 for area

A = (int) 2x^3 - 4x^2

= 1/2(2^4) - 4/3(2^3)

= 8 - 10.666

Re: Two curves, centroid, integration

Are you even reading my posts? I already told you that, for x between 0 and 2, 4x^2> 2x^3. The area is the integral of 4x^2- 2x^3, NOT "2x^3- 4x^2". In fact, I have already told you that the area is 8/3= 32/3- 8.

Re: Two curves, centroid, integration

Yes thanks for that. Continuing on

finding centroid of X

1/(8/3) (int) x[2(x^3) - 4(x^2)]

= 2(x^4) - 4(x^3)

= 2/5(x^5) - 1/2(x^4)

= 2/5(2^5) - 1/2(2^4) = 6.4-8

using boundaries of 2 and -2, being plugged into X, the 2 and -2 with exponents on them (5 and 4 respectively) turn into the same thing so you end up with (6.4 - 8) - (6.4 - 8) which is zero and incorrect.

Re: Two curves, centroid, integration

You are **still** integrating $\displaystyle 2x^3- 4x^2$ when I have told you **twice** that $\displaystyle 4x^2> 2x^3$. You need to use $\displaystyle 4x^2- 2x^3$.

The boundaries are NOT "-2 and 2". The two curves intersect at x= 0 and at x= 2.