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Draw a 3 m x 3 m square. On the diagonal is a line corner to corner: y=x. Under that is a line enclosing a area with the function y=x^3/9.

 

How is the steps to integrate y bar: (integ $y dA) /(integ $ dA)

difficulty advanced
EinsteinAlbert  Mar 17, 2015

Best Answer 

 #2
avatar+26412 
+10

Here's another interpretation of the question:

 Average y

 

However, this interpretation makes the Y(x) = x somewhat irrelevant, so it's quite possible this is not what was wanted!

.

Alan  Mar 18, 2015
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11+0 Answers

 #1
avatar+81142 
+5

I assume you want to find the area between y= x and y =x^3/9

We have

3                                                           

∫ x - x^3/9 dx   =  

0

[x^2/2  - (1/36)x^4] from 0 to 3  = [3^2/2 - (1/36)3^4 ]= 9/2 - 81/36 = [162 - 81] / 36 = 81/36 = 9/4 =  2.25 sq units

 

  

CPhill  Mar 17, 2015
 #2
avatar+26412 
+10
Best Answer

Here's another interpretation of the question:

 Average y

 

However, this interpretation makes the Y(x) = x somewhat irrelevant, so it's quite possible this is not what was wanted!

.

Alan  Mar 18, 2015
 #3
avatar+21 
0

Where does y2+((y1-y2)/2) dy = dA come into the workings shown by CPhill? 

EinsteinAlbert  Mar 21, 2015
 #4
avatar+21 
0

Also was his answer showing the equation for numerator and denominator or just the numerator? Seems like the later.

EinsteinAlbert  Mar 21, 2015
 #5
avatar+21 
0

I want the centroid location x bar and y bar of the uncommon area. I don't think it has been answered right.

EinsteinAlbert  Mar 21, 2015
 #6
avatar+91510 
+5


hi Albert,   

 

My answers keep disappearing - I shall try again  

 

I did not answer this in the first place because I did not understand the question.

DO YOU  want to find the area between the curves    $$y=x \;\;and\;\; y=\frac{x^3}{9}$$     for x=0 to x=3   ?

(I think that this was CPhill's assumption)

[You were logged on when I published this :/  ]

Melody  Mar 21, 2015
 #7
avatar+26412 
+5

If it's the coordinates of the centroid of the area between the two curves that you want, then see the following:

 

 Centroid coordinates:

.

Alan  Mar 21, 2015
 #8
avatar+81142 
+5

What the H**L???

How was I supposed to know WHAT you wanted???

I certainly didn't see the word "centroid" anywhere in your question...!!!

I thought you were wanting the area between the curves {silly assumption on my part...}

If you want a specific thing......maybe you should INCLUDE that specific thing when you post.....!!!!

 

  

CPhill  Mar 21, 2015
 #9
avatar+91510 
0

Hear, hear Chris!  I agree !!

We are not paid to be mindreaders - oh, I just remembered, we are not paid at all!     LOL

I still have to assimilate what a centroid is exactly.   

Melody  Mar 21, 2015
 #10
avatar+81142 
+5

Melody, if we have a plane of equal thickness,  the centroid can be thought of as the "balance point" of the plane. 

 

  

CPhill  Mar 21, 2015
 #11
avatar+21 
0

I should have specified the area between. 

Y bar and equation for the location of the centroid in the Y axis is writen in my question. All answers were fine and helpful. The inside area makes sense and makes a good assumption CPhill. 

And how much posts have you all answered,  if you had a question you could ask and simply wait for a reply or ask for terms such as y bar to be clarified. We are all at fault here but sarcasm ain't friendly.  

I can understand it now from answers Cheers.

EinsteinAlbert  Mar 21, 2015

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