Find centroid two-dimensional region

  • Thread starter octowilli
  • Start date
  • #1
10
0

Homework Statement



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

[tex]y = x^3[/tex][tex]x + y = 10[/tex][tex]y = 0[/tex]

Homework Equations



[tex]\bar x = \frac{1}{A}\int^b_a xf(x) dx[/tex][tex]\bar y = \frac{1}{A}\int^b_a \frac{1}{2}(f(x))^2 dx[/tex]
Where A is the area of the region containing the centroid.

The Attempt at a Solution



First thing I did was find the intersection of the curves [tex]x^3 = 10-x \Rightarrow x = 2 \Rightarrow P(2,8)[/tex]
Then I put things in terms of y to integrate to find the area A of the region.[tex]y = x^3 \Rightarrow x = \sqrt[3] y[/tex][tex]x + y = 10 \Rightarrow x = 10 - y[/tex][tex]A = \int^8_0 ((10-y)-(\sqrt[3] y)) dy = 36[/tex]
So [tex]\bar x = \frac{1}{A}\int^b_a yf(y) dy = \frac{1}{36}\int^8_0 y((10-y)-(\sqrt[3] y))dy = \frac{496}{189} \approx 2.63[/tex][tex]\bar y = \frac{1}{A}\int^b_a \frac{1}{2}(f(y))^2 dy = \frac{1}{36}\int^8_0 \frac{1}{2}((10-y)^2-(\sqrt[3] y)^2)dy = \frac{584}{135} \approx 4.33[/tex]
So the centroid is at [tex]\left(\frac{496}{189},\frac{584}{135}\right)[/tex]
What I'm getting here isn't correct. Thanks for commenting!
 

Answers and Replies

  • #2
HallsofIvy
Science Advisor
Homework Helper
41,833
964

Homework Statement



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

[tex]y = x^3[/tex][tex]x + y = 10[/tex][tex]y = 0[/tex]

Homework Equations



[tex]\bar x = \frac{1}{A}\int^b_a xf(x) dx[/tex][tex]\bar y = \frac{1}{A}\int^b_a \frac{1}{2}(f(x))^2 dx[/tex]
Where A is the area of the region containing the centroid.

The Attempt at a Solution



First thing I did was find the intersection of the curves [tex]x^3 = 10-x \Rightarrow x = 2 \Rightarrow P(2,8)[/tex]
Then I put things in terms of y to integrate to find the area A of the region.[tex]y = x^3 \Rightarrow x = \sqrt[3] y[/tex][tex]x + y = 10 \Rightarrow x = 10 - y[/tex][tex]A = \int^8_0 ((10-y)-(\sqrt[3] y)) dy = 36[/tex]
This is completely wrong. If you look at a graph of x+y= 10 and [itex]y= x^3[/itex], then you will see that, for each y, x ranges from 0 on the left to
1)[itex] y^{1/3}[/itex] for y between 0 and 8 and
2) [itex]10 - y[/itex] for y between 8 and 10.
The area is given by
[tex]\int_{y= 0}^8 y^{1/3}dy+ \int_{y= 8}^{10} (10- y) dy[/tex].

Much simpler, and what most people would do, is to integrate with respect to x. For each x, between 0 and 2, y ranges from [itex]x^3[/itex] to [itex]10- x[/itex] so the area would be
[tex]\int_{x=0}^2(10- x- x^3) dx[/tex]

So [tex]\bar x = \frac{1}{A}\int^b_a yf(y) dy = \frac{1}{36}\int^8_0 y((10-y)-(\sqrt[3] y))dy = \frac{496}{189} \approx 2.63[/tex][tex]\bar y = \frac{1}{A}\int^b_a \frac{1}{2}(f(y))^2 dy = \frac{1}{36}\int^8_0 \frac{1}{2}((10-y)^2-(\sqrt[3] y)^2)dy = \frac{584}{135} \approx 4.33[/tex]
So the centroid is at [tex]\left(\frac{496}{189},\frac{584}{135}\right)[/tex]
What I'm getting here isn't correct. Thanks for commenting!
 
  • #3
10
0
Thank you for responding, HallsofIvy!

I think I'm still confused. I thought this was the region I needed to find the centroid of:

wh9tq.png


Which has an area [tex]A = \int^2_0 x^3 dx + \int^{10}_2 (10-x) dx = 36 [/tex]
 
  • #4
HallsofIvy
Science Advisor
Homework Helper
41,833
964
Oh, blast! I misead "y= 0" as "x= 0".
 
  • #5
10
0
I had another look and figured it out. I guess when I did this in terms of y my x and y coordinates got flipped. The values of the integral are correct, but the coordinates are backwards, so the centroid is at[tex]\left(\frac{584}{135},\frac{496}{189}\right)[/tex]
 

Related Threads on Find centroid two-dimensional region

  • Last Post
Replies
2
Views
1K
  • Last Post
Replies
1
Views
3K
Replies
2
Views
20K
  • Last Post
Replies
10
Views
2K
Replies
7
Views
14K
Replies
2
Views
3K
  • Last Post
Replies
1
Views
1K
Replies
1
Views
3K
  • Last Post
Replies
10
Views
1K
Top