Thread:

1. Find the centroid

$x=y^2, y=x-2$

I have the formulas

$<br /> \bar{x}=\frac{1}{A}\int_{a}^b x[f(x)-g(x)]\,dx<br />$
and
$<br /> \bar{y}=\frac{1}{A}\int_{a}^b \frac{1}{2}[(f(x))^{2}-(g(x))^{2}]\,dx$

I found A= 9/2

Now I do not know how to write $x=y^2$ as f(x) function, since $x=y^2$would be $f(x) = y = +-\sqrt{x}$

$g(x) = x-2$

Thanks for any help.

2. suggestion

Originally Posted by DBA
$x=y^2, y=x-2$

I have the formulas

$<br /> \bar{x}=\frac{1}{A}\int_{a}^b x[f(x)-g(x)]\,dx<br />$
and
$<br /> \bar{y}=\frac{1}{A}\int_{a}^b \frac{1}{2}[(f(x))^{2}-(g(x))^{2}]\,dx$

I found A= 9/2

Now I do not know how to write $x=y^2$ as f(x) function, since $x=y^2$would be $f(x) = y = +-\sqrt{x}$

$g(x) = x-2$

Thanks for any help.
i would substitute the obtained value and calculate for both the positive and the negative values.

3. Sorry, but I do not understand what you mean. Calculate two times $\bar {x}$ and two times $\bar {y}$ ?

How do I end up with one centroid point then?

4. Originally Posted by DBA
Sorry, but I do not understand what you mean. Calculate two times $\bar {x}$ and two times $\bar {y}$ ?

How do I end up with one centroid point then?
i drew the graph. one of the points lie on the complex plane . your triangle has the following points (0,0),(2,sqrt(2)),(2,i.sqrt(2)). so you cannot find the centroid on the real plane. or if you want to find out then some imaginary part will stay behind.

5. Thanks for your answer. We did not have this type of questions with imaginary numbers.
So, I am not familiar how graph this.

Is there a possibility to express the formulas for
$\bar{x}$ and $\bar{y}$ in terms of y (so instead of dx, we would use dy)?
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