Centroid help

**thegame189** · March 6th 2008, 12:03 PM

y=e^(5x)
y=0
x=0
x=.2

Find the exact coordinates of the centroid.

I got the area to be (e-1)/5.

I got the x coordinate to be 1/(5e-5).

I'm having trouble with the y coordinate. Any help?

**wingless** · March 6th 2008, 12:47 PM

These are the formulas for the centroids of the area under the curve f(x) or f(y):
(a, b are the bounds of the curve for x and c, d are the bounds for y)

$ \displaystyle{\begin{array}{l|l} \text{For} ~f(x) ~& ~ \text{For} ~f(y) \\ \hline \\ M_x = \frac{1}{2}\int^{b}_{a} [f(x)]^2 ~dx ~& ~ M_x = \int^{d}_{c} y~f(y) ~dy \\ \\ M_y = \int^{b}_{a} x~f(x) ~dx ~& ~ M_y = \frac{1}{2}\int^{d}_{c} [f(y)]^2 ~dy \\ \\ A = \text{Area}= \int^{b}_{a} f(x)~dx ~& ~ A = \text{Area}= \int^{d}_{c} f(y)~dy \\ \end{array}}$

Then,
$\begin{array}{ll}\bar{x} = \frac{M_y}{A} & \bar{y} = \frac{M_x}{A}\\\end{array}$

The origin is at $O\left ( \bar{x} , \bar{y }\right )$

As our function is a function of x, we'll use the formulas on the left side of the table.

Firstly, this is the graph:

$y=f(x) = e^{5x}$

Find $M_x, M_y, A$

$M_x = \frac{1}{2}\int^{b}_{a} [f(x)]^2 ~dx$
$M_x = \frac{1}{2}\int^{0.2}_{0} e^{10x} ~dx$
$M_x = \frac{e^2-1}{20}$

$M_y = \int^{b}_{a} x~f(x) ~dx$
$M_y = \int^{0.2}_{0} xe^{5x} ~dx$
$M_y = 0.04$

$A = \int^{b}_{a} f(x) ~dx$
$A = \int^{0.2}_{0} e^{5x} ~dx$
$A = \frac{e-1}{5}$

Hence,

$\begin{array}{l} \bar{x} = \frac{M_y}{A} \\ \bar{y} = \frac{M_x}{A}\\\end{array}$

$\begin{array}{l} \bar{x} = \frac{0.04}{\frac{e-1}{5}} \\ \bar{y} = \frac{\frac{e^2-1}{20}}{\frac{e-1}{5}}\\\end{array}$

$\begin{array}{l}\bar{x} = \frac{1}{5e-5} \\ \bar{y} = \frac{e+1}{4}\\\end{array}$

So the centroid is at $O~\left ( ~\frac{1}{5e-5} ~, ~\frac{e+1}{4}~\right )$

**thegame189** · March 6th 2008, 12:52 PM

I forgot to factor the e^2-1 to cancel with the e+1 on the bottom. Thanks a bunch! You guys are always so helpful!

The grader said that it was wrong. Maybe there is an error in the system... I'll have to talk to my teacher since I'm pretty sure it's right.

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