MHB Can someone help me solve a integral problem.

[Cornshe]

Hello,

I have a question. It is below.

A design can be modeled on a curve equivalent to the graph y= 2x^2 +6 between X=0 and X=2.

Find the centroid of the feature.

I have had a go and am getting X= 1.38 and Y = 6.28

Is this correct? If not can someone show me where i have gone wrong please.

 

[I like Serena]

Hello,

I have a question. It is below.

A design can be modeled on a curve equivalent to the graph y= 2x^2 +6 between X=0 and X=2.

Find the centroid of the feature.

I have had a go and am getting X= 1.38 and Y = 6.28

Is this correct? If not can someone show me where i have gone wrong please.

Hi Cornshe! Welcome to MHB! (Smile)

If we look at a drawing of the graph, we can estimate where the centroid should be.
In particular, Y=6.28 is way to low, and X=1.38 is on the high side.

The formula for the centroid is:
\begin{cases}
L &= \int_0^2 \sqrt{1+(y')^2} dx &= \int_0^2 \sqrt{1+(4x)^2} dx \\
x_{centroid} &= \frac 1 L \int_0^2 x \sqrt{1+(y')^2} dx &= \frac 1 L \int_0^2 x \sqrt{1+(4x)^2} dx \\
y_{centroid} &= \frac 1 L \int_0^2 y \sqrt{1+(y')^2} dx &= \frac 1 L \int_0^2 (2x^2 +6) \sqrt{1+(4x)^2} dx
\end{cases}

Is that what you did?
Can you evaluate those? (Wondering)