Question

Find the centroid of the following:

Answer 1

Here's what I did....
\[\huge \bar{y}= \frac{\int\limits y dA}{\int\limits dA}\]
\[\huge \frac{\int\limits_0^1 y\sqrt{4-y}dy}{\int\limits_{-2}^{2} 1-\frac{1}{4}x^2dx}\]
\[\huge=0.342m\]

Answer 2

**They only want y centroid

Answer 3

I would use
\[\LARGE \bar{y} = \frac{1}{A}*\int_{-2}^{2}\frac{1}{2}*\left(1-\frac{1}{4}x^2\right)^2dx\]
where A is the area of the blue region

http://tutorial.math.lamar.edu/Classes/CalcII/CenterOfMass.aspx

When I used that formula I didn't get 0.342. I got something slightly larger.

Answer 4

@jim_thompson5910 It's for a statics course, so I have to do it using this particular equation. I'm not sure what I'm doing wrong

Answer 5

If you used your version of the formula, it would actually be this

\[\huge \frac{\int\limits_0^1 4y\sqrt{1-y}dy}{\int\limits_{-2}^{2} 1-\frac{1}{4}x^2dx}\]

Answer 6

Oh I see my stupid error... Also I think you meant 2**, not 4

Answer 7

if you solve y = 1-(1/4)x^2 for x, you get

x = 2*sqrt(-y+1) or x = -2*sqrt(-y+1)

using symmetry, we know the two halves of the region are equal

so we can integrate from y = 0 to y = 1 and then double the area

so 2*2*sqrt(-y+1) = 4*sqrt(-y+1)

Answer 8

Ohh I didn't realize you already factored that in

Answer 9

Thank you!

Answer 10

what final answer did you get?

Answer 11

~0.400m

Answer 12

yep 2/5 = 0.4

Answer 13

X)