Area and centroid

**dadon** · May 2nd 2006, 08:41 AM

Hi All,

a) Find the area bounded by the curve y=x^2 and the line y=4. Determine the co-ordinates of the centroid of the area.

(I know about the centroid equations but do not know how to apply to this question. Also I think I know the area bounded equation $\int\pi{y^2}dx$ but not to sure if you have to apply it to this question)

b) A uniform wire is bent in the form of a three quarter circle (picture below). Determine the co-ordinates of the mass centre.

(Also what kind of topic is this?)

I'm sure you guys can help.

Thanks

**earboth** · May 2nd 2006, 10:15 AM

Originally Posted by dadon

Hi All,

a) Find the area bounded by the curve y=x^2 and the line y=4. Determine the co-ordinates of the centroid of the area.

(I know about the centroid equations but do not know how to apply to this question. Also I think I know the area bounded equation $\int\pi{y^2}dx$ but not to sure if you have to apply it to this question)

b) A uniform wire is bent in the form of a three quarter circle (picture below). Determine the co-ordinates of the mass centre....

Hello,

to a)

The coordinates of a centroid which is bounded "at the top" by the curve f and is bounded "at the bottom" by the curve g can be calculated like this:

$x_C=\frac{\int^{a}_{b}{(x\cdot (f(x)-g(x) )) dx}}{\int^{a}_{b}{(f(x)-g(x))dx}}$ and

$y_C=\frac{\int^{a}_{b}{(f(x))^2-(g(x))^2 ) dx}}{2 \cdot \int^{a}_{b}{(f(x)-g(x))dx}}$

With your problem
f(x) = 4
g(x) = x^2
b = -2
a = 2

Plug in all values you got, calculate the integrals and you'll get $C \left(0 , \frac{12}{5} \right)$

I've attached a diagram to show the situation.

Greetings

EB

**dadon** · May 2nd 2006, 01:18 PM

HEY! thanks for the reply!

for yc I keep getting 6/5 rather then 12/5

heres what I do:

TOP

$y_c =\left( \frac{x^5}{5} \right)$

Limit -2 and 2

Therefore:-

$\frac{(2)^5}{5} - \frac{(-2)^5}{5} = 6.4$

BOTTOM

$y_c =\left( \frac{x^3}{3} \right)$

Limit -2 and 2

Therefore:-

$\frac{(2)^3}{3} - \frac{(-2)^3}{3} = \frac{16}{3}$

6.4 / 16/3 = 6/5 ?

lol I don't know where I am going wrong!

**earboth** · May 2nd 2006, 08:19 PM

Originally Posted by dadon

HEY! thanks for the reply!

lol I don't know where I am going wrong!

Hello,

TOP

$y_c =\left( \frac{x^5}{5} \right)$

Limit -2 and 2

Therefore:-

$\frac{(2)^5}{5} - \frac{(-2)^5}{5} = 6.4 \ \ \longleftarrow \mbox{this here looks funny to me}$

Greetings

EB

**dadon** · May 3rd 2006, 12:02 AM

Thanks for that earboth!

Sorry my mistake should have been 12.8 not 6.4 and hence gives the answer 12/5.

Kind Regards,

dadon

**dadon** · May 3rd 2006, 01:11 PM

oh sorry I forgot to ask where it says area bounded to the curve do I just integrate and plug in the limits (2 & -2) to get the area.

Thanks

If that is the case I get 16/3

**ThePerfectHacker** · May 3rd 2006, 06:51 PM

Originally Posted by dadon

oh sorry I forgot to ask where it says area bounded to the curve do I just integrate and plug in the limits (2 & -2) to get the area.

Thanks

If that is the case I get 16/3

This is case of top curve minus bottom curve.
The top curve is $y=4$
The bottom curve is $y=x^2$
From $-2\leq x\leq 2$ because they intersect at $-2,2$ .
Thus,
$\int^2_{-2} 4-x^2 dx=\left 4x-\frac{x^3}{3} \right|^2_{-2}=4(2)-\frac{8}{3}-4(-2)+\frac{-8}{3}=$ $16-\frac{16}{3}=\frac{32}{3}$

**dadon** · May 4th 2006, 01:16 PM

Thanks guys for that!

And for correcting my mistakes. Hopefully I will learn from them.

Kind Regards,

dadon

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