# Thread: Area and centroid

1. ## Area and centroid

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 $\displaystyle \int\pi{y^2}dx$ 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

2. 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 $\displaystyle \int\pi{y^2}dx$ 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:

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

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

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

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

I've attached a diagram to show the situation.

Greetings

EB

3. ## re:

HEY! thanks for the reply!

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

heres what I do:

TOP

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

Limit -2 and 2

Therefore:-

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

BOTTOM

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

Limit -2 and 2

Therefore:-

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

6.4 / 16/3 = 6/5 ?

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

4. Originally Posted by dadon
HEY! thanks for the reply!

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

Hello,

TOP

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

Limit -2 and 2

Therefore:-

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

Greetings

EB

5. ## re:

Thanks for that earboth!

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

Kind Regards,

6. ## re:

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

7. 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 $\displaystyle y=4$
The bottom curve is $\displaystyle y=x^2$
From $\displaystyle -2\leq x\leq 2$ because they intersect at $\displaystyle -2,2$.
Thus,
$\displaystyle \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}=$$\displaystyle 16-\frac{16}{3}=\frac{32}{3}$

8. Thanks guys for that!

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

Kind Regards,