# Finding center of mass

Printable View

• May 29th 2010, 12:14 PM
Em Yeu Anh
Finding center of mass
Q: Determine the mass and center of mass of the lamina for region D if D is bounded by $y=x^2$ and $x=y^2$; $p(x,y)=5\sqrt{x}$

Could someone assist me with setting up the limits on the integral for finding the mass?
• May 29th 2010, 01:03 PM
skeeter
Quote:

Originally Posted by Em Yeu Anh
Q: Determine the mass and center of mass of the lamina for region D if D is bounded by $y=x^2$ and $x=y^2$; $p(x,y)=5\sqrt{x}$

Could someone assist me with setting up the limits on the integral for finding the mass?

$dA = (\sqrt{x} - x^2) \, dx$

$dm = \rho \, dA = 5\sqrt{x}(\sqrt{x} - x^2) \, dx
$

$m = 5\int_0^1 x - x^{\frac{5}{2}} \, dx$
• May 29th 2010, 01:27 PM
AllanCuz
Quote:

Originally Posted by Em Yeu Anh
Q: Determine the mass and center of mass of the lamina for region D if D is bounded by $y=x^2$ and $x=y^2$; $p(x,y)=5\sqrt{x}$

Could someone assist me with setting up the limits on the integral for finding the mass?

Well,

$C.O.M = ( \bar x , \bar y )$

Where,

$\bar x = \frac{ \iint_D x P(x,y) dA }{ \iint_D P(x,y) dA}$

$= \frac{ 5 \int_0^1 x \sqrt{x} dx \int_{x^2}^{ \sqrt{x} }dy }{5 \int_0^1 \sqrt{x} dx \int_{x^2}^{ \sqrt{x} } dy }$

$= \frac{ 5 \int_0^1 x \sqrt{x} ( \sqrt{x} - x^2 ) dx }{ 5 \int_0^1 \sqrt{x} ( \sqrt{x} - x^2 ) dx }$

$= \frac{ 5 \int_0^1 (x^2 - x^{ \frac{7}{2} } ) dx }{ 5 \int_0^1 ( x - x^{ \frac{5}{2} } ) dx }$

And,

$\bar y = \frac{ \iint_D y P(x,y) dA }{ \iint_D P(x,y) dA}$

$= \frac{ 5 \int_0^1 \sqrt{x} dx \int_{x^2}^{ \sqrt{x} } y dy }{5 \int_0^1 \sqrt{x} dx \int_{x^2}^{ \sqrt{x} } dy }$

$= \frac{ 5 \int_0^1 \sqrt{x} ( \frac{ x^2 }{2} - \frac{x^4 }{2} ) dx }{ 5 \int_0^1 \sqrt{x} ( \sqrt{x} - x^2 ) dx }$

$= \frac{ \frac{5}{2} \int_0^1 ( x^{ \frac{5}{2} } - x^{ \frac{9}{2} } ) dx }{ 5 \int_0^1 ( x - x^{ \frac{5}{2} } ) dx }$

If you don't know double integrals yet that's fine, just compute them as the single integrals that I've reduced them to. This will serve as your C.O.M.

Your mass, as Skeeter pointed out is the denominator in the above calculations. So,

$Mass = 5 \int_0^1 ( x - x^{ \frac{5}{2} } ) dx$