Answer to Question #142081 in Calculus for Promise Omiponle

Question

Find the mass and center of mass of the lamina that occupies the region bounded by y=x+ 2 and y=x^2, where the density at any point is given by σ(x,y) = 3x^2.

Answer 1

The intersection between the curves $y=x+2$ and $y=x^2$ is,

$x+2=x^2$

$x^2-x-2=0$

$x^2-2x+x-2=0$

$x(x-2)+1(x-2)=0$

$(x+1)(x-2)=0$

$x=-1,2$

So, the point of intersection between the curves are $(-1,1),(2,4)$

The sketch of the region of lamina is as shown in the figure below:

The mass of the lamina is evaluated as,

$m=\iint_{D}\sigma(x,y)dA$

$=\iint_{D}3x^2dA$

$=\int_{-1}^{2}\int_{x^2}^{x+2}3x^2dydx$

$=\int_{-1}^{2}3x^2(x+2-x^2)dx$

$=\int_{-1}^{2}(3x^3+6x^2-3x^4)dx$

$=[3(\frac{x^4}{4})+6(\frac{x^3}{3})-3(\frac{x^5}{5})]_{-1}^{2}$

$=\frac{45}{4}+2(9)-\frac{99}{5}$

$=\frac{189}{20}$

Therefore, the mass of the lamina is $m=\frac{189}{20}$

The $x$ -coordinates of the center of mass is evaluated as,

$x_{bar}=\frac{1}{m}\iint_{D}x\sigma(x,y)dA$

$=\frac{20}{189}\iint_{D}x(3x^2)dA$

$=\frac{20}{189}\int_{-1}^{2}\int_{x^2}^{x+2}3x^3dydx$

$=\frac{20}{189}\int_{-1}^{2}3x^3(x+2-x^2)dx$

$=\frac{20}{189}\int_{-1}^{2}(3x^4+6x^3-3x^5)dx$

$=\frac{20}{189}[3(\frac{x^5}{5})+6(\frac{x^4}{4})-3(\frac{x^6}{6})]_{-1}^{2}$

$=\frac{20}{189}(\frac{99}{5}+\frac{45}{2}-\frac{63}{2})$

$=\frac{20}{189}(\frac{54}{5})$

$=\frac{8}{7}$

The $y$ -coordinate of the center of mass is evaluated as,

$y_{bar}=\frac{1}{m}\iint_{D}y\sigma(x,y)dA$

$=\frac{20}{189}\iint_{D}y(3x^2)dA$

$=\frac{20}{189}\int_{-1}^{2}\int_{x^2}^{x+2}3x^2ydydx$

$=\frac{20}{189}\int_{-1}^{2}3x^2\frac{1}{2}[((x+2)^2-x^4)]dx$

$=\frac{20}{189}(\frac{3}{2})\int_{-1}^{2}x^2(x^2+4x+4-x^4)dx$

$=\frac{20}{189}(\frac{3}{2})\int_{-1}^{2}(x^4+4x^3+4x^2-x^6)dx$

$=\frac{20}{189}(\frac{3}{2})[\frac{x^5}{5}+4(\frac{x^4}{4})+4(\frac{x^3}{3})-\frac{x^7}{7}]_{-1}^{2}$

$=\frac{20}{189}(\frac{3}{2})[\frac{33}{5}+15+12-\frac{129}{7}]$

$=\frac{20}{189}(\frac{3}{2})(\frac{531}{35})$

$=\frac{118}{49}$

Therefore, the $y$ -coordinate of the center of mass is $y_{bar}=\frac{118}{49}$

So, the coordinates of the center of mass is $(x_{bar},y_{bar})=(\frac{8}{7},\frac{118}{49})$