Answer to Question #186546 in Calculus for Jethro

Question #186546

Consider a lamina that occupies the region D bounded by the parabola x = 1 - y2 and the coordinate axes in the first quadrant with density function p (x, y) = y. Find the center of mass.


1
Expert's answer
2021-05-07T14:32:39-0400

Given, the region D is bounded by the parabola x = 1 - y2 and the coordinate axes in the first quadrant with density function p (x, y) = y.

Moments,

Mx=∫∫yp(x,y)dA=∫x=01∫y=01yp(x,y)dA=∫x=01∫y=01y2dydx=∫x=01[y33]01dydx=13∫x=01dx=13[x]01=13M_x=\int\int yp(x,y)dA\newline=\int_{x=0}^1 \int_{y=0}^1 yp(x,y)dA\newline=\int_{x=0}^1 \int_{y=0}^1 y^2dydx\newline=\int_{x=0}^1[ \frac{y^3}{3}]_{0}^1dydx\newline=\frac{1}{3}\int_{x=0}^1dx\newline=\frac{1}{3}[ x]_{0}^1\newline=\frac{1}{3}\newline

My=∫∫xp(x,y)dA=∫x=01∫y=01xydydx=∫x=01x[y22]01dydx=12∫x=01xdx=12[x22]01=14M_y=\int\int xp(x,y)dA\newline=\int_{x=0}^1 \int_{y=0}^1 xydydx\newline=\int_{x=0}^1x[ \frac{y^2}{2}]_{0}^1dydx\newline=\frac{1}{2}\int_{x=0}^1xdx\newline=\frac{1}{2}[ \frac{x^2}{2}]_{0}^1\newline=\frac{1}{4}

And,

Mass,m is given by,

m=∫∫p(x,y)dA=∫x=01∫y=01ydydx=∫x=01[y22]01dydx=12∫x=01dx=12[x]01=12m=\int\int p(x,y)dA\newline=\int_{x=0}^1 \int_{y=0}^1 ydydx\newline=\int_{x=0}^1[ \frac{y^2}{2}]_{0}^1dydx\newline=\frac{1}{2}\int_{x=0}^1dx\newline=\frac{1}{2}[ x]_{0}^1\newline=\frac{1}{2}

Let (x,y) be the center of mass.

y=Mym=1412=12x=Mxm=1312=23y=\frac{M_y}{m}\newline=\frac{\frac{1}{4}}{\frac{1}{2}}\newline=\frac{1}{2}\newline x=\frac{M_x}{m}\newline=\frac{\frac{1}{3}}{\frac{1}{2}}\newline=\frac{2}{3}

Thus, center mass is (23,12).(\frac{2}{3},\frac{1}{2}).

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