MvCalc41

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Here density f(x,y)=xy\,

Mass M=\iint_R f(x,y)dx dy=\int_0^4\int_0^2 xy dx dy=\int_0^4[\frac{xy^2}{2}]_0^2 dx\,

=\int_0^4 2x dx=16\,

Let x_c,y_c\, be the centre of gravity of R then,

x_c=\frac{1}{M}\iint_R xf(x,y)dx dy=\frac{1}{16}\int_0^4\int_0^2 x(xy) dx dy\,

=\frac{1}{16}\int_0^4 x^2[\frac{y^2}{2}]_0^2 dx=\frac{1}{8}\int_0^4 x^2dx=\frac{8}{3}\,

y_c=\frac{1}{M}\iint_R yf(x,y) dx dy=\int_0^4\int_0^2 y(xy)dx dy=\frac{1}{16}\int_0^4 x[\frac{y^3}{3}]_0 ^2 dx=\frac{1}{6}\int_0^4 xdx=\frac{4}{3}\,

Moment of inertia relative to x axis

I_x=\iint_R y^2f(x,y)dx dy=\int_0^4\int_0^2 y^2(xy)dx dy=\int_0^4 x[\frac{y^4}{4}]_0^2 dx=4\int_0^4 x dx=4[\frac{x^2}{2}]_0^4=32\,

Similarly,I_y=\iint_R x^2f(x,y) dx dy=\int_0^4\int_0^2 x^2(xy)dx dy=\int_0^4 x^3[\frac{y^2}{2}]_0^2dx=2\int_0^4 x^3 dx=2[\frac{x^4}{4}]_0^4=128\,

I_0=I_x+I_y=160\,

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