MvCalc41

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

Mass M=\iint _{R}f(x,y)dxdy=\int _{0}^{4}\int _{0}^{2}xydxdy=\int _{0}^{4}[{\frac  {xy^{2}}{2}}]_{0}^{2}dx\,

=\int _{0}^{4}2xdx=16\,

Let x_{c},y_{c}\, be the centre of gravity of R then,

x_{c}={\frac  {1}{M}}\iint _{R}xf(x,y)dxdy={\frac  {1}{16}}\int _{0}^{4}\int _{0}^{2}x(xy)dxdy\,

={\frac  {1}{16}}\int _{0}^{4}x^{2}[{\frac  {y^{2}}{2}}]_{0}^{2}dx={\frac  {1}{8}}\int _{0}^{4}x^{2}dx={\frac  {8}{3}}\,

y_{c}={\frac  {1}{M}}\iint _{R}yf(x,y)dxdy=\int _{0}^{4}\int _{0}^{2}y(xy)dxdy={\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^{2}f(x,y)dxdy=\int _{0}^{4}\int _{0}^{2}y^{2}(xy)dxdy=\int _{0}^{4}x[{\frac  {y^{4}}{4}}]_{0}^{2}dx=4\int _{0}^{4}xdx=4[{\frac  {x^{2}}{2}}]_{0}^{4}=32\,

Similarly,I_{y}=\iint _{R}x^{2}f(x,y)dxdy=\int _{0}^{4}\int _{0}^{2}x^{2}(xy)dxdy=\int _{0}^{4}x^{3}[{\frac  {y^{2}}{2}}]_{0}^{2}dx=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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