Integrais Duplas!

Essa questão é do Thomas?

Para encontrarmos o centro de massa, realizaremos os cálculos abaixo:

\(\begin{align} & p(x,y)=k({{x}^{2}}+{{y}^{2}}) \\ & \\ & \int_{0}^{a}{\int_{0}^{a-x}{p(x,y)=}}\int_{0}^{a}{\int_{0}^{a-x}{5({{x}^{2}}+{{y}^{2}})dydx}} \\ & M=\frac{1}{2}\int_{0}^{a}{\int_{0}^{a}{k({{x}^{2}}+{{y}^{2}})dxdy}} \\ & M=\frac{k}{2}\int_{0}^{a}{\left( \frac{{{x}^{3}}}{3}+x{{y}^{2}} \right)}dy \\ & M=\frac{k}{2}\int_{0}^{a}{\left( \frac{{{a}^{3}}}{3}+a{{y}^{2}} \right)}dy \\ & M=\frac{k}{2}\left( \frac{{{a}^{3}}y}{3}+\frac{a{{y}^{3}}}{3} \right)_{0}^{a} \\ & M=\frac{k}{2}\left( \frac{{{a}^{4}}}{3}+\frac{{{a}^{4}}}{3} \right) \\ & M=\frac{k{{a}^{4}}}{3} \\ \end{align}\ \)

Portanrto, o centro de massa será \(\boxed{M = \frac{{k{a^4}}}{3}}\).