Calcule a integral dupla abaixo do paraboloide z = 3x2 + y2 e acima da região delimitada por y = x e x = y2 – y.

A = Z R Z s 1 + ( ∂z ∂x ) 2 + ( ∂z ∂y ) 2dxdy = Z R Z s 1 + x 2 4(x 2 + y 2) + y 2 4(x 2 + y 2) dxdy = Z R Z r 5 4 dxdy = r 5 4 Z 2π 0 Z 1 1 2 rdrdθ = 3 4 π r 5 4 .

Para calcular a integral dupla, realizaremos os cálculos abaixo:

\(\begin{align} & \int_{{}}^{{}}{\int_{{}}^{{}}{(x,y)=}}\int_{0}^{1}{\int_{0}^{1}{3{{x}^{2}}+{{y}^{2}}}dxdy} \\ & \int_{{}}^{{}}{\int_{{}}^{{}}{(x,y)=}}\int_{0}^{1}{\left( \frac{3{{x}^{3}}}{3}+x{{y}^{2}} \right)_{0}^{1}dy} \\ & \int_{{}}^{{}}{\int_{{}}^{{}}{(x,y)=}}\int_{0}^{1}{\left( \frac{3}{3}+{{y}^{2}} \right)dy} \\ & \int_{{}}^{{}}{\int_{{}}^{{}}{(x,y)=}}\int_{0}^{1}{\left( 1+{{y}^{2}} \right)dy} \\ & \int_{{}}^{{}}{\int_{{}}^{{}}{(x,y)=}}\left( 1y+\frac{{{y}^{3}}}{3} \right)_{0}^{1} \\ & \int_{{}}^{{}}{\int_{{}}^{{}}{(x,y)=}}\left( 1+\frac{{{1}^{3}}}{3} \right) \\ & \int_{{}}^{{}}{\int_{{}}^{{}}{(x,y)=}1+\frac{1}{3}} \\ & \int_{{}}^{{}}{\int_{{}}^{{}}{(x,y)=}\frac{4}{3}} \\ \end{align}\ \)

Portanto, o valor da integral dupla será \(\boxed{I = \frac{4}{3}}\).