Para encontrarmos a integral da superficie, realizaremos os calculos abaixo:

\(\begin{align} & f(x,y,z)={{x}^{2}}+{{z}^{2}}=25 \\ & \\ & \int_{{}}^{{}}{\int_{{}}^{{}}{f(x,y,z)}=\int_{{}}^{{}}{\int_{{}}^{{}}{{{x}^{2}}-{{z}^{2}}-25}}dxdy} \\ & \int_{{}}^{{}}{\int_{{}}^{{}}{{{x}^{2}}-{{z}^{2}}-25}}dxdy=\int_{0}^{3}{\int_{0}^{2}{{{x}^{2}}-{{z}^{2}}-25}}dxdy \\ & \int_{0}^{3}{\int_{0}^{2}{{{x}^{2}}-{{z}^{2}}-25}}dxdy=\int_{0}^{3}{\left( \frac{{{x}^{3}}}{3}-x{{z}^{2}}-25x \right)}_{0}^{2} \\ & \int_{0}^{3}{\left( \frac{8}{3}-2{{z}^{2}}-50 \right)}=\left( \frac{8z}{3}-\frac{2{{z}^{3}}}{3}-50z \right)_{0}^{3} \\ & \left( \frac{8z}{3}-\frac{2{{z}^{3}}}{3}-50z \right)_{0}^{3}=\frac{24}{3}-\frac{54}{3}-150 \\ & \left( \frac{8z}{3}-\frac{2{{z}^{3}}}{3}-50z \right)_{0}^{3}=-10-150 \\ & \left( \frac{8z}{3}-\frac{2{{z}^{3}}}{3}-50z \right)_{0}^{3}=-160 \\ & A=160 \\ \end{align}\ \)

Portanto, a área da superfície será de A=160.