Answer:
[tex]\int\limits^1_0 \int^1_0{z} \, dA[/tex]
Step-by-step explanation:
[tex]\int\limits^1_0\int^1_0 {9-4x^2-5y^2dy} \, dx[/tex]
[tex]\int\limits^1_0 {9y-4x^2y-\frac{5y^3}{3}|_0 ^1 } \, dx[/tex]
[tex]\int\limits^1_0 {22/3 -4x^2} \, dx[/tex]
You finish the leftover.