x²=8cos²t, y²=8sin²t, so x²+y²=8, which is a circle centre (0,0) and radius 2√2.
I guess the lines should be x=3 and y≥3. y≥3 is a region bounded by the line y=3, so we only need to consider the boundary of the region. I guess that the missing part is that the axes form another boundary, otherwise the area would be infinite. What we then have is a square with vertices (0,0), (0,3), (3,3) and (3,0) and the positive quadrant of the circle (0≤t≤π/2). The area of this quadrant is a quarter the area of the circle =π(2√2)²/4=(8/4)π=2π square units. The area of the square is 3²=9 square units, so the area between the circle, the lines and the axes is 9-2π=2.7168 square units approx.