Problem 8 asks you to approximate the requested double-integral when you're given the level curves. The level curves look something like:
The idea is to break the region up into rectangles, calculate the height of the function at the midpoint of the rectangle, and then approximate the volume above that rectangle by assuming the function has the same height throughout. For example, let's break it up into a single rectangle, and look at the midpoint (marked in blue).
The height at the midpoint is 11, so let's assume that the function is uniformly 11 throughout the entire region. Then our approximation for the volume would be 11 ⋅ 1 ⋅ 1 = 11.
If we break up the region into four rectangles, each with sides of length 1/2, then we'd get a picture like:
Here the approximation is a little finer, but it requires more calculation. Clockwise from top-left, the heights at the midpoints are about 12.8, 10.9, 9.9, and 11 (according to my drawing). Thus we calculate the volume above each rectangle, clockwise from top-left, is 3.2, 2.725, 2.475, 2.75. Adding these up, we get an approximation for the total area of 11.15.
Of course, you can continue this process and do as many rectangles as you want. If you really want to, I welcome you to try sixteen regions:
As always, feel free to email me if you still have questions on this solution or others.