Have you tried to draw the picture? The first equation is a cylinder whose xy-trace is a circle of radius 1 centered at (0,0). Draw that circle then extend it up and down the z-axis. Then draw the sphere. The inner region will be the cylinder with a rounded top and bottom. The outer region will be a sphere missing the parts of it at the very top that were "taken on" by the cylinder.
After drawing the picture, think about in which planes you will get one-to-one projections. for the sphere with the tops of it missing, you will get a one-to-one projection in the xy-plane if you only consider the top half. think about collapsing that sphere with the missing tops into the xy-plane, literally smashing it from 3D into 2D..it will look like a washer. The inner circlular boundary of the washer region will be the circle of radius one centered at the origin, the outer circle washer region will be a circle of radius 6 centered at the origin. Remember that we are only considering the top half of the sphere to allow for a one-to-one projection...so after you evaluate the surface integral you will need to multiply the result by two to get the surface area of the top and bottom.
There's a lot more steps when it comes to actually setting up and evaluating the integral, you have to consider the and that might be a little problematic.
Another thing, those surfaces are really really nice. We know formulas for the surface areas of cylinders and spheres...so if you're not required to set up a surface integral, maybe using some formulas from geometry would be a lot easier. However, I am assuming this is not the case since you are posting in the calculus section.
And...the title of your post is surface areas, so I'm assuming that in your post when you say area, you mean surface area.
Hope this got you started,