In your first question, the cross sections are square. Now, the dimensions are delineated by the equation for a circle centered at the origin with a radius of one. Try and picture this in your mind's eye. Visualization is key to these problems. The bottom leg of the square is the height above the x-axis given by (1-x^2)^.5 and below the x-axis by -(1-x^2)^.5 This boundary is that of a circle and therefore symmetric about the x- axis. Therefore, the length of the bottom leg of the square cross-section has a length of 2(1-x^2)^.5 If that is the length of one side of the square, what is its area? Obviously, 4(1-x^2). Now, you have your area function. Simply integrate along the x-axis from -1 to 1 and there is your answer.
Second question. You can rewrite the equation sure. However, you'll need the equation in terms of the variable whose axis you will revolve about. If you are revolving about the y axis, then your equation needs to be in the form x= ..... not y=..... otherwise you'll end up with a mix of x and y in the integral and you can't have that. (Not yet anyway...wait until next semester!) So, keep it as x = y^(3/2) that's your integral. In these problems, you often need to hunt for the boundaries of integration. They are usually easy to find. Here you have x = 0 and y = 2. Well, we need everything in terms of y. So, when x = 0 what does y equal? 0 Square the integrand so as to construe it as a radius of an infinitesimal circular cross section and integrate (y^(3/2))^2 from 0 to 2. multiply by pi and bingo.
The third. Often, you will see this type of volume of revolution. The first two types discussed above are relatively simple. This third, however, is of a third, somewhat more sophisticated type. There are shapes that when revolved, produce cavities. This is one of those. If I am picturing this graph correctly in my mind's eye, this solid does not produce a completely solid shape. There is an opening like that of brass instrument. These volumes are calculated thus: square both functions as to construe each as the radius of the circular cross section of infinitesimal width. Take the squared "lower" function (the function lower on the graph grid) and subtract it from the squared "higher" function. This is your integral. You are calculating the volume created by the higher function minus that of the lower. Integrate and multiply by pi.