The objective of this exercise is to calculate the volume of a given solid by integrating through specific limits.
We have two constants, a=2 and b=1.
(a) For part (a), we are required to set up an integral for the volume of the solid. According to the given formula, we have to integrate the square root of the square of radius 'r' (in this case 'a') subtracted by the square of variable 'x' from -r to r. Here, the factor of 16 * pi * R is a scale factor that depends on the actual problem. The volume is thus given by:
\[16 * pi * a * \int_{-a}^{a} \sqrt{a^2 - x^2} dx\]
When the above expression is computed for a=2, it gives a result equal to approximately 201.06*pi.
(b) Similar to part (a), in part (b) the volume of another solid is calculated, where the radius 'r' is now replaced by 'b', and the limits of integration change to 0 to r. The volume is thus given by:
\[16 * pi * b * \int_{0}^{b} \sqrt{b^2 - y^2} dy\]
When this expression is computed for b=1, it evaluates to approximately 12.57*pi.
Please note, in order to obtain numerical results, integral calculus and possibly special mathematical software would be required. As this is a conceptual solution, it is advised to verify and practice the computation with appropriate methods and tools.
Mathematical expressions are advised to be evaluated thoroughly to ensure the correct understanding and application. Following the detailed steps strategically will enable you to solve mathematical problems systematically.