Problem 4: For part (a), by symmetry the integral of z^3 + \sin y over the unit ball is 0 and by geometric interpretation of integral, the integral of 3 over the unit ball is 3 times the volume of the unit ball, which is 4\pi. Most students got this part right. For part (b), without the help of symmetry, it is hopeless to integrate z^3 + \sin y under any of the coordinates. For the integral of 3 over the unit ball, the spherical coordinates would be the best option.
Problem 5: Some students forgot to write down the names of the surfaces or write down the wrong name for part (b). Few students did not simplify the trigonometric functions in the answer for part (b).
Problem 8: Most problems appeared when students tried to use cylindrical coordinates. The straightforward way to break the volume into two integrals, one of which is \int_0^{2\pi}\int_0^{\sqrt{2}}\int_0^r rdzdrd\theta and the other of which is \int_0^{2\pi}\int_{\sqrt{2}}^2\int_0^{\sqrt{4-r^2}}rdzdrd\theta. Another approach is to consider the complement of the solid, whose volume can be easily written as \int_0^{2\pi}\int_0^{\sqrt{2}}\int_r^{\sqrt{4-r^2}} rdzdrd\theta. Then subtract this volume from the volume of half of the ball.
Problem 9: The most common mistake happened when students were trying to describe the solid in terms of the spherical coordinates. Few students claimed that \theta changes from 0 to 2\pi, \phi changes from 0 to \pi / 6 and \rho changes from 0 to a.. However, the correct description of the solid should be as follows. \theta changes from 0 to \pi / 6, \phi changes from 0 to \pi and \rho changes from 0 to a.