Multivariable Calculus Question of the Day

1. Spherical, because the outer boundary is a sphere, so $\rho$ is constant.
2. Spherical, because the object is symmetric with respect to rotations about the z-axis.
3. Cylindrical, because the inner boundary is a cylinder, simplifying the bounds on $r$. (correct answer)
4. Cartesian, using the washer method by integrating cross-sectional areas.

Explanation: The object has a clear rotational symmetry about the z-axis, suggesting cylindrical or spherical coordinates. The boundaries are the cylinder $r=a$ and the sphere $r^2+z^2 = (2a)^2$. In cylindrical coordinates, the limits are straightforward: $\theta$ from $0$ to $2\pi$, $r$ from $a$ to $2a$, and $z$ from $-\sqrt{4a^2-r^2}$ to $\sqrt{4a^2-r^2}$. The volume element is $r\,dz\,dr\,d\theta$. This is manageable. (A) In spherical coordinates, the outer boundary $\rho=2a$ is simple, but the inner boundary $r=a$ becomes $\rho\sin\phi = a$, or $\rho=a\csc\phi$. This makes the lower limit for $\rho$ a function of $\phi$, which is more complicated than the cylindrical setup. (B) While the object has this symmetry, this reason alone is insufficient; cylindrical coordinates exploit this symmetry more effectively for this particular geometry. (D) The washer method is a valid technique from single-variable calculus, but it is not a 3D coordinate system. The question asks for the best coordinate system for a triple integral.