This one is pretty simple, we can see at the maximum angle, r is along the diagonal within the
1st quadrant, so that means y = x at any given point. That will our max term in terms of x and
the lower bound should just be y when theta = 0 which is also just 0. Note, I used the fact that
the maximum x should also be equals to the maximum r in order to find bounds for x. Since
again, theta only spans the first quadrant, x lower is bound is at 0 instead of (-)maximum x like
we usually see in this sort of problem.
Explanation Below
This one is harder for sure. Rho and theta are pretty easy to solve, but we need a bit of trickery
to solve for phi. We can use the fact that the top bound of z can be rewritten in the form
x^2 + y^2 + z^2 = 4z and that the maximum angle phi should be when the two planes intersect.
Therefore, since the previous equation is a rewriting of the upper bound, we can keep in mind
that z is also = sqrt(x^2 +y^2) and rewrite the left side of the equation as 2z^2. This is able to be
done since z is both equals to the top and bottom equation at the same time, so we can
substitute into or out of one equation as needed. Solving for this tells us that z = 2 at any
intersection. Since phi is at its max at intersection, and rho will be at its maximum value
(rho = 4cos(phi)) at any point that intersects with the upper bound, we can use this info to say 2
= (rho)cos(phi) (as it should be at any point) and plug in our maximum value for rho. We then
say 2 = 4cos^2(phi). Again, z is at its intersection, which is where phi maxes, and this
interception is always going to be at a point along the upper bound, which means rho is at its
maximum as well, and we know the value of z at this point.
So if we solve for phi here, we should get our maximum phi which ends up being either
pi/4 or 7pi/4. Conventionally, 0 <= phi <= pi, so we’ll say that phi maximum = pi/4 since the other
value would just be redundant (it gives the same answer.) The minimum for phi will be 0 since
these functions intersect with the z-axis somewhere(there’s a point at (0,0))