To eliminate the remaining six regions by logical analysis

Question

1. Consider the bounded 3-D region R bounded by the planes z = x, z = -x, and the curved surface z = 1 - y²

a. This is a rather cryptic way to describe the region. Which side of these planes and curved surface are we talking about? There are 8 possibilities, unless you're told which way the inequalities run

i. Show that the region z ≤ x, z ≤ -x, z ≤ 1 - y² includes the half-line (x,y,z) = t(0, 0, -1), t ≥ 0. Does this mean that R has to be one of the other seven?

ii. Show that for every point (x,y,z) in the region, R1 = {(x,y,z) : z ≥ x, z ≥ -x, z ≤ 1 - y²,}. We have z ≤ 1, z ≥ 0, -1 ≤ y ≤ 1, x ≤ 1, and x ≥ -1. Hint: for starters, z ≤ 1 because otherwise z cannot be ≤ 1 - y² . Next, how could z be less than 0, when z must be at least as the greater of x and -x? And so it goes

b. To eliminate the remaining six regions by logical analysis would be rather a pain. Here's a picture.

Using the picture of R = R1, or using the inequalities defining R1, find the volume of R, and find z.

(2)  Set up an integral for the area of the curved bounding surface of the R = R1 from problem 1. You'll have to figure out what the shadow, in the x-y plane, of that curved bounding surface is. It looks a little bit like a lemon.