Is this double integral over D positive, negative, or zero? R is the region inside the unit circle centered at the origin, R is its right half

Question

I've some confusions about this textbook question:

1. Firstly, isn't this question describing a unit sphere (and not a unit circle)? Because I don't see how a 2D circle could have both a "right" and a "bottom" half...

2. Assuming R is the unit sphere, then I can at least visualize the regions D and B. But the questions still make no sense to me. This chapter covered functions of two variables, yet every question here is integrating a function of a single variable...

3. I tried graphing, for example, $y^3+y^5$ using GeoGebra, but obviously nothing comes up since it doesn't make sense to graph a 2D function in 3D.

4. Question 9 makes no sense in a different way. If R is a sphere, then it doesn't make sense to think about whether a function is positive or negative (i.e. above or below) with respect to it.

I clearly have some very significant misunderstandings here and I think I'm just missing a couple key points. Even if you can't clear up all my misunderstandings, any help whatsoever is greatly appreciated!

Answer 1

1. As Joe has already mentioned in the comments, this is about the unit circle in $\Bbb R^2$. The Right half is where $x \ge 0$, the Bottom half is where $y \le 0$.
2. Actually, the first one is integrating a function of no variables, but I'm betting you didn't have a problem with that. You know it just means a function $f(x, y) = 1$ for all $x$ and $y$. So even though no variables are seen in it, we can still consider it a function of $x$ and $y$. The same thing is true about all the rest. Problems 9 and 10 are about the function $f(x,y) = 5x$ for all $x, y$. Problems 11 and 12 are about the function $f(x,y) = y^3 + y^5$ for all $x,y$. They may be constant with respect to one of the variables, but we can still consider that a function of two variables.
3. You didn't set it up right in GeoGebra. It should graph a surface which is flat in one direction and follows the 5th degree polynomial in the other.
4. $R$ is not a sphere. Of course, this partly because of the misunderstanding of (1). But $R$ is not even a circle. It is a half-circle. Problem 9 says nothing about whether the function is positive or negative. It asks whether the integral is positive or negative, or zero. And by the integral, they mean the value that you get upon performing the integration. Recall that integration is by definition a method for producing a particular number from the function and the set $R$. They want you to tell what the sign of that number will be.

Answer 2

For each of the given integrls the region is part of a unit disk and the integrand on that region is either positive, negative or mixed.

For integrals of problems $8$ and $9$ the integrand is positive so the integral will be positive.

For integrals of problems $12$ and $14$ the integrand is negative so the integral will be negative.

For problems $10$, $11$, and $13$ the integrand takes both positve and negative values but the symmetrical nature of region implies that you have equal values of positve and negtive so the integra is $0$