Center of mass question I am not sure how to set the limits for this question:
Let (x¯, y¯) be the center of mass of the triangle with vertices at (−2, 0), (0, 1), (2, 0) and
uniform density δ = 1.
a)  Write an integral formula for y¯. Do not evaluate the integral(s), but write explicitly the
integrand and limits of integration.
b) Find x¯
For a, I did
1/2*$\iint_{-2}^{2} int_{0.5x+1}^{-0.5x+1}y \,dydx$
So the limits for x are -0.5x+1 and 0.5x+1, and y are 2 and -2.
In the correct answer:
1/2*$\iint_{0}^{1} int_{2y-2}^{2-2y}y \,dxdy$
So the limits for x are 2-2y and 2y-2 and y are 1 and 0.
Why are my limits incorrect?
 A: When picking integration bounds, the point is just that your bounds should describe the correct triangle.
Your amounts to saying: For each $x$ satisfying $-2 \le x \le 2$, I'll cover all $y$ satisfying $1+x/2 \le y \le 1-x/2$. Here is a picture of the points that satisfy your conditions:

It's kind of related to the triangle you want (uses some of the same boundary lines) but it's not the correct triangle.

You could fix your answer by instead saying: For each $x$ satisfying $-2 \le x \le 0$, we need to cover all $y$ satisfying $0 \le y \le 1+x/2$, and in addition for each $x$ satisfying $0 \le x \le 2$ we need to cover all $y$ satisfying $0 \le y \le 1 - x/2$. That would give the answer
$$\frac 1 2 \left(\int_{-2}^0 \int_0^{1+x/2} y dy dx + \int_0^2 \int_0^{1-x/2} y dy dx\right)$$
The above is a correct answer to the problem, but it's kind of messy: If we choose $x$ first and $y$ afterward (i.e. making $x$ the outer integral), then the maximum value of $y$ is defined by a different line depending on whether $x<0$ or $x>0$, which is why we needed to break into two integrals.
We can get a simpler-looking answer by instead choosing $y$ first and $x$ afterward (making $y$ the outer integral). In that case, the triangle we want has $y$ values $0 \le y \le 1$, and for each $y$ value we need to cover all $x$ satisfying $2y-2 \le x \le 2-2y$. That gives the answer
$$\frac 1 2 \int_0^1 \int_{2y-2}^{2-2y} y dx dy.$$
Note this version only has one integral to compute, and note that the $dx$ integral is on the outside this time.
Both of the formulas I've given are correct answers to this problem. If you needed to actually compute the final value $y^-$, you could use whichever integral formula you prefer; they'll both come out to the same result.
