The solid region in the first octant bounded by the cylinder $x^2+y^2=4$, the plane $z=y$, the $xy$ plane and the $yz$ plane.
Here is my thought process:
My first question is where does the plane y=z and the cylinder $x^2+y^2=4$ intersect? So I plugin y=z into the cylinder equation:
$x^2+z^2=4$ Then I solved for $z = \sqrt{4-x^2}$ Now I have the integrand for my double integral. The base of the solid is a quarter circle with
$0 \le x \le 2,0 \le y \le 2$
So finally my integral becomes:
$\int_0^2 \int_0^2 (4-x^2)dy dx$
Does this look correct?
Not nearly correct, unfortunately. You are integrating over a square, not over a quarter circle. And the integrand isn't right either (not even if you include the square root sign that you seem to have lost).
Let $E$ be the quarter circle: $$ E = \bigl\{ (x,y)\in\mathbf{R}^2: x\ge 0, \, y \ge 0, \, x^2+y^2 \le 4 \bigr\} . $$ The point $(x,y,z)$ belongs to your region $D$ if and only if $(x,y)\in E$ and $0\le z \le y$; that is, the $z$ coordinate ranges from the $xy$ plane $z=0$, which is the “bottom” of the solid, to the plane $z=y$, which is the “lid”.
So the volume of your region $D$ is $$ \iiint_D dxdydz = \iint_E \left(\int_{z=0}^y dz\right) dxdy . $$ To compute $\iint_E$, it's easiest to use polar coordinates.