Here is the graph on $xy-plane$: https://gifyu.com/image/ecLk
Here is the given: Let $S$ be the solid in the first octant that is bounded by the $xy$-plane, $xz$-plane, $2y=x$ and $z=4−x^2−4y^2$
When the solid $S$ is projected on the $xy$-plane. What to do to label the boundary curves and the points of intersection? When graphed, it has been split by $2y=x$ line, do I split it into 2 regions then get the integration formula of both and add? How will I know if it is Type 1 or 2 if not specified?
I think if someone can help with these questions, I can go on answering the actual problems 'cause the beginning part where I am really stuck.
I need to set up an iterated double integral in rectangular coordinates that is equal to the volume of the solid $S$. So, to set them up I need the boundary curves and intersec pts. and to know if it is a Type 1 region (vertical strip) or Type 2 (Horizontal Strip) though I do not know how to get/determine them. Any help or clue is much appreciated.
Using $2y=x$ and/or the $x$ and $y$ intercepts of the ellipse, distinguish between the $x$ and $y$-axes. Remember, the $2y=x$ plane bounds the region of interest, not cut across it. The $xz$-plane contains the $x$-axis, so that is another boundary; the $y$-axis is not. This region in the first octant is also bounded by the given paraboloid.
You can integrate $dydx$ if every line perpendicular to the $x$-axis goes from $y=f(x)$ to $y=g(x)$; $dxdy$ is possible when every line parallel to the $x$-axis goes from $x=f^{-1}(y)$ to $x=g^{-1}(y)$. Only one of these does not require splitting up the region of integration.