A question in my Calculus book states:
"Find the volume of the solid in the first octant bounded by the cylinder $z=9-y^2$ and the plane $x=2$"
When this answer was being covered in class, it was decided that the double integral to find this should be:
$\int_{0}^{3} \int_{0}^{2} 9-y^2 dxdy $
The process I understand well, but I am not quite following 'why' or 'how' the boundaries for the integral for $y$ was determined.
In my notes I have written: $$ \begin{aligned} 0 &=9-y^2 \\ y^2&=9\\ y&=3 \end{aligned} $$
But this still doesn't seem quite clear to me as to why it is done. Why is z set to 0 to get this? Any help would be appreciated.
Have you thought about the graphs of these? In the yz- plane, z= 9- y^2 is a parabola with vertex at y= 0, z= 9 opening downward. Since there is no "x" in the formula, x can be anything so that is a parbolic cylinder extending infinitely along the x- axis. Since this is in the first quadrant, the planes x= 0 and z= 0 are part of the boundary and we are told that x= 2 is a boundary. Since one boundary is z= 0, that will intersect the cylinder z= 9- y^2 where 0= 9- y^2. That is the equation you have which gives them y= 3 and y= -3. The cylinder cuts the plane z= 0 in those two lines, extended parallel to the x- axis. So the projection of the cylinder into its base, the z= 0 plane, is the rectangle bounded by the lines x= 0, x= 2, y= 3, and y= -1. That is the region you integrate over.