I would be grateful if anyone could help me understand a concept in double integrals.
I'm attaching a screenshot from a MV Calc textbook. I'm interested in problems 35 and 36.
I have been taught (I thought) that when you have two "z = " curves, you should solve for their intersection and this will give you the level curve on the xy plane.
I have the answer key to these problems. In problem 36, this is exactly what is done. You get a line, y = 1, which forms the upper bound of the region.
But on problem 35, this was NOT the process done. Instead, the region was created only by the two "y = " functions. When you find the intersection of the two planes, I get a line, y = -8/3 + x. But this apparently does not have any impact on the region.
Hmmm…..
I thought you needed to find a level curve of the "z = " function on the xy plane. Indeed, problem 36 did this. Why didn't problem 35?
Here are the photos:
Best Answer
The planes can be rewritten as $z_1=2-x-y$ and $z_2=2x+2y+10$. Then
$$z_2-z_1=3x+3y+8\,,$$
so if $-1\le x,y\le 1$, then $z_2-z_1\ge-3-3+8=2$. If $\langle x,y,z\rangle$ is in side the vertical cylinder between $y=1-x^2$ and $y=x^2-1$, clearly $-1\le x,y\le 1$, so the $z_2$ plane lies entirely above the $z_1$ plane within that cylinder: the intersection of the planes is outside the cylinder and has no effect on the volume in question, which can therefore be obtained by integrating $z_2-z_1$ over the region in the $xy$-plane bounded by the two parabolas.