The problem involves using a triple integral to find the volume of a region defined by the inequalities below the plane x+2y+2z=4, above the plane z=2x, and constrained to the first octant.
The discussion is ongoing, with participants sharing their thoughts on the intersections of the planes and their implications for setting up the integral. There is a focus on understanding the relationships between the planes and their intersections with the coordinate axes.
Participants express uncertainty about the limits of integration and the intersections of the planes, indicating a need for clarification on these aspects. There are reminders about forum posting rules regarding patience and avoiding "bumping" threads.
Patience, please.Timebomb3750 said:Any assistance would be greatly appreciated. Thanks.
Timebomb3750 said:Homework Statement
Use a triple integral to find the volume of the region. Below x+2y+2z=4, above z=2x, in the first octant.
Homework Equations
V=∫∫∫dV=∫∫∫dxdydz
The Attempt at a Solution
I have no clue where to begin as to finding those darn limits to integrate with. I'm sure I can evaluate the integral just fine, but I need help finding limits.
Where do the planes, x+2y+2z=4, and, z=2x, intersect?Timebomb3750 said:Well, after plotting those two equations into my mac grapher app, it seems my y-limits could be from 0 to 2. But I'm unsure as to finding my x and z limits.
SammyS said:Where do the planes, x+2y+2z=4, and, z=2x, intersect?
Where does each of those planes intersect the coordinate axes?
I should have asked, "Where does each of those planes intersect the coordinate planes?"Timebomb3750 said:Well, z=2x goes through the entire the y-axis, but doesn't intersect any other axes. x+2y+2z=4 intersects axes at x=4, y=2, and z=2.
The two planes intersect at 2y+5x=4.
But what's your point? What do I get out of this?