From Plane Equations to Cartesian Bounds
A solid bounded by planes is described by linear inequalities in \(x\), \(y\), and \(z\). The equations of the planes identify possible boundaries, but they do not by themselves tell us which side of each plane belongs to the solid. Nor do they immediately give integration limits in a chosen order. To obtain valid Cartesian bounds, we must determine the region, find where its boundary planes meet, and inspect its projection and vertical fibers.
This geometric work is essential: a bound that looks plausible can still include points outside the solid or omit points inside it. The method in this tutorial turns plane constraints into bounds that can be checked directly against the original inequalities.
A general plane has an equation \(ax+by+cz=d\), where \(a,b,c\) are not all zero. One side of the plane is described by \(ax+by+cz\le d\), and the other by \(ax+by+cz\ge d\). A solid is bounded only when its inequalities confine the points in every direction. For example, \(x\ge0\), \(y\ge0\), and \(z\ge0\) describe an unbounded octant, not a bounded solid.
Find the Intersections Before Writing the Limits
The vertices of a three-dimensional solid occur where boundary planes meet. To locate a candidate vertex, solve three independent plane equations simultaneously. A plane may also meet another plane in an edge, so solving pairs of equations is useful when identifying projected boundaries. A candidate intersection belongs to the solid only if it satisfies all of the defining inequalities.
Once the boundary has been identified, choose an integration direction. For bounds in the order \(dz\,dy\,dx\), imagine holding \(x\) and \(y\) fixed. The points of the solid above that fixed pair form a vertical fiber. If the fiber is nonempty, its lower and upper endpoints come from boundary planes solved for \(z\). The set of \((x,y)\) values for which such a fiber exists is the projection onto the \(xy\)-plane.
Why Projection Gives Correct Bounds
Suppose the constraints that involve \(z\) can be rewritten as lower bounds \(z\ge g_i(x,y)\) and upper bounds \(z\le h_j(x,y)\). For a fixed \((x,y)\), all those constraints hold precisely when \(z\) is at least every lower bound and at most every upper bound. Thus the fiber is described by \(L(x,y)\le z\le U(x,y)\), where \(L\) is the greatest of the lower bounds and \(U\) is the least of the upper bounds. A fiber exists exactly when \(L(x,y)\le U(x,y)\), together with any constraints that involve only \(x\) and \(y\).
Solve each nonvertical boundary plane for \(z\), noting whether it gives a lower or upper limit.
Require that the lower limit be no greater than the upper limit, and retain all constraints on \(x\) and \(y\).
Describe \(D\) using one variable's range and then the other variable's range, splitting the region if a single set of bounds does not cover it.
Verify that every point within the proposed bounds satisfies the original inequalities and that each point of the solid is represented.
This reasoning also justifies the corresponding iterated integral. If \(f\) is integrable on the solid and the bounds describe it without omission or excess, then the integral can be written with the inner variable ranging over each fiber and the outer variables ranging over the projection. The geometry must be correct before the integral is evaluated.
Worked Example: A Tetrahedron Under an Oblique Plane
Worked Example: A Tetrahedron Under an Oblique Plane
Describe the solid bounded by the coordinate planes and \(x+2y+z=4\), using bounds in the order \(dz\,dy\,dx\).
The coordinate-plane constraints are \(x\ge0\), \(y\ge0\), and \(z\ge0\). The solid lies below the oblique plane, so its remaining inequality is \(x+2y+z\le4\). Solving for \(z\) gives
For this interval to be nonempty, \(4-x-2y\ge0\), so the projection satisfies \(x\ge0\), \(y\ge0\), and \(x+2y\le4\). Its boundary meets the axes at \(x=4\) and \(y=2\). Therefore \(0\le x\le4\), and for each fixed \(x\), \(0\le y\le(4-x)/2\). The bounds are
Check the vertices by intersecting triples of boundary planes. The coordinate planes meet at \((0,0,0)\). Setting \(y=z=0\) in the oblique plane gives \(x=4\), so another vertex is \((4,0,0)\). Setting \(x=z=0\) gives \(2y=4\), or \(y=2\), giving \((0,2,0)\). Setting \(x=y=0\) gives \(z=4\), giving \((0,0,4)\). Each point satisfies all four inequalities, and the projected triangle has the expected intercepts \((4,0)\), \((0,2)\), and \((0,0)\).
As a further check, the proposed bounds imply \(z\ge0\) and \(x+2y+z\le4\). Conversely, any point satisfying the four original inequalities has \(0\le x\le4\), \(0\le y\le(4-x)/2\), and \(0\le z\le4-x-2y\). Thus the bounds describe exactly the solid.
Worked Example: A Wedge with a Sloping Top
Worked Example: A Wedge with a Sloping Top
Consider the solid bounded by \(x=0\), \(x=2\), \(y=0\), \(z=0\), and \(y+z=4\). Find bounds in the order \(dz\,dy\,dx\).
The inequalities are \(0\le x\le2\), \(y\ge0\), \(z\ge0\), and \(y+z\le4\). Solving the last inequality for \(z\) gives \(z\le4-y\), so the vertical fiber is \(0\le z\le4-y\). It exists when \(4-y\ge0\), hence \(0\le y\le4\). The \(x\)-coordinate is independently restricted to \(0\le x\le2\). Therefore
To check the intersections, the planes \(y=0\), \(z=0\), and \(y+z=4\) produce the triangular cross-section in the \(yz\)-plane with vertices \((y,z)=(0,0),(4,0),(0,4)\). The planes \(x=0\) and \(x=2\) place this same cross-section at the two endpoints of the \(x\)-direction. The corresponding six vertices are \((0,0,0)\), \((2,0,0)\), \((0,4,0)\), \((2,4,0)\), \((0,0,4)\), and \((2,0,4)\). Each obeys \(0\le x\le2\), \(y,z\ge0\), and \(y+z\le4\).
Every point inside the stated limits satisfies the inequalities because \(z\le4-y\) is equivalent to \(y+z\le4\). In the other direction, the inequalities force \(x\) and \(y\) into their stated ranges and force \(z\) between \(0\) and \(4-y\). The bounds are therefore exact.
Worked Example: A Solid Between Two Sloping Planes
Worked Example: A Solid Between Two Sloping Planes
Describe the solid defined by \(x\ge0\), \(y\ge0\), \(z\ge x+y\), and \(z\le6-x-y\), using the order \(dz\,dy\,dx\).
The lower and upper planes are already solved for \(z\). For a vertical fiber to exist, the lower value must not exceed the upper value:
Adding \(x+y\) to both sides gives \(2x+2y\le6\), or \(x+y\le3\). Together with \(x,y\ge0\), this is the triangular projection with vertices \((0,0)\), \((3,0)\), and \((0,3)\). Hence one description is
The two sloping planes meet where \(x+y=3\) and \(z=3\). On the projected triangle, the lower and upper values agree along that edge. At the projected vertex \((0,0)\), the vertical endpoints are \(z=0\) and \(z=6\). At \((3,0)\), both endpoints equal \(3\), and at \((0,3)\), both endpoints also equal \(3\). Thus the solid has vertices \((0,0,0)\), \((0,0,6)\), \((3,0,3)\), and \((0,3,3)\). Substitution confirms that each satisfies the four inequalities.
Finally, if \((x,y,z)\) is within the proposed bounds, it has \(x,y\ge0\), \(z\ge x+y\), and \(z\le6-x-y\), so it belongs to the solid. Conversely, any point in the solid must satisfy \(x+y\le3\) by comparing its lower and upper \(z\)-constraints; it therefore lies over the stated triangle and has the stated vertical limits.
A Common Mistake: Losing a Plane Coefficient
When solving for a projected boundary, every coefficient matters. In the first example, the plane is \(x+2y+z=4\). A common incorrect step is to treat the projection boundary as \(x+y\le4\), forgetting the coefficient \(2\). That mistake would give \(0\le y\le4-x\). At \(x=0\), it permits \(y=3\), but the point \((0,3,0)\) violates the actual plane inequality because \(0+2(3)+0=6>4\). It is outside the solid.
The correction is to set the vertical lower endpoint no higher than the upper endpoint: \(0\le4-x-2y\). Rearranging gives \(2y\le4-x\), so the correct limit is \(0\le y\le(4-x)/2\). Substituting a boundary value into the original plane inequality is a quick way to catch this kind of error.
Why These Bounds Matter
Cartesian bounds connect a geometric description to a triple integral. A correct description identifies the solid one fiber at a time and makes clear which plane supplies each endpoint. The same solid may have different valid descriptions in different orders; changing the order may require a different projection or splitting the projected region. The earlier material on choosing integration orders and splitting solids gives tools for that situation, while the central geometric task remains the same: establish the region represented by the bounds.
When setting up bounds from planes, use these checks:
- Confirm that the half-space inequalities select the intended side of every plane.
- Find boundary intersections and test candidate vertices against all constraints.
- Derive the projection by requiring that each fiber be nonempty.
- Substitute proposed endpoints into the original plane equations or inequalities.
- Check that the resulting region is bounded; finitely many inequalities alone do not guarantee boundedness.
Careful projection and substitution prevent errors that can otherwise remain hidden in a long iterated integral. In particular, the bounds must describe a nonempty, bounded solid, not merely a collection of plane constraints that happens to include some of its points.
Check Your Understanding
Use plane intersections and projections to answer the following questions.
- For the solid \(x,y,z\ge0\), \(2x+y+z\le6\), what is its projection onto the \(xy\)-plane?
- For the same solid, write bounds in the order \(dz\,dy\,dx\).
- In the wedge bounded by \(x=0\), \(x=1\), \(y=0\), \(z=0\), and \(2y+z=8\), what is the upper \(z\)-bound?
- For the region \(x,y\ge0\), \(z\ge2x+y\), \(z\le8-x-2y\), derive the inequality defining its projection.
- Why must a plane-bounded solid be required to be bounded in addition to being described by finitely many linear inequalities?