Why Shell Bounds Come from the Region
In a shell integral, the variable of integration tracks the position of a thin slice parallel to the axis of rotation. The shell bounds must therefore describe where the actual region exists in that coordinate—not merely where two formulas happen to intersect.
For vertical shells, the slice position is \(x\), so the relevant set of values is the region's projection onto the \(x\)-axis. For horizontal shells, the slice position is \(y\), so the relevant set is its projection onto the \(y\)-axis. Once that projection is known, its endpoints and any gaps determine the integration limits.
This distinction matters when curves are extended beyond the boundary segments that enclose the region, when a boundary has a turning point, or when the region has more than one separated piece.
From Projection to Shell Limits
If the region is connected, its projection onto either coordinate axis is an interval. For example, suppose two \(x\)-values \(a\) and \(b\) occur in the projection, with \(a<b\). If an intermediate value \(c\) did not occur, the region would have points on both sides of the vertical line \(x=c\), but none on that line. The parts with \(x<c\) and \(x>c\) would then separate the region into two nonempty pieces, contradicting connectedness. Thus every value between \(a\) and \(b\) also occurs.
A bounded closed region has a bounded, closed coordinate projection. If the region is connected, that projection is an interval. If it is not connected, the projection can have gaps, and the integral must cover its separate pieces rather than the entire interval between the smallest and largest coordinates.
For vertical shells, the projection \(P_x\) supplies the possible \(x\)-values; for horizontal shells, \(P_y\) supplies the possible \(y\)-values. The shell height is found from the region's boundary at each such coordinate. A candidate intersection helps identify a possible endpoint, but it is relevant only if it lies on the actual boundary of the chosen region.
A Reliable Endpoint-Finding Process
Vertical shells use \(x\); horizontal shells use \(y\). The projection in that coordinate is the set of candidate integration values.
Solve equations for boundary-curve intersections. Also check intersections with coordinate boundaries, any endpoints given in the problem, and coordinate extrema along each boundary curve.
Check each point against the region's inequalities or specified boundary segments. A solution of equations for extended curves is not automatically part of the region.
Write its interval or separate intervals. Confirm that every shell position between stated endpoints intersects the region.
Solving an equation may create candidates that need special checking. For instance, squaring both sides can introduce solutions that do not satisfy the original equation. Likewise, two curve extensions may intersect outside the boundary segments used to define the region. Coordinate extrema require a separate check: a boundary can turn at an endpoint of its projection even though it does not intersect another boundary curve there.
Worked Example: Intersections Set the Bounds
Worked Example: Intersections Set the Bounds
Let \(R\) be the region between \(y=x^2\) and \(y=2x\), and use vertical shells about the \(y\)-axis. Find the shell limits and set up the volume integral.
First find the intersections:
The curves meet at \(x=0\) and \(x=2\). Between those values, \(2x-x^2\geq 0\), so the line \(y=2x\) is above the parabola \(y=x^2\). The region is precisely the set \(0\leq x\leq2\), \(x^2\leq y\leq2x\); its \(x\)-projection is \([0,2]\).
At each \(x\), the shell height is \(2x-x^2\), and its radius is \(x\). Therefore:
Evaluating gives:
At both endpoints the curves meet, so the shell height is zero: \(2(0)-0^2=0\) and \(2(2)-2^2=0\). Also, the height \(2x-x^2=x(2-x)\) is nonnegative throughout \([0,2]\), as required for the stated top-minus-bottom expression.
Worked Example: Rejecting an Extraneous Intersection
Worked Example: Rejecting an Extraneous Intersection
Let \(R\) be the triangular region bounded by \(x=0\), \(y=x\), and \(y=2-x\). Find its \(x\)-projection and set up its volume when rotated about the \(y\)-axis using vertical shells.
The two sloping lines intersect where \(x=2-x\), so \(2x=2\) and \(x=1\). The line \(x=0\) meets \(y=x\) at \((0,0)\) and \(y=2-x\) at \((0,2)\). These three vertices show that the triangle extends from \(x=0\) to \(x=1\). In this interval its lower boundary is \(y=x\), and its upper boundary is \(y=2-x\).
There is another tempting calculation: setting \(y=0\) in \(y=2-x\) gives \(x=2\). But \(y=0\) is not a boundary line of this triangle, and the point \((2,0)\) is not on its boundary. It is only an intersection of the extended line \(y=2-x\) with the \(x\)-axis. It must not be used as a shell endpoint.
The correct projection is \([0,1]\), with shell height \((2-x)-x=2-2x\) and radius \(x\). The volume setup and evaluation are:
As an endpoint check, the height is \(2\) at \(x=0\) and \(0\) at \(x=1\), matching the triangle's vertical edge and its top vertex. It remains nonnegative on the entire interval because \(2-2x\geq0\) for \(0\leq x\leq1\).
Boundary Turning Points Also Matter
Not every endpoint of a projection comes from two boundary curves meeting. A smooth curve can turn in the coordinate being projected. For example, the circle \(x^2+y^2=1\) reaches its greatest and least \(x\)-values at \((1,0)\) and \((-1,0)\). These are coordinate extrema on the curve, and they determine the endpoints of the disk's \(x\)-projection.
Worked Example: A Projection Endpoint at a Curve Extremum
Let \(R\) be the disk \(x^2+y^2\leq1\). Find its \(x\)-projection and set up the volume obtained by rotating the disk about the \(y\)-axis using shells.
On the circular boundary, \(y^2=1-x^2\). Real \(y\)-values exist exactly when \(1-x^2\geq0\), which is equivalent to \(-1\leq x\leq1\). Thus \(P_x=[-1,1]\). Its endpoints arise where the circle has an \(x\)-coordinate extremum: setting \(y=0\) in the circle equation gives \(x^2=1\), hence \(x=\pm1\).
For shells about the \(y\)-axis, using the whole interval \([-1,1]\) would count each radius twice: the slices at \(x\) and \(-x\) generate the same circular shell. The disk is symmetric about the \(y\)-axis, so use its right half, whose projection is \([0,1]\). There the shell height is \(2\sqrt{1-x^2}\), and its radius is \(x\).
Set \(u=1-x^2\), so \(du=-2x\,dx\). When \(x=0\), \(u=1\); when \(x=1\), \(u=0\). Therefore:
The antiderivative checks because \(\frac{d}{du}\left(\frac{2}{3}u^{3/2}\right)=u^{1/2}\). The endpoint check is geometric: the vertical slice at \(x=1\) has zero height, while the slice at \(x=0\) has height \(2\). The bound \(x=1\) is essential even though no second boundary curve intersects the circle there.
When a Projection Has Gaps
A region does not have to be connected merely because it is bounded and closed. For example, consider the union of two rectangles, \(0\leq x\leq1,\ 0\leq y\leq1\) and \(3\leq x\leq4,\ 0\leq y\leq1\). Its \(x\)-projection is \([0,1]\cup[3,4]\), not the whole interval \([0,4]\). For vertical shells about the \(y\)-axis, integrate over those two pieces:
The shells occupy radii from \(0\) to \(1\) and from \(3\) to \(4\), so these two sets of shells do not overlap. Integrating across the gap from \(1\) to \(3\) would instead add shells that do not come from the region.
Common Mistake: Solving Intersections Without Checking the Region
A common error is to solve equations for all boundary curves and use every resulting coordinate as an endpoint. In the triangular example, that would include \(x=2\), obtained by intersecting the extension of \(y=2-x\) with \(y=0\). But \((2,0)\) is not on the triangle. Including it would replace the correct interval \([0,1]\) with an interval that contains positions outside the region.
The correction is to test each candidate against the actual region and its boundary segments. Then check for extrema along each boundary curve and confirm that the proposed intervals cover exactly the coordinate values occupied by the region. For a connected region, a proposed projection interval must contain no gaps; for a region with disconnected projection, integrate over each separate piece.
Check Your Understanding
For each question, identify the relevant coordinate projection before deciding on shell limits.
- For the region between \(y=x^2\) and \(y=4\), what equations identify the candidate endpoints of the \(x\)-projection, and which solutions belong to the bounded region?
- Why does the intersection of an extended boundary curve with a coordinate axis not necessarily give a shell endpoint?
- What additional boundary features should be checked besides intersections and explicitly specified endpoints?
- A bounded closed region consists of two separated rectangles. Must its coordinate projection be one interval? Explain what this means for the integral limits.
- For the disk \(x^2+y^2\leq1\), why is \(x=1\) an endpoint of the \(x\)-projection even though it is a smooth-curve extremum?
