Why Inverse Boundaries Matter for Horizontal Shells
When a region is rotated about the \(x\)-axis, horizontal slices are parallel to the axis of rotation. Rotating a thin horizontal slice produces a cylindrical shell. Its radius is the slice’s distance from the \(x\)-axis, while its height is the total horizontal length of the part of the slice that lies inside the region.
That horizontal length is not always easy to read from the original equations. A boundary may be written as \(y=f(x)\), while a horizontal slice asks which \(x\)-values occur at a particular height \(y\). The practical step is to solve the boundary equation for \(x\) in terms of \(y\). If a curve is not one-to-one on the relevant interval, it may have multiple inverse branches, and the horizontal slice may need to be described in separate pieces.
Inverse Branches and Horizontal Slice Length
Suppose a boundary curve is given by \(y=f(x)\). To use it as a horizontal boundary, solve \(y=f(x)\) for \(x\). An inverse function \(x=f^{-1}(y)\) is defined only when \(f\) is one-to-one on the interval being considered. If \(f\) is not one-to-one, restrict it to intervals where it is one-to-one; each restriction gives an inverse branch.
At a fixed height \(y\), the region’s horizontal slice may be one interval \([x_{\text{left}}(y),x_{\text{right}}(y)]\). Its length is \(x_{\text{right}}(y)-x_{\text{left}}(y)\). It may instead consist of several disjoint intervals. In that case, add the lengths of the intervals that actually lie in the region; do not include the gaps between them.
For a region lying on or above the \(x\)-axis, a shell at height \(y\) has radius \(y\). If its horizontal length is \(H(y)\), its circumference is \(2\pi y\), so the shell-volume formula is:
The bounds must cover the region’s range of \(y\)-values. The formula applies when the horizontal slices describe the region’s material without counting any gaps as part of a shell. If the slice structure or its boundary formulas change at a height, split the integral there.
Why the Shell Formula Works
A thin horizontal strip at height \(y\), with thickness \(\Delta y\) and horizontal length \(H(y)\), sweeps out an approximately cylindrical layer when rotated about the \(x\)-axis. Its radius is approximately \(y\), its circumference is approximately \(2\pi y\), and its thickness is \(\Delta y\). Its volume is therefore approximately \(2\pi yH(y)\Delta y\).
Adding these approximations over thin strips gives a Riemann sum. As the strip thickness tends to zero, the sum becomes the integral of the shell volume per unit height:
This explains why both factors matter: \(y\) determines the shell radius, and \(H(y)\) measures the actual horizontal amount of region rotated at that height.
Worked Example: A Parabola with Two Inverse Branches
Worked Example: A Parabola with Two Inverse Branches
Find the volume obtained by rotating the region bounded by \(y=x^2\) and \(y=4\) about the \(x\)-axis.
Step 1: Find the height range. The lowest point of the region is at \(y=0\), and the top boundary is \(y=4\). Thus \(0\le y\le4\).
Step 2: Express the boundaries as \(x\)-values. Solving \(y=x^2\) gives \(x=-\sqrt y\) and \(x=\sqrt y\). These are the two inverse branches of the parabola. At each height, the slice runs from the left branch to the right branch, so
Step 3: Set up and evaluate the shell integral.
Check: The derivative of \(\frac{2}{5}y^{5/2}\) is \(y^{3/2}\), as required. The height \(2\sqrt y\) is nonnegative over \(0\le y\le4\), and the volume is positive.
The original parabola is not one-to-one over the whole region’s \(x\)-range. Using only one inverse branch would omit half of each horizontal slice. Finding both branches before subtracting is what gives the full shell height.
Worked Example: Inverse Branches of a Downward-Opening Parabola
Worked Example: Inverse Branches of a Downward-Opening Parabola
Find the volume formed by rotating the region between \(y=0\) and \(y=4x-x^2\), for \(0\le x\le4\), about the \(x\)-axis.
Step 1: Find the height range. The curve reaches its maximum at \(x=2\), where \(y=4(2)-2^2=4\). The region therefore has \(0\le y\le4\).
Step 2: Solve for the inverse branches. Starting with \(y=4x-x^2\), rearrange and complete the square:
For each \(y\) from \(0\) to \(4\), the left boundary is \(2-\sqrt{4-y}\) and the right boundary is \(2+\sqrt{4-y}\). Hence
Step 3: Evaluate the integral. The shell formula gives
Use \(u=4-y\), so \(y=4-u\) and \(dy=-du\). When \(y=0\), \(u=4\); when \(y=4\), \(u=0\). Therefore
Thus \(V=4\pi\cdot\frac{128}{15}=\frac{512\pi}{15}\).
Check: An antiderivative of \(y\sqrt{4-y}\) is \(-\frac{8}{3}(4-y)^{3/2}+\frac{2}{5}(4-y)^{5/2}\). Differentiating gives \(y\sqrt{4-y}\). Its change from \(y=0\) to \(y=4\) is \(0-(-128/15)=128/15\), matching the integral.
When a Horizontal Slice Has Separate Pieces
Two inverse branches do not always form the two ends of one continuous interval. The region itself determines which parts of a horizontal line are included. For example, a horizontal slice through the region below a U-shaped curve may have a left piece and a right piece with a gap between them. Add the lengths of those pieces separately.
Worked Example: A Horizontal Slice with a Gap
Rotate the region \(0\le x\le2\), \(0\le y\le(x-1)^2\), about the \(x\)-axis. Find its volume.
Step 1: Determine the height range. On \(0\le x\le2\), the curve \((x-1)^2\) ranges from \(0\) to \(1\). Therefore \(0\le y\le1\).
Step 2: Find all \(x\)-intervals in a horizontal slice. The condition for a point to be in the region is \(y\le(x-1)^2\). For fixed \(y\), this means \(|x-1|\ge\sqrt y\), or
After also imposing \(0\le x\le2\), the slice consists of \([0,1-\sqrt y]\) and \([1+\sqrt y,2]\). Their lengths are \(1-\sqrt y\) and \(1-\sqrt y\), so the total shell height is
Step 3: Set up and evaluate the volume.
Check: Differentiating \(\frac{y^2}{2}-\frac25y^{5/2}\) gives \(y-y^{3/2}\). The evaluated integral is positive, as it must be for a nonzero region.
A Common Mistake: Counting the Gap
In the last example, the leftmost and rightmost \(x\)-values are \(0\) and \(2\). A tempting but incorrect shell height is therefore \(2-0=2\). That treats the gap between \(1-\sqrt y\) and \(1+\sqrt y\) as if it were part of the region.
Using that incorrect height would give \(2\pi\int_0^1 2y\,dy=2\pi\), which is not the volume of the rotated region. The correct height counts only the two included intervals: \(2(1-\sqrt y)\). The corresponding volume is \(\frac{2\pi}{5}\).
A Reliable Setup Procedure
Before integrating, identify the actual intervals of the region at each height. If an inverse branch or the number of intervals changes, split the \(y\)-range at that height and write a separate expression for each piece.
Use the region’s lowest and highest \(y\)-values.
Identify the applicable inverse branch or branches on the height range.
At a fixed \(y\), determine every interval of \(x\)-values that is actually in the region.
Use the sum of interval lengths as \(H(y)\); exclude any gaps.
Use \(2\pi\int yH(y)\,dy\), then verify the antiderivative and the sign of the result.
Check Your Understanding
For each question, identify the inverse branches and the actual horizontal slice intervals before writing the shell height.
- For the region bounded by \(y=x^2\) and \(y=9\), what are the left and right \(x\)-boundaries at height \(y\)?
- For \(y=6x-x^2\) on \(0\le x\le6\), find both inverse branches and the horizontal shell height for \(0\le y\le9\).
- For \(0\le x\le2\), \(0\le y\le(x-1)^2\), why is \(H(y)\) not \(2\) when \(0<y<1\)?
- If a horizontal slice consists of three disjoint intervals, how should its shell height be calculated?
- State the horizontal shell-volume formula for a region above the \(x\)-axis and define each factor.
