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Horizontal Shells for a Region Bounded in Y

Use horizontal strips to identify shell radius and height, set the correct y-bounds, and compute volumes for regions whose boundaries are given as functions of y.

Intermediate 10 min read

What You'll Learn

  • Identify the left and right boundaries of a region at each height.
  • Choose y-bounds from the region’s endpoints and curve intersections.
  • Build a horizontal-shell integral about the x-axis.
  • Compute volumes for regions bounded by lines and by a line and a parabola.
  • Recognize when horizontal descriptions avoid solving for y in terms of x.
  • Check shell heights, signs, and antiderivatives.

Why Describe a Region by Its Horizontal Slices?

Some bounded regions are most naturally described by saying how far they extend horizontally at each height. If the left and right edges are given as \(x=x_L(y)\) and \(x=x_R(y)\), a horizontal strip has length \(x_R(y)-x_L(y)\). When that strip is revolved about the \(x\)-axis, it sweeps out a thin cylindrical shell.

The volume calculation depends on two measurements: the shell’s radius, which is its distance from the axis of rotation, and its height, which is the horizontal length of the strip. Here we consider regions lying on or above the \(x\)-axis, so the radius at height \(y\) is \(y\). The resulting integral is with respect to \(y\).

Definition: Suppose a region has one horizontal component at every height \(y\) from \(a\) to \(b\), where \(0\leq a<b\), and its left and right boundaries are \(x_L(y)\) and \(x_R(y)\), with \(x_L(y)\leq x_R(y)\). When this region is revolved about the \(x\)-axis, its horizontal-shell volume is $$ V=2\pi\int_a^b y\bigl(x_R(y)-x_L(y)\bigr)\,dy. $$ Here \(y\) is the shell radius and \(x_R(y)-x_L(y)\) is the shell height.

The formula applies directly when the region has a single horizontal interval at each height and the stated boundaries describe that interval throughout \([a,b]\). If the left or right boundary changes, the setup must be divided into appropriate pieces. The restriction \(a\geq 0\) also keeps the radius equal to \(y\); handling a region that crosses the axis requires additional care.

Why the Shell Formula Works

Consider a thin horizontal strip at height \(y\), with length \(h=x_R(y)-x_L(y)\) and thickness \(\Delta y\). Its rotation about the \(x\)-axis produces a thin shell with radius approximately \(y\), height approximately \(h\), and radial thickness \(\Delta y\). The surface area of a cylinder with radius \(y\) and height \(h\) is \(2\pi yh\), so the shell’s volume is approximately \(2\pi yh\,\Delta y\).

The small correction in this approximation can be seen by comparing two cylinders. If the inner radius is \(r\), the outer radius is \(r+\Delta y\), and the height is \(h\), their volume difference is

$$ \pi(r+\Delta y)^2h-\pi r^2h = \bigl(2\pi rh+\pi h\Delta y\bigr)\Delta y. $$

The second term inside the parentheses contributes \(\pi h(\Delta y)^2\), which becomes negligible relative to the leading term as the strip thickness tends to zero. For a curved boundary, the strip height can also vary slightly across its thickness; that variation is accounted for by taking thinner strips and using a representative height for each one.

1
Partition the y-interval.
Divide \([a,b]\) into thin intervals of thickness \(\Delta y_i\), and choose a representative height \(y_i^*\) in each interval.
2
Estimate each shell.
At \(y_i^*\), the radius is \(y_i^*\) and the height is \(h(y_i^*)=x_R(y_i^*)-x_L(y_i^*)\). Its volume is approximately \(2\pi y_i^*h(y_i^*)\Delta y_i\).
3
Take the limit.
Adding the estimates gives a Riemann sum. As the largest strip thickness tends to zero, the sum approaches \(2\pi\int_a^b y h(y)\,dy\), provided the boundaries are continuous on the interval.

This argument explains both the factors in the formula and why the integration variable is \(y\): each small piece is a horizontal strip, so its thickness is measured in the vertical direction.

Worked Example: A Region Bounded by Two Lines

Worked Example: A Triangular Region Revolved About the x-Axis

Let a region be bounded by \(y=0\), \(x=1+y\), and \(x=5-y\). Find the volume when it is revolved about the \(x\)-axis.

First find the height interval. The two sloping boundaries meet when

$$ 1+y=5-y \quad\Longrightarrow\quad 2y=4 \quad\Longrightarrow\quad y=2. $$

The bottom edge is \(y=0\), so the bounds are \(0\leq y\leq 2\). On this interval, the left boundary is \(x_L(y)=1+y\) and the right boundary is \(x_R(y)=5-y\), since

$$ (5-y)-(1+y)=4-2y\geq 0 \qquad (0\leq y\leq 2). $$

Thus the shell radius is \(y\), and the shell height is \(4-2y\). Set up and evaluate the integral:

$$ \begin{aligned} V &=2\pi\int_0^2 y(4-2y)\,dy\\ &=2\pi\int_0^2 (4y-2y^2)\,dy\\ &=2\pi\left[2y^2-\frac{2}{3}y^3\right]_0^2\\ &=2\pi\left(8-\frac{16}{3}\right)\\ &=2\pi\cdot\frac{8}{3} =\frac{16\pi}{3}. \end{aligned} $$

Check by differentiating the antiderivative: \(\frac{d}{dy}(2y^2-\frac{2}{3}y^3)=4y-2y^2=y(4-2y)\), the integrand before the factor \(2\pi\). The height is nonnegative on the entire interval and is zero at \(y=2\), where the boundaries meet. Therefore the volume is \(\frac{16\pi}{3}\) cubic units.

Worked Example: A Parabola and a Vertical Line

Worked Example: Shells from \(x=y^2\) and \(x=4\)

Consider the region bounded by \(x=y^2\), \(x=4\), and \(y=0\). Find the volume when it is revolved about the \(x\)-axis.

The curved and vertical boundaries meet where \(y^2=4\). Since the region lies above \(y=0\), the relevant intersection is \(y=2\). At each height from \(0\) to \(2\), the left boundary is \(x_L(y)=y^2\) and the right boundary is \(x_R(y)=4\). Hence the shell height is \(4-y^2\).

$$ \begin{aligned} V &=2\pi\int_0^2 y(4-y^2)\,dy\\ &=2\pi\int_0^2 (4y-y^3)\,dy\\ &=2\pi\left[2y^2-\frac{y^4}{4}\right]_0^2\\ &=2\pi(8-4)\\ &=8\pi. \end{aligned} $$

For a check, \(\frac{d}{dy}(2y^2-\frac{y^4}{4})=4y-y^3=y(4-y^2)\), as required. Also, \(4-y^2\geq 0\) for \(0\leq y\leq 2\), and the shell height is zero at the intersection \(y=2\). The volume is \(8\pi\) cubic units.

When Horizontal Descriptions Simplify the Bounds

A useful reason to use horizontal shells is that a region may already be described by simple equations for \(x\) in terms of \(y\). In that case, the y-bounds come directly from the region’s lowest and highest points, while the shell height comes from subtracting the left curve from the right curve.

For comparison, in the preceding example the parabolic boundary \(x=y^2\) can be rewritten on the upper branch as \(y=\sqrt{x}\). The region’s vertical description is \(0\leq y\leq\sqrt{x}\) for \(0\leq x\leq4\). That description requires solving for \(y\), whereas the horizontal description uses \(x=y^2\) directly. The horizontal-shell setup therefore avoids introducing a square root into its boundary description.

Worked Example: A Region Between \(x=y^2\) and \(x=2y\)

Find the volume obtained by revolving the region enclosed by \(x=y^2\) and \(x=2y\), for the enclosed part above the \(x\)-axis.

Find the intersections by setting the two expressions for \(x\) equal:

$$ y^2=2y \quad\Longrightarrow\quad y(y-2)=0 \quad\Longrightarrow\quad y=0\text{ or }y=2. $$

Thus the y-bounds are \(0\) and \(2\). To identify the left and right boundaries, compare the curves on this interval:

$$ 2y-y^2=y(2-y)\geq 0 \qquad (0\leq y\leq 2). $$

Therefore \(x_L(y)=y^2\), \(x_R(y)=2y\), and the shell height is \(2y-y^2\). The volume is

$$ \begin{aligned} V &=2\pi\int_0^2 y(2y-y^2)\,dy\\ &=2\pi\int_0^2 (2y^2-y^3)\,dy\\ &=2\pi\left[\frac{2}{3}y^3-\frac{y^4}{4}\right]_0^2\\ &=2\pi\left(\frac{16}{3}-4\right)\\ &=\frac{8\pi}{3}. \end{aligned} $$

The derivative check gives \(\frac{d}{dy}(\frac{2}{3}y^3-\frac{y^4}{4})=2y^2-y^3=y(2y-y^2)\). The height is nonnegative throughout the interval and zero at both intersections. Hence the volume is \(\frac{8\pi}{3}\) cubic units.

This example illustrates the value of keeping the region’s natural orientation. Its boundaries are already expressed as \(x\)-functions of \(y\), and the intersection heights give the shell bounds directly. Rewriting the curves in terms of \(y\) as functions of \(x\) is possible, but is not needed for the horizontal-shell calculation.

A Common Height Error

The shell height is always the right boundary minus the left boundary. Reversing the subtraction gives a negative height and therefore an impossible negative volume. In the last example, at \(y=1\), the boundaries are \(x_L(1)=1\) and \(x_R(1)=2\). The incorrect subtraction gives

$$ x_L(1)-x_R(1)=1-2=-1, $$

which would produce a negative shell contribution \(2\pi(1)(-1)\,dy\). The correct height is \(x_R(1)-x_L(1)=2-1=1\), giving a positive contribution \(2\pi(1)(1)\,dy\). Before integrating, check the order of the boundaries at a height inside the interval, and verify that the proposed shell height is nonnegative throughout.

Checklist for a Horizontal-Shell Integral

  • Confirm that the axis of rotation is horizontal and that the horizontal strips are parallel to it.
  • Find the lowest and highest y-values occupied by the region.
  • Identify the left and right boundary at each height, checking whether their roles change.
  • Use distance to the axis for the radius; for a region above the \(x\)-axis, the radius is \(y\).
  • Use right boundary minus left boundary for the shell height.
  • Integrate \(2\pi\) times radius times height with respect to \(y\), then check the antiderivative and the sign of the height.

When the horizontal bounds and boundary functions are simple, the shell integral often follows with little algebra. Careful identification of the interval, radius, and left-to-right height is what makes the setup reliable.

Check Your Understanding

For each question, identify the bounds and shell dimensions before evaluating any integral.

  1. For the region between \(x=y^2\) and \(x=3y\) above the \(x\)-axis, find the intersection heights and state which curve is left on the enclosed interval.
  2. A region has \(x_L(y)=y+1\), \(x_R(y)=5-y\), and \(0\leq y\leq2\). What are the shell radius and shell height about the \(x\)-axis?
  3. Why is the shell height \(x_R(y)-x_L(y)\), rather than \(x_L(y)-x_R(y)\)?
  4. For the region bounded by \(x=y^2\), \(x=4\), and \(y=0\), state the shell radius, shell height, and y-bounds.
  5. What feature of the curves \(x=y^2\) and \(x=2y\) makes horizontal bounds convenient for their shell-volume calculation?

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