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Volumes by Cylindrical Shells · Tutorial 383 of 998

Horizontal Shells from Curves Written as x=g(y)

Use right-minus-left boundary functions to find horizontal shell heights and calculate volumes by integrating with respect to y.

Intermediate 9 min read

What You'll Learn

  • Identify the left and right boundaries of a horizontal slice from equations written as x=g(y)
  • Express shell height as the right boundary minus the left boundary
  • Set up a shell integral using the horizontal thickness dy
  • Choose the shell radius for a region above or below the x-axis
  • Check the boundary order and verify a completed integral
  • Recognize why a region crossing the axis needs special care

Reading Horizontal Shells from \(x=g(y)\)

When a region is described by equations of the form \(x=g(y)\), each boundary gives an \(x\)-coordinate for a chosen height \(y\). A horizontal slice runs from its left endpoint to its right endpoint, so its length is found by subtracting the left \(x\)-coordinate from the right \(x\)-coordinate.

For a horizontal shell about the \(x\)-axis, that slice becomes a cylindrical shell. Its height comes from the horizontal length of the region, its radius comes from the slice’s distance to the axis, and its thickness is measured in the \(y\)-direction. The central task here is to read the height correctly from the two boundary functions.

Definition: Suppose a region has horizontal slices extending from the left boundary \(x=x_L(y)\) to the right boundary \(x=x_R(y)\) for \(a\le y\le b\). If \(x_L(y)\le x_R(y)\) throughout that interval, the horizontal shell height is \(h(y)=x_R(y)-x_L(y)\).

Boundary Order Determines the Subtraction

The expressions \(x_L(y)\) and \(x_R(y)\) name positions, not functions that can be subtracted in either order. At a fixed value of \(y\), the right boundary has the greater \(x\)-coordinate, so the horizontal distance between the endpoints is

$$ \text{horizontal length} = \text{right endpoint}-\text{left endpoint} = x_R(y)-x_L(y). $$

If the curves are given as \(x=g_1(y)\) and \(x=g_2(y)\), first identify which function is farther right on the stated \(y\)-interval. Then use that function as \(x_R\); the other is \(x_L\). A difference that is negative at some height is a warning that the stated boundary order is wrong there, or that the region has not been described by one consistent pair of boundaries on the whole interval.

The interval \(a\le y\le b\) is taken as given in the examples below. In an application, it must cover the part of the region being revolved, and the left-right order must hold throughout that interval. Determining bounds from curve intersections is a separate step.

From Slice Length to Shell Integral

For rotation about the \(x\)-axis, a horizontal slice at height \(y\) lies a distance \(|y|\) from the axis. If the region lies entirely above the axis, the radius is \(r(y)=y\). If it lies entirely below, the radius is \(r(y)=-y\). In either case, \(r(y)=|y|\), and the shell height is \(x_R(y)-x_L(y)\).

Shell formula: If a region lies entirely on one side of the \(x\)-axis, allowing its boundary to touch the axis, and its horizontal slices run from \(x_L(y)\) to \(x_R(y)\) for \(a\le y\le b\), then the shell volume is \(V=2\pi\int_a^b |y|\bigl(x_R(y)-x_L(y)\bigr)\,dy\), provided the left-right order holds across the interval.

The one-side condition matters. If a region extends both above and below the \(x\)-axis, horizontal slices at heights \(y\) and \(-y\) can generate shells with the same radius. Integrating across both sides using the displayed formula can count some volume twice. The formula as stated is for a region entirely on one side; do not apply it across an axis-crossing region without a separate treatment of the overlap.

Why the Formula Works

Consider a thin horizontal strip at height \(y\), with thickness \(\Delta y\) and length \(h(y)=x_R(y)-x_L(y)\). When rotated about the \(x\)-axis, it forms a thin cylindrical shell. Its radius is approximately \(|y|\), its circumference is approximately \(2\pi|y|\), and its height is \(h(y)\). Thus its volume is approximately circumference times height times thickness.

$$ \Delta V \approx 2\pi|y|\,h(y)\,\Delta y. $$

Divide the \(y\)-interval into increasingly thin strips and add their shell volumes. The resulting sums approach the definite integral of the shell-volume contribution per unit \(y\). This gives the formula, provided the shells represent the region without duplicate volume. The assumption that the region is on one side of the axis ensures that strips at different \(y\)-values have different radii, apart from a possible boundary at radius zero.

Worked Examples

Worked Example: A Region Above the Axis with a Linear Boundary

A region has \(0\le y\le 2\), left boundary \(x_L(y)=y\), and right boundary \(x_R(y)=4\). Find its volume when revolved about the \(x\)-axis.

First check the order of the boundaries. On \(0\le y\le2\), \(y\le2<4\), so \(x_L(y)=y\) is left of \(x_R(y)=4\). The shell height is

$$ h(y)=x_R(y)-x_L(y)=4-y. $$

The region is above the axis, so \(r(y)=y\). Therefore

$$ \begin{aligned} V &=2\pi\int_0^2 y(4-y)\,dy\\ &=2\pi\int_0^2(4y-y^2)\,dy\\ &=2\pi\left[2y^2-\frac{y^3}{3}\right]_0^2\\ &=2\pi\left(\left(2(2)^2-\frac{(2)^3}{3}\right)-0\right)\\ &=2\pi\left(8-\frac{8}{3}\right)\\ &=2\pi\cdot\frac{16}{3} =\frac{32\pi}{3}. \end{aligned} $$

Check the antiderivative: \(\frac{d}{dy}(2y^2-y^3/3)=4y-y^2\), which is the integrand. The evaluated endpoint difference is \(16/3\), so the shell factor gives \(2\pi(16/3)=32\pi/3\). The volume is \(\frac{32\pi}{3}\) cubic units.

Worked Example: Curved Left and Right Boundaries

A region lies between \(x_L(y)=y^2\) and \(x_R(y)=4-y\) for \(0\le y\le1\). Find the volume formed by revolving it about the \(x\)-axis.

Check the order: on this interval, \(y^2\le1\), while \(4-y\ge3\). Thus \(y^2\) is the left boundary and \(4-y\) is the right boundary. The shell height is

$$ h(y)=(4-y)-y^2=4-y-y^2. $$

Since \(0\le y\le1\), the radius is \(y\). Substituting the height into the shell formula gives

$$ \begin{aligned} V &=2\pi\int_0^1 y(4-y-y^2)\,dy\\ &=2\pi\int_0^1(4y-y^2-y^3)\,dy\\ &=2\pi\left[2y^2-\frac{y^3}{3}-\frac{y^4}{4}\right]_0^1\\ &=2\pi\left(2-\frac13-\frac14\right)\\ &=2\pi\left(\frac{24}{12}-\frac{4}{12}-\frac{3}{12}\right)\\ &=2\pi\cdot\frac{17}{12} =\frac{17\pi}{6}. \end{aligned} $$

Differentiate the antiderivative: \(\frac{d}{dy}(2y^2-y^3/3-y^4/4)=4y-y^2-y^3\), as required. At \(y=1\), the bracket is \(17/12\); at \(y=0\), it is \(0\). The resulting volume is \(\frac{17\pi}{6}\) cubic units.

Worked Example: A Region Below the Axis

A region has \(-2\le y\le0\), left boundary \(x_L(y)=0\), and right boundary \(x_R(y)=y+2\). Find the volume when it is revolved about the \(x\)-axis.

For \(-2\le y\le0\), \(y+2\ge0\), so the stated left and right boundaries are correctly ordered. The height is

$$ h(y)=(y+2)-0=y+2. $$

This region lies below the axis, so the radius is \(|y|=-y\). Hence

$$ \begin{aligned} V &=2\pi\int_{-2}^0(-y)(y+2)\,dy\\ &=2\pi\int_{-2}^0(-y^2-2y)\,dy\\ &=2\pi\left[-\frac{y^3}{3}-y^2\right]_{-2}^0\\ &=2\pi\left(0-\left(-\frac{(-2)^3}{3}-(-2)^2\right)\right)\\ &=2\pi\left(0-\left(\frac{8}{3}-4\right)\right)\\ &=2\pi\cdot\frac{4}{3} =\frac{8\pi}{3}. \end{aligned} $$

Check the antiderivative: \(\frac{d}{dy}(-y^3/3-y^2)=-y^2-2y\), the expanded integrand. Its value at \(-2\) is \(8/3-4=-4/3\), and at \(0\) it is \(0\), giving an integral of \(4/3\). The volume is \(\frac{8\pi}{3}\) cubic units.

A Common Boundary-Order Mistake

Suppose the boundaries in the first example are \(x=y\) and \(x=4\) on \(0\le y\le2\). A common mistake is to subtract in the order the equations happen to be listed:

$$ h_{\mathrm{wrong}}(y)=y-4. $$

This produces a negative “height” throughout the interval; for instance, at \(y=1\), it gives \(1-4=-3\). A horizontal length cannot be negative. The correction is to compare the \(x\)-coordinates: \(y\le2<4\), so \(y\) is the left boundary and \(4\) is the right boundary. Therefore

$$ h(y)=4-y, $$

which gives a positive length, such as \(4-1=3\) at \(y=1\). Always determine the boundary order before subtracting.

Why the Axis Condition Is Essential

For a region contained above or below the \(x\)-axis, distinct horizontal slices have distinct distances from the axis, so the shells fit together without the same shell being counted from opposite sides. A region that crosses the axis does not have this property: slices at opposite heights can rotate into the same cylindrical shell. For example, the rectangle \(0\le x\le1\), \(-1\le y\le1\) becomes a cylinder of radius \(1\) and length \(1\), with volume \(\pi\). Using \(2\pi\int_{-1}^1|y|\,dy\) would give \(2\pi\), counting the volume twice.

Key takeaway: For horizontal shells described by \(x=g(y)\), determine the right and left \(x\)-boundaries at each height and use \(h(y)=x_R(y)-x_L(y)\). For rotation about the \(x\)-axis, pair that height with radius \(|y|\) and thickness \(dy\); apply the full-interval formula only when the region lies on one side of the axis.

Check Your Understanding

For each question, identify the boundary order before setting up any integral.

  1. A region has \(x_L(y)=y+1\), \(x_R(y)=5\), and \(0\le y\le2\). What is its shell height?
  2. A region has boundaries \(x=y^2\) and \(x=3\) for \(0\le y\le1\). Which is the left boundary, and what is the height?
  3. For a region below the \(x\)-axis, what expression gives the radius of a horizontal shell about that axis?
  4. Why can the one-side shell formula double-count volume for a region that crosses the \(x\)-axis?
  5. What does a negative value for a proposed shell height tell you to check?

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