Vertical Shells About the Y-Axis

Why Vertical Shells Fit Rotation About the Y-Axis

To find the volume created by rotating a region about the y-axis, imagine slicing the region into thin vertical strips. Each strip sweeps around the axis and forms a thin cylindrical shell. Its distance from the axis determines the shell's radius; the strip's vertical length determines its height.

This viewpoint turns a three-dimensional volume into an integral in $x$. The key is to keep the three parts of the shell distinct: circumference around the axis, height along the strip, and thickness across the strip.

Key idea: For a vertical strip at horizontal position $x$, rotated about the y-axis, the shell radius is $x$. If the strip extends from the x-axis up to $y=f(x)$, its height is $f(x)$.

Setting Up the Shell Integral

Consider a region bounded by the graph $y=f(x)$, the x-axis, and the vertical lines $x=a$ and $x=b$. Assume $0\leq a<b$, and that $f$ is continuous and nonnegative on $[a,b]$. These conditions ensure that each vertical strip has one height, $f(x)$, and that its distance from the y-axis is $x$.

A strip of small width $\Delta x$ at $x$ forms a shell with approximate circumference $2\pi x$, height $f(x)$, and thickness $\Delta x$. Its approximate volume is therefore circumference times height times thickness. Summing these contributions and taking the limit gives the volume.

Theorem: If a region under $y=f(x)$ and above the x-axis, for $a\leq x\leq b$, is rotated about the y-axis, where $0\leq a<b$ and $f$ is continuous and nonnegative on $[a,b]$, then the volume is $$ V=2\pi\int_a^b x f(x)\,dx. $$ Here $x$ is the shell radius, $f(x)$ is the shell height, and $dx$ represents the shell thickness in the limiting sum.

In particular, the factor $x$ is not part of the height: it accounts for how far the shell lies from the axis. A strip farther from the y-axis sweeps out a larger circumference, while a strip closer to the axis sweeps out a smaller one.

Why the Formula Works

The shell integral follows from adding the volumes of thin shells. For a shell with constant height $h$, inner radius $r$, and thickness $\Delta x$, the volume is the height times the area of the circular ring at its end. That ring has outer radius $r+\Delta x$ and inner radius $r$, so its area is $\pi((r+\Delta x)^2-r^2)$.

\begin{aligned} \text{shell volume} &=\pi\big((r+\Delta x)^2-r^2\big)h\\ &=\big(2\pi r h\big)\Delta x+\pi h(\Delta x)^2. \end{aligned}

The first term is circumference times height times thickness. The second term contains $(\Delta x)^2$, so its total contribution vanishes as the widths of the partition intervals approach zero. For a region whose height varies with $x$, continuity lets us approximate the height on each sufficiently narrow strip by its value at a sample point. The resulting sums approach the integral of $2\pi x f(x)$.

V=\lim_{\max \Delta x_i\to 0} \sum_i 2\pi x_i^* f(x_i^*)\Delta x_i =2\pi\int_a^b x f(x)\,dx.

This also explains why the method uses vertical strips here: their distance from the y-axis and their height can both be read directly from the horizontal coordinate $x$.

Worked Examples

Worked Example: A Parabolic Region

Rotate the region under $y=4-x^2$, above the x-axis, and between $x=0$ and $x=2$ about the y-axis. The function is nonnegative on this interval, since $4-x^2\geq 0$ for $0\leq x\leq 2$.

At a position $x$, the radius is $x$ and the height is $4-x^2$. Thus

\begin{aligned} V &=2\pi\int_0^2 x(4-x^2)\,dx\\ &=2\pi\int_0^2 (4x-x^3)\,dx\\ &=2\pi\left[2x^2-\frac{x^4}{4}\right]_0^2\\ &=2\pi\left(\left(2(2)^2-\frac{(2)^4}{4}\right) -\left(2(0)^2-\frac{(0)^4}{4}\right)\right)\\ &=2\pi(8-4)\\ &=8\pi. \end{aligned}

Check the antiderivative: $\frac{d}{dx}(2x^2-x^4/4)=4x-x^3=x(4-x^2)$, the required integrand. The endpoint evaluation gives an integral of $4$, so multiplying by $2\pi$ gives $8\pi$ cubic units.

Worked Example: A Triangular Region

Rotate the region under $y=3-x$, above the x-axis, and between $x=0$ and $x=3$ about the y-axis. The graph meets the x-axis at $x=3$, and $3-x\geq0$ throughout the interval.

The shell at $x$ has radius $x$ and height $3-x$. Therefore

\begin{aligned} V &=2\pi\int_0^3 x(3-x)\,dx\\ &=2\pi\int_0^3 (3x-x^2)\,dx\\ &=2\pi\left[\frac{3x^2}{2}-\frac{x^3}{3}\right]_0^3\\ &=2\pi\left(\left(\frac{3(3)^2}{2}-\frac{(3)^3}{3}\right) -\left(\frac{3(0)^2}{2}-\frac{(0)^3}{3}\right)\right)\\ &=2\pi\left(\frac{27}{2}-9\right)\\ &=2\pi\left(\frac{9}{2}\right)\\ &=9\pi. \end{aligned}

Check: the derivative of $3x^2/2-x^3/3$ is $3x-x^2=x(3-x)$. The integral is $9/2$, so the volume is $9\pi$ cubic units.

Worked Example: A Region Under a Square-Root Curve

Rotate the region under $y=\sqrt{x}$, above the x-axis, and between $x=0$ and $x=4$ about the y-axis. The radius is $x$, and the height is $\sqrt{x}$, so the shell integrand is $2\pi x\sqrt{x}$.

\begin{aligned} V &=2\pi\int_0^4 x\sqrt{x}\,dx\\ &=2\pi\int_0^4 x^{3/2}\,dx\\ &=2\pi\left[\frac{2}{5}x^{5/2}\right]_0^4\\ &=2\pi\left(\frac{2}{5}(4)^{5/2} -\frac{2}{5}(0)^{5/2}\right)\\ &=2\pi\left(\frac{2}{5}\cdot 32\right)\\ &=\frac{128\pi}{5}. \end{aligned}

For the endpoint value, $4^{5/2}=(\sqrt{4})^5=2^5=32$. Check the antiderivative: $\frac{d}{dx}(\frac{2}{5}x^{5/2})=x^{3/2}=x\sqrt{x}$ for $x\geq0$. Thus the volume is $\frac{128\pi}{5}$ cubic units.

A Common Setup Error

A frequent mistake is to leave out the shell radius and use only the height in the integrand. For the parabolic region in the first example, that incorrect setup would be

2\pi\int_0^2 (4-x^2)\,dx =2\pi\left[4x-\frac{x^3}{3}\right]_0^2 =\frac{32\pi}{3}.

This calculation is incorrect because it treats every shell as if its circumference were $2\pi$, regardless of its distance from the axis. The actual circumference at position $x$ is $2\pi x$, not $2\pi$. The corrected setup includes $x$: $2\pi\int_0^2 x(4-x^2)\,dx=8\pi$. Its integrand accounts for both the shell's height and its changing circumference.

Takeaway: Before integrating, identify the shell radius, height, and thickness separately. About the y-axis, a vertical strip at $x\geq0$ has radius $x$; for a region between the x-axis and $y=f(x)$, its height is $f(x)$.

When to Use This Setup

The vertical-shell formula is appropriate when each vertical strip through the region gives one segment from the x-axis to the graph, the interval lies on or to the right of the y-axis, and the region is revolved about that axis. Under these assumptions, different vertical positions produce shells with the radii and heights used in the integral.

Check the geometry before calculating. The function should be nonnegative across the stated interval, and the bounds should describe the entire region being rotated. If a strip's height or distance to the axis is not what the formula assumes, the integral must be reconsidered rather than applied automatically.

The method matters because it translates a rotational volume into a one-variable integral that directly reflects the geometry. A reliable setup is often the most important part: once the radius, height, and limits are correct, the remaining work is integration and careful evaluation.

Check Your Understanding

For each question, identify the shell geometry before attempting the calculation.

For the region under $y=5-x$ from $x=0$ to $x=5$, rotated about the y-axis, what are the radius and height at $x$?
Write, but do not evaluate, the shell integral for the region under $y=x^2$ from $x=0$ to $x=2$, rotated about the y-axis.
Evaluate the integral in question 2 and check your antiderivative by differentiation.
Why does the shell integrand include a factor of $x$ when the axis of rotation is the y-axis?
What geometric information is lost if $2\pi x f(x)$ is replaced by $2\pi f(x)$?

Vertical Shells About the Y-Axis

What You'll Learn

Why Vertical Shells Fit Rotation About the Y-Axis

Setting Up the Shell Integral

Why the Formula Works