From a Graph to a Shell Integral
A shell integral begins with a geometric reading of the region, not with an equation chosen from memory. The graph tells you which direction to slice, which boundaries each slice meets, how far the slice is from the rotation axis, and over what coordinate values the region extends.
This matters because a plausible-looking integral can still use the wrong boundary, radius, or limits. A careful graph reading separates these decisions before any integration begins.
What to Read from the Graph
For a vertical axis, such as the \(y\)-axis or a line \(x=a\), use vertical slices. A vertical slice at coordinate \(x\) sweeps out a cylindrical shell when the region rotates. Its radius is a horizontal distance, and its height is a vertical length.
For a horizontal axis, such as the \(x\)-axis or a line \(y=b\), use horizontal slices. A horizontal slice at coordinate \(y\) sweeps out a shell. Its radius is a vertical distance, and its height is a horizontal length.
In either orientation, read the limits from the coordinate values covered by the region. If the slice meets different boundary segments in different parts of the graph, split the integral at the coordinate where the boundary changes. Earlier material on shell heights between curves and switching boundaries gives the corresponding boundary comparisons; here the emphasis is on extracting them from the graph.
The condition that the region lies on one side of the axis is important. If the region crosses the axis, slices on opposite sides can rotate into the same shells. Adding both sides as if they produced separate shells counts overlapping volume twice. A later section addresses this issue directly.
A Short Derivation of the Shell Setup
Consider a thin vertical slice of width \(\Delta x\), height approximately \(h(x)\), and distance \(r(x)\) from a vertical axis. Rotating the slice makes a thin cylindrical shell. Its circumference is approximately \(2\pi r(x)\), so its volume is approximately circumference times height times thickness:
Adding these contributions across the region produces a Riemann sum. As the slice widths approach zero, the sum approaches the integral of the shell contributions, provided the boundary dimensions are continuous or piecewise continuous and the region stays on one side of the axis. The horizontal-slice argument is the same, with \(\Delta y\) replacing \(\Delta x\).
The graph-reading task is therefore to identify the correct coordinate interval and the two geometric functions in the integrand.
Worked Example: Vertical Shells from a Parabolic Region
Worked Example: Vertical Shells from a Parabolic Region
The region is bounded by \(y=4-x^2\), the \(x\)-axis, and the \(y\)-axis. It is rotated about the \(y\)-axis. Read the graph information and find the shell integral and volume.
Step 1: Choose the slice direction. The axis is vertical, so use vertical slices. A vertical slice at \(x\) is parallel to the axis and creates a shell.
Step 2: Read the limits. The region begins at the \(y\)-axis, \(x=0\). Its right endpoint is where the parabola meets the \(x\)-axis: $$ 4-x^2=0,\qquad x^2=4,\qquad x=2 $$ on the part of the graph in the region. Thus \(0\le x\le2\).
Step 3: Read the radius and height. The distance from the slice at \(x\) to the \(y\)-axis is \(r(x)=x\). The slice runs from the lower boundary \(y=0\) to the upper boundary \(y=4-x^2\), so \(h(x)=(4-x^2)-0=4-x^2\).
Step 4: Set up and evaluate. $$ \begin{aligned} V &=2\pi\int_0^2 x(4-x^2)\,dx\\ &=2\pi\int_0^2(4x-x^3)\,dx\\ &=2\pi\left[2x^2-\frac{x^4}{4}\right]_0^2\\ &=2\pi\left((8-4)-(0-0)\right)\\ &=8\pi. \end{aligned} $$ The volume is \(8\pi\) cubic units.
Check: Differentiating the antiderivative gives $$ \frac{d}{dx}\left(2x^2-\frac{x^4}{4}\right)=4x-x^3=x(4-x^2), $$ which matches the integrand. The limits also match the graph’s horizontal extent, from \(0\) to \(2\).
Worked Example: Horizontal Shells from a Graph
Worked Example: Horizontal Shells from a Graph
A triangular region has vertices \((0,0)\), \((0,2)\), and \((2,2)\). It is rotated about the horizontal line \(y=-1\). Find the shell integral and volume.
Step 1: Choose the slice direction. The axis \(y=-1\) is horizontal, so use horizontal slices. The graph shows the region extending vertically from \(y=0\) to \(y=2\), giving limits \(0\le y\le2\).
Step 2: Read the horizontal boundaries. At a fixed height \(y\), the slice starts at the left edge \(x=0\) and ends at the slanted side connecting \((0,0)\) to \((2,2)\). That side has \(x=y\), so the shell height is \(h(y)=y-0=y\).
Step 3: Read the radius. The axis is \(y=-1\), below the entire region. A slice at height \(y\) is therefore \(y-(-1)=y+1\) units from the axis.
Step 4: Set up and evaluate. $$ \begin{aligned} V &=2\pi\int_0^2(y+1)y\,dy\\ &=2\pi\int_0^2(y^2+y)\,dy\\ &=2\pi\left[\frac{y^3}{3}+\frac{y^2}{2}\right]_0^2\\ &=2\pi\left(\frac{8}{3}+2\right)\\ &=2\pi\cdot\frac{14}{3}\\ &=\frac{28\pi}{3}. \end{aligned} $$ The volume is \(\frac{28\pi}{3}\) cubic units.
Check: The derivative of the antiderivative is $$ \frac{d}{dy}\left(\frac{y^3}{3}+\frac{y^2}{2}\right)=y^2+y=y(y+1), $$ the required radius-times-height expression. At the bottom of the region the radius is \(1\), and at the top it is \(3\), consistent with the graph and axis.
When a Boundary Changes, Split the Integral
A graph may show one side of the region as several joined curve segments. Even if the region is connected, the expression for the slice height can change at a joining point. Read the graph from left to right (or bottom to top for horizontal slices), and split the integral wherever the slice meets a different boundary formula.
Worked Example: A Graph with a Changing Lower Boundary
A region lies between \(y=3\) and a lower boundary made of two segments: \(y=x\) for \(0\le x\le1\), and \(y=(3-x)/2\) for \(1\le x\le3\). The region is rotated about the \(y\)-axis. Find its shell integral and volume.
Step 1: Choose the slice direction and limits. The axis is vertical, so take vertical slices. The graph extends from \(x=0\) to \(x=3\), and the lower boundary changes at \(x=1\).
Step 2: Read the radius and heights. The radius is \(r(x)=x\), since the region is to the right of the \(y\)-axis. On \(0\le x\le1\), the height is \(3-x\). On \(1\le x\le3\), the height is $$ 3-\frac{3-x}{2}=\frac{3+x}{2}. $$
Step 3: Write a piecewise shell integral. $$ V=2\pi\left(\int_0^1x(3-x)\,dx+\int_1^3x\left(\frac{3+x}{2}\right)\,dx\right). $$
Step 4: Evaluate each part. $$ \begin{aligned} \int_0^1x(3-x)\,dx &=\int_0^1(3x-x^2)\,dx\\ &=\left[\frac{3x^2}{2}-\frac{x^3}{3}\right]_0^1\\ &=\frac32-\frac13=\frac76. \end{aligned} $$ For the second part, $$ \begin{aligned} \int_1^3x\left(\frac{3+x}{2}\right)\,dx &=\int_1^3\frac{3x+x^2}{2}\,dx\\ &=\left[\frac{3x^2}{4}+\frac{x^3}{6}\right]_1^3\\ &=\left(\frac{27}{4}+\frac{27}{6}\right) -\left(\frac34+\frac16\right)\\ &=\frac{45}{4}-\frac{11}{12} =\frac{31}{3}. \end{aligned} $$ Therefore, $$ V=2\pi\left(\frac76+\frac{31}{3}\right) =2\pi\left(\frac{69}{6}\right) =23\pi. $$
Check: On the first interval the derivative of \(\frac{3x^2}{2}-\frac{x^3}{3}\) is \(3x-x^2=x(3-x)\). On the second interval the derivative of \(\frac{3x^2}{4}+\frac{x^3}{6}\) is \(\frac{3x+x^2}{2}\), exactly the second shell integrand. The split at \(x=1\) follows from the graph’s boundary change.
A Pitfall: A Region That Crosses the Axis
Distance is always nonnegative, so it may seem that using an absolute value for the shell radius solves every issue when a region crosses the axis. It does not. The problem is that slices on opposite sides of the axis can sweep out the same radial shells.
For example, take the rectangle \([-1,1]\times[0,1]\) and rotate it about the \(y\)-axis. A tempting but incorrect calculation is $$ 2\pi\int_{-1}^{1}|x|(1)\,dx =2\pi\left(\int_{-1}^{0}(-x)\,dx+\int_0^1x\,dx\right) =2\pi\left(\frac12+\frac12\right)=2\pi. $$ This counts each shell twice: the slice at \(x=r\) and the slice at \(x=-r\) rotate into the same shell.
To count the actual solid once, describe it by its radial distance \(r\) from the axis. For every \(0\le r\le1\), the shell has height \(1\), regardless of which side supplied the slice. Thus $$ V=2\pi\int_0^1r(1)\,dr =2\pi\left[\frac{r^2}{2}\right]_0^1 =\pi. $$ The result is the volume of a cylinder of radius \(1\) and height \(1\). Differentiating \(r^2/2\) gives \(r\), confirming the integral evaluation.
A Reliable Graph-Reading Routine
Decide whether it is vertical or horizontal, and note whether it is a coordinate axis or a shifted line.
Use a vertical slice for a vertical axis and a horizontal slice for a horizontal axis.
The radius is the perpendicular distance to the axis; the height is the slice length across the region.
Use the coordinate extent of the graph, splitting wherever the slice’s boundary description changes.
Confirm that the region stays on one side of the axis for a direct shell integral, then check the integrand against the graph and the evaluated integral against its antiderivative.
A graph with labeled coordinates can determine the integral’s structure even when the boundary is not written as an explicit formula. However, exact evaluation requires enough information to express the radius and height as functions or otherwise evaluate their accumulated contribution. A sketch without a scale or boundary data may support a qualitative setup but not a precise numerical volume.
Why the Graph Reading Matters
The shell method converts a two-dimensional region into a family of circular layers. The integral is only as reliable as the geometric information put into it. A wrong slice direction changes the coordinate of integration; a wrong boundary changes the height; and a mistaken distance changes the radius. Drawing one typical slice and labeling all three quantities before integrating is an efficient way to catch these errors.
Pay particular attention to shifted axes: distance is measured from the specified line, not automatically from the origin. Also inspect the graph for boundary changes and for portions on both sides of the axis. Those checks determine whether one integral is enough and whether the direct shell formula applies.
Check Your Understanding
For each question, identify the slice orientation and describe the shell dimensions before doing any integration.
- A region lies between \(y=0\) and \(y=9-x^2\) for \(0\le x\le3\), and is rotated about the \(y\)-axis. What are the shell radius, height, and limits?
- A region has horizontal extent from \(x=1\) to \(x=5\), with top boundary \(y=4\) and bottom boundary \(y=1\), and is rotated about \(y=0\). What shell coordinate, radius, and height should be used?
- A graph’s lower boundary changes formula at \(x=2\). What should you check before deciding whether to use one integral or two?
- Why does using the absolute value of \(x\) as a radius not by itself justify integrating across both sides of the \(y\)-axis?
- A region lies entirely to the left of \(x=3\) and is rotated about \(x=3\). How is the shell radius read from a vertical slice at \(x\)?
