Polynomial Boundaries and Shell Volumes
Polynomial curves often create regions whose boundaries can be identified by solving an equation such as \(x^2=2x\) or \(x^3=x\). Once the interval of the region is known, the shell method turns its polynomial boundary lengths into an integral for volume.
The essential work is to describe each shell accurately: determine where the shells begin and end, how far each shell is from the axis of rotation, and how tall it is. A polynomial formula does not remove the need to check which curve lies above or to the right on the interval being used.
Turning Polynomial Boundaries into Shell Dimensions
For rotation about a vertical line \(x=a\), use vertical strips. A strip at \(x\) produces a shell whose radius is the horizontal distance to the axis, \(r(x)=|x-a|\). Its height is the vertical distance across the region: typically the upper boundary minus the lower boundary. Thus, when those boundaries are \(y=f(x)\) and \(y=g(x)\), with \(f(x)\geq g(x)\), the height is \(f(x)-g(x)\).
For rotation about a horizontal line \(y=b\), use horizontal strips. A strip at \(y\) produces a shell of radius \(r(y)=|y-b|\). Its height is the horizontal distance across the region: typically the right boundary minus the left boundary. If the boundaries are \(x=R(y)\) and \(x=L(y)\), with \(R(y)\geq L(y)\), the height is \(R(y)-L(y)\).
These radius and height formulas are distances, so they must be nonnegative. If the axis crosses the region, the absolute value in the radius and any change in which boundary is nearer may require splitting the integral. In the examples below, the chosen axes do not cross the regions, so one radius formula applies throughout each interval.
Worked Example: Two Polynomial Curves Rotated About the \(y\)-Axis
Worked Example: Two Polynomial Curves Rotated About the \(y\)-Axis
Find the volume obtained by rotating the region between \(y=2x\) and \(y=x^2\) about the \(y\)-axis. Assume lengths are measured in units.
1. Find the bounds and identify the boundary order. The intersections satisfy \(2x=x^2\), so \(x(x-2)=0\) and \(x=0\) or \(x=2\). On \(0\leq x\leq2\), the difference is \(2x-x^2=x(2-x)\geq0\), so \(2x\) is above \(x^2\).
2. Determine the shell dimensions. Vertical strips make shells about the vertical \(y\)-axis. At \(x\), the radius is \(x\), the height is \(2x-x^2\), and the thickness is \(dx\).
3. Evaluate and check. An antiderivative of \(2x^2-x^3\) is \(\frac{2}{3}x^3-\frac14x^4\), because its derivative is \(2x^2-x^3\). Therefore,
The result is \(\frac{8\pi}{3}\) cubic units. The shell dimensions have the expected signs throughout \(0\leq x\leq2\), and differentiating the antiderivative confirms the integrand.
Why the Shell Integral Works
The shell formula follows by approximating a curved solid with many thin cylindrical shells. For a shell of radius \(r\), height \(h\), and small thickness \(\Delta x\), its volume is approximately its curved surface area, \(2\pi rh\), multiplied by its thickness. This gives \(2\pi rh\Delta x\). Adding those contributions over the interval produces a Riemann sum; in the limit, that sum is the definite integral.
Use strips parallel to the axis: vertical strips for a vertical axis and horizontal strips for a horizontal axis.
Measure the radius from the strip to the axis, and measure the height across the region in the direction perpendicular to the strip.
Integrate \(2\pi\) times radius times height over the interval covered by the shells.
This is the reason the formula is not simply an integral of the region’s area. Each shell contributes according to both its circumference and its height, and the integral sums these contributions through the thickness of the solid.
Worked Example: Horizontal Shells from a Polynomial Boundary
Worked Example: Horizontal Shells from a Polynomial Boundary
The region is enclosed by \(x=y^2\), \(x=4\), and \(y=0\). Find its volume when it is rotated about the \(x\)-axis.
1. Find the \(y\)-bounds. The curve \(x=y^2\) meets \(x=4\) when \(y^2=4\). The region above \(y=0\) therefore extends from \(y=0\) to \(y=2\).
2. Determine the shell dimensions. Horizontal strips are parallel to the \(x\)-axis. For \(0\leq y\leq2\), the right boundary is \(x=4\) and the left boundary is \(x=y^2\), since \(4-y^2\geq0\). The shell radius is \(y\), and its height is \(4-y^2\).
3. Evaluate and check. The antiderivative \(2y^2-\frac14y^4\) differentiates to \(4y-y^3\). Thus,
The volume is \(8\pi\) cubic units. The integrand uses the horizontal width of the region, not its vertical height.
Shifted Axes: Measure Radius from the Actual Axis
When the axis is shifted, the shell radius must include that shift. For a vertical axis \(x=a\), the radius is \(|x-a|\), not simply \(x\). For a horizontal axis \(y=b\), it is \(|y-b|\). The height is still measured across the region, so changing the axis can change the radius without changing the height or bounds.
Worked Example: Rotating About a Shifted Vertical Line
Rotate the region between \(y=2x\) and \(y=x^2\), for \(0\leq x\leq2\), about the vertical line \(x=-1\). Find the volume.
1. Reuse the polynomial region’s dimensions. The intersection equation \(2x=x^2\) gives bounds \(0\) and \(2\). As before, \(2x\geq x^2\) on this interval, so the shell height is \(2x-x^2\).
2. Measure the new radius. The vertical axis is one unit left of the \(y\)-axis. Thus the distance from \(x\) to \(x=-1\) is \(x+1\), which is positive on the interval.
3. Evaluate and check. An antiderivative is \(-\frac14x^4+\frac13x^3+x^2\), whose derivative is \(-x^3+x^2+2x\). Therefore,
The volume is \(\frac{16\pi}{3}\) cubic units. The shell height remains the same as in the first example, but the larger radii increase the volume.
Worked Example: A Cubic Curve and a Line
Worked Example: A Cubic Curve and a Line
In the first quadrant, consider the region between \(y=x\) and \(y=x^3\). Find the volume when it is rotated about the \(y\)-axis.
1. Find the relevant intersections. Solve \(x=x^3\): \(x(x-1)(x+1)=0\). The intersections in the first quadrant are \(x=0\) and \(x=1\). On \(0\leq x\leq1\), \(x-x^3=x(1-x^2)\geq0\), so \(y=x\) is above \(y=x^3\).
2. Set up the shells. Vertical strips give radius \(x\), height \(x-x^3\), and bounds \(0\leq x\leq1\).
3. Evaluate and check. The antiderivative \(\frac13x^3-\frac15x^5\) differentiates to \(x^2-x^4\). Hence,
The volume is \(\frac{4\pi}{15}\) cubic units. Both the shell height and the radius are positive over the interior of the interval.
A Common Radius Mistake
For the shifted-axis example, a tempting but incorrect setup is to use radius \(x\) because the shells are vertical. That would give
which is the volume for rotation about \(x=0\), not \(x=-1\). The axis is at \(x=-1\), so the correct distance is \(x-(-1)=x+1\). Using that radius gives \(\frac{16\pi}{3}\) cubic units. The correction is to locate the axis first and compute a distance from it; the variable \(x\) alone is not automatically the radius.
Interpreting and Checking the Result
Polynomial shell integrals are especially useful when the region’s vertical or horizontal widths are easy to express as differences of polynomials. Intersections provide natural bounds, while the ordering of the curves determines the sign of the height. Before integrating, check that radius and height are nonnegative on the chosen interval and that the strip orientation matches the axis.
Volume has units of length cubed. If the original coordinates are measured in centimeters, for example, the calculated volume is in cubic centimeters. Keep exact values such as \(\frac{16\pi}{3}\) when possible; if a decimal is requested, round only at the end and state the precision.
Check Your Understanding
For each question, identify the relevant intersections and shell dimensions before evaluating any integral.
- For the region between \(y=3x\) and \(y=x^2\) in the first quadrant, find the bounds and shell height for rotation about the \(y\)-axis.
- For the region between \(x=y^2\) and \(x=9\), with \(y\geq0\), state the bounds and shell height for rotation about the \(x\)-axis.
- When rotating a region with vertical strips about \(x=-2\), what distance gives the shell radius at a point \(x\geq0\)?
- Explain why the order of two polynomial curves must be checked on the full interval before using their difference as shell height.
- What units should a shell-volume result have if all boundary coordinates are measured in meters?
