Finding Shell Bounds from Intersections

Why Intersections Determine Shell Bounds

For a shell integral, the limits must describe the full interval over which each shell slice belongs to the intended region. When the region is enclosed by two curves, their intersections often supply the endpoints: at an intersection, the boundaries meet, and the height or length between them is zero.

Solving for intersections is only part of the task. You must also check which boundary is on top (for vertical shells) or on the right (for horizontal shells) between the endpoints. That check confirms that the proposed shell height has the correct sign and describes the region you mean to rotate.

Key idea: Solve for boundary intersections to find candidate endpoints, then test the boundaries between those endpoints to confirm the shell-height formula and its interval.

From Intersections to an Interval

For vertical shells, the integration variable is \(x\). If the boundaries are given by \(y=f(x)\) and \(y=g(x)\), their intersection \(x\)-coordinates satisfy \(f(x)=g(x)\). For horizontal shells, the variable is \(y\); when boundaries are given by \(x=p(y)\) and \(x=q(y)\), intersection \(y\)-coordinates satisfy \(p(y)=q(y)\).

The solutions are candidates, not automatically the complete limits. The region may also have a stated restriction, such as \(x\geq 0\), or may be bounded by a line rather than by a second curve. Once the candidate endpoints are identified, choose a test value between them. Comparing the boundary values there tells you which curve supplies the top, bottom, right, or left edge.

Definition: A shell-height formula is valid on an interval when the two boundary expressions used in its difference describe the intended region throughout that interval, in the stated order. For example, \(f(x)-g(x)\) is a nonnegative vertical height when \(f(x)\geq g(x)\) throughout the interval.

When the axis of rotation is vertical and lies to the left of the region, a vertical slice at \(x\) has radius equal to its horizontal distance from the axis. When the axis is horizontal and lies below the region, a horizontal slice has radius equal to its vertical distance from the axis. In either case, the shell method uses \(2\pi\) times radius times height, integrated over the interval of slice positions.

Why the Intersection Check Works

Suppose a bounded region lies between continuous curves \(y=f(x)\) and \(y=g(x)\) for \(a\leq x\leq b\), with \(f(a)=g(a)\) and \(f(b)=g(b)\). If \(f(x)\geq g(x)\) throughout this interval, a vertical slice has height \(f(x)-g(x)\). Rotating such slices about a vertical axis to the left of the region forms cylindrical shells. The shell at \(x\) has the radius given by its distance to the axis and the height \(f(x)-g(x)\). Adding the volumes of thin shells across the interval gives the integral of \(2\pi\) times radius times height.

The intersections identify where the two curve boundaries meet; the comparison between them confirms the height expression between those meeting points. If the boundary order reverses inside the proposed interval, the difference cannot describe a nonnegative height over that whole interval. The interval or the boundary description must then be reconsidered.

V=\int_a^b 2\pi(\text{radius})(\text{height})\,dx

This formula is applied only after identifying the relevant interval and confirming that the radius and height describe the shells of the intended region. The intersection equations do not replace those checks.

Worked Examples

Worked Example: A Region Between a Line and a Parabola

Find the shell-method volume when the region between \(y=x\) and \(y=x^2\) is rotated about the \(y\)-axis.

First solve the intersection equation:

x=x^2 \quad\Longrightarrow\quad x^2-x=0 \quad\Longrightarrow\quad x(x-1)=0.

The curves meet at \(x=0\) and \(x=1\), so the candidate interval is \([0,1]\). Test \(x=\frac12\): the line has value \(\frac12\), while the parabola has value \(\frac14\). Thus the line is above the parabola there. Since their difference is \(x-x^2=x(1-x)\), it is nonnegative throughout \([0,1]\). The shell height is \(x-x^2\).

The axis is \(x=0\), so a vertical slice at \(x\) has radius \(x\). Therefore:

\begin{aligned} V &=2\pi\int_0^1 x(x-x^2)\,dx\\ &=2\pi\int_0^1 (x^2-x^3)\,dx\\ &=2\pi\left[\frac{x^3}{3}-\frac{x^4}{4}\right]_0^1\\ &=2\pi\left(\frac13-\frac14\right)\\ &=2\pi\left(\frac{1}{12}\right) =\frac{\pi}{6}. \end{aligned}

Check the antiderivative: \(\frac{d}{dx}\left(\frac{x^3}{3}-\frac{x^4}{4}\right)=x^2-x^3\). The bounds also match the region’s endpoints, where the shell height is zero.

Worked Example: Intersections with a Shifted Axis

A region lies between \(y=2x\) and \(y=x^2\). Find its shell-method volume when rotated about \(x=-1\).

The intersection equation is:

2x=x^2 \quad\Longrightarrow\quad x^2-2x=0 \quad\Longrightarrow\quad x(x-2)=0.

Thus the intersections have \(x\)-coordinates \(0\) and \(2\). At \(x=1\), the line gives \(2\) and the parabola gives \(1\), so the line is above the parabola. Also, \(2x-x^2=x(2-x)\geq 0\) for \(0\leq x\leq 2\). Hence the interval is \([0,2]\), and the shell height is \(2x-x^2\).

The axis \(x=-1\) lies to the left of the region, so the radius is \(x-(-1)=x+1\). The volume is:

\begin{aligned} V &=2\pi\int_0^2 (x+1)(2x-x^2)\,dx\\ &=2\pi\int_0^2 (-x^3+x^2+2x)\,dx\\ &=2\pi\left[-\frac{x^4}{4}+\frac{x^3}{3}+x^2\right]_0^2\\ &=2\pi\left(-4+\frac83+4\right)\\ &=2\pi\left(\frac83\right) =\frac{16\pi}{3}. \end{aligned}

Check by differentiation: the derivative of \(-\frac{x^4}{4}+\frac{x^3}{3}+x^2\) is \(-x^3+x^2+2x\), the expanded integrand. The radius is positive across the interval: it ranges from \(1\) to \(3\).

Worked Example: Symmetric Intersections and a Shifted Axis

Consider the region between \(y=4-x^2\) and \(y=x^2\). Find the volume formed by rotating it about \(x=-2\).

Set the curve expressions equal:

4-x^2=x^2 \quad\Longrightarrow\quad 2x^2=4 \quad\Longrightarrow\quad x^2=2 \quad\Longrightarrow\quad x=-\sqrt{2},\ \sqrt{2}.

The candidate interval is \([-\sqrt{2},\sqrt{2}]\). At \(x=0\), the first curve has value \(4\) and the second has value \(0\). Their difference is \(4-2x^2\), which is nonnegative throughout the interval because \(x^2\leq 2\). The height is therefore \(4-2x^2\). Since \(-2<-\sqrt{2}\), the axis lies to the left of the whole region, and the radius is \(x+2\).

\begin{aligned} V &=2\pi\int_{-\sqrt{2}}^{\sqrt{2}}(x+2)(4-2x^2)\,dx\\ &=2\pi\int_{-\sqrt{2}}^{\sqrt{2}}(-2x^3-4x^2+4x+8)\,dx. \end{aligned}

The terms \(-2x^3\) and \(4x\) are odd, so their integrals cancel over the symmetric limits. The remaining terms give:

\begin{aligned} V &=2\pi\int_{-\sqrt{2}}^{\sqrt{2}}(-4x^2+8)\,dx\\ &=2\pi\left[-\frac{4x^3}{3}+8x\right]_{-\sqrt{2}}^{\sqrt{2}}\\ &=2\pi\left(\frac{16\sqrt{2}}{3}-\left(-\frac{16\sqrt{2}}{3}\right)\right)\\ &=2\pi\cdot\frac{32\sqrt{2}}{3} =\frac{64\pi\sqrt{2}}{3}. \end{aligned}

For the check, an antiderivative of the expanded integrand is \(-\frac{x^4}{2}-\frac{4x^3}{3}+2x^2+8x\); differentiating it gives \(-2x^3-4x^2+4x+8\). The computed integral is positive, as expected from positive radii and heights throughout the interval.

A Common Mistake: Losing an Intersection

A frequent error is to divide an intersection equation by a variable without checking whether that variable could be zero. For the first example, dividing \(x=x^2\) by \(x\) gives \(1=x\), and can make it appear that \(x=1\) is the only intersection. But division by \(x\) is not valid at \(x=0\), which is also a solution.

x=x^2 \quad\Longrightarrow\quad x(x-1)=0 \quad\Longrightarrow\quad x=0\ \text{or}\ x=1.

The correction is to factor the equation and keep every solution. Both endpoints are needed to describe the bounded region. Then test between them to verify that the chosen height, \(x-x^2\), is nonnegative.

A Reliable Bounds Check

Write the boundary equations.
Use the same variable for both expressions: \(y=f(x)\) and \(y=g(x)\) for vertical shells, or \(x=p(y)\) and \(x=q(y)\) for horizontal shells.

Solve for intersections.
Find all relevant solutions of \(f(x)=g(x)\) or \(p(y)=q(y)\), without discarding solutions through division.

Confirm the intended interval.
Check that the endpoints correspond to the region being rotated and respect any stated domain restrictions.

Compare the boundaries inside the interval.
Use a test value and, where helpful, the sign of their difference to establish which boundary is above or to the right.

Write the shell dimensions.
Use the confirmed boundary order for height and the distance from the slice to the axis for radius.

Why the Bounds Matter

The limits determine which shells are included in the volume. Incorrect bounds can omit part of the region or include shells that do not belong to it. Even correct intersection coordinates are not enough if they are paired with a height formula that reverses sign or describes the wrong side of the region.

Before integrating, make three checks: the endpoints come from the relevant boundaries, the chosen height is nonnegative between them, and the radius is the distance to the stated axis. These checks connect the algebraic intersection equations to the geometry of the solid.

Check Your Understanding

For each question, identify the relevant intersections and explain how you would verify the shell height on the interval.

For the curves \(y=3x\) and \(y=x^2\), what equation gives the candidate vertical-shell bounds?
Why should you test a value between two intersection coordinates before choosing a shell-height difference?
What solution can be lost by dividing the equation \(x=x^2\) by \(x\)?
If horizontal shells are used for boundaries \(x=p(y)\) and \(x=q(y)\), what equation gives their intersection \(y\)-coordinates?
For a region to the right of the axis \(x=-3\), what is the radius of a vertical shell at \(x\)?