Why Trigonometric Boundaries Need Care
When a region is bounded by trigonometric curves, locating their intersections is part of setting up the volume integral. Sine and cosine can cross more than once, and their relative order can change across the interval. A single signed difference may therefore fail to represent the region's height everywhere.
For vertical shells about the y-axis, each shell's radius is determined by its horizontal distance from the axis, while its height is the vertical distance between the region's upper and lower boundaries. The intersection points determine where that height must be reconsidered.
The Shell Setup for Regions to the Right of the y-Axis
Suppose a region lies between \(x=a\) and \(x=b\), where \(0\leq a<b\), and is bounded vertically by \(y=f(x)\) and \(y=g(x)\). Rotating it around the y-axis produces cylindrical shells. At position \(x\), the shell radius is \(x\), and its height is the vertical distance between the two boundaries.
When the curves are \(y=f(x)\) and \(y=g(x)\), their height is \(|f(x)-g(x)|\). In practice, it is usually clearer to find where \(f(x)=g(x)\), test their order between intersections, and write the integral piece by piece.
Why the Shell Formula Works
Take a thin vertical strip at \(x\) with width \(\Delta x\) and height \(h(x)\). When revolved around the y-axis, it forms a thin cylindrical shell. Its radius is approximately \(x\), its circumference is approximately \(2\pi x\), and its height is approximately \(h(x)\). Its volume is therefore approximately circumference times height times thickness.
Adding these approximations over a partition of the interval gives a Riemann sum. As the largest strip width approaches zero, the sum approaches the integral of \(2\pi xh(x)\). If the height changes formula at an intersection, the integral must be divided at that point so that each subinterval uses the correct height.
A Reliable Intersection-and-Sign Process
Trigonometric identities can turn a curve-intersection equation into a simpler equation, but the stated domain still matters: retain only solutions that lie within it. After ordering the intersection points, test a point in each open interval. This identifies the upper curve there and makes the shell height explicit.
Set the boundary functions equal and solve on the stated interval.
Include the interval endpoints and any interior intersections.
In each subinterval, determine which curve is above the other.
Multiply the nonnegative height by \(2\pi x\), integrate on each subinterval, and add the results.
Worked Examples
Worked Example: Sine and Cosine Switch Order
The region between \(y=\sin x\) and \(y=\cos x\) for \(0\leq x\leq\frac{\pi}{2}\) is revolved around the y-axis. Find its volume.
First solve the intersection equation:
On \(0\leq x<\frac{\pi}{4}\), testing \(x=0\) gives \(\sin 0=0<1=\cos 0\), so the height is \(\cos x-\sin x\). On \(\frac{\pi}{4}<x\leq\frac{\pi}{2}\), testing \(x=\frac{\pi}{2}\) gives \(\sin x>\cos x\), so the height is \(\sin x-\cos x\). Therefore:
For the first integral, an antiderivative is \(F(x)=x(\sin x+\cos x)+\cos x-\sin x\), since
Thus its value is
For the second integral, an antiderivative is \(G(x)=-x(\sin x+\cos x)+\sin x-\cos x\), because
Its value is
Adding the two pieces gives
The two antiderivatives differentiate to their respective integrands, and the split point is the only interior intersection on the stated interval. Since \(\sqrt{2}>1\), the final volume is positive.
Worked Example: A Sine Curve and a Horizontal Boundary
The region between \(y=\sin x\) and \(y=\frac12\) on \(0\leq x\leq\pi\) is revolved around the y-axis. Find the volume of the region enclosed between the intersection points.
Solve for the intersections on the specified domain:
Between these intersections, \(\sin x>\frac12\); for example, \(\sin(\frac{\pi}{2})=1>\frac12\). Thus the shell height is \(\sin x-\frac12\), and the volume is
An antiderivative of the integrand inside the integral is
as differentiation verifies:
Evaluate both endpoints:
Subtracting gives
Therefore,
The derivative check confirms the antiderivative, and the height is positive throughout the interval between the intersections. The evaluated integral, and hence the volume, is positive.
Worked Example: Two Sine Curves with Multiple Intersections
The region between \(y=\sin x\) and \(y=\sin 2x\) for \(0\leq x\leq\pi\) is revolved around the y-axis. Find the volume.
Set the functions equal and use the double-angle identity:
On \([0,\pi]\), this gives \(x=0,\pi\) from \(\sin x=0\), and \(x=\frac{\pi}{3}\) from \(\cos x=\frac12\). For \(0<x<\frac{\pi}{3}\), testing \(x=\frac{\pi}{6}\) shows \(\sin 2x>\sin x\). For \(\frac{\pi}{3}<x<\pi\), testing \(x=\frac{\pi}{2}\) shows \(\sin x>\sin 2x\). The shell height therefore changes at \(\frac{\pi}{3}\):
An antiderivative for the first integrand is
Differentiation gives
Consequently, the first integral equals
An antiderivative for the second integrand is
Indeed,
Its definite integral is
Adding both parts and multiplying by \(2\pi\) gives
Both antiderivatives differentiate to the required integrands, and the three intersection points on the closed interval are \(0,\frac{\pi}{3},\pi\). The two positive-height integrals give a positive total volume.
Common Pitfall: Keeping a Signed Difference
A frequent error is to subtract the same curve from the other over the entire interval, even though they switch order. In the first example, using \(\sin x-\cos x\) from \(0\) to \(\frac{\pi}{2}\) gives
This is not the volume: on \(0\leq x<\frac{\pi}{4}\), \(\sin x-\cos x<0\), but a shell height cannot be negative. The signed integral lets that negative part cancel some of the positive part. The correction is to split at \(x=\frac{\pi}{4}\), use \(\cos x-\sin x\) on the first interval, and use \(\sin x-\cos x\) on the second.
Why the Intersection Work Matters
Trigonometric equations can have several solutions, so solving an equation without checking the stated domain can add irrelevant split points or miss a required one. The sign test is equally important: knowing where curves meet does not by itself tell you which curve is above between those points.
Before evaluating a shell integral, check that each interval has a nonnegative height, that the radius matches the axis of rotation, and that every interior boundary switch is included. These checks prevent a correct integration technique from being applied to the wrong integrand.
Check Your Understanding
For each question, identify intersections and boundary order before writing a volume integral.
- On \(0\leq x\leq\pi\), where do \(y=\sin x\) and \(y=\frac{\sqrt{2}}{2}\) intersect?
- For the region between \(y=\cos x\) and \(y=\sin x\) on \(0\leq x\leq\frac{\pi}{2}\), which curve is above on each side of their interior intersection?
- Write, but do not evaluate, the shell integral for the region between \(y=\sin x\) and \(y=\frac12\) from \(x=\frac{\pi}{6}\) to \(x=\frac{5\pi}{6}\), revolved around the y-axis.
- Why must the region between \(y=\sin x\) and \(y=\sin 2x\) on \([0,\pi]\) be split when setting up its shell integral?
- What error occurs if a signed difference of boundary functions is used as the shell height across an interval where the curves switch order?
