From a Region to a Volume Integral
A volume-of-revolution problem begins with a plane region, but the integral is determined by more than the equations of its boundary curves. You must identify the interval covered by the region, choose slices perpendicular to the axis of rotation, and measure each slice’s radii from that axis. These choices are especially important when the axis is shifted away from a coordinate axis.
This workshop combines those decisions in regions bounded by lines, parabolas, and circles. The central habit is to describe a typical slice in words before writing its integral: where does the slice begin and end, what happens when it rotates, and how far are its boundaries from the axis?
For rotation about a horizontal line, vertical slices are perpendicular to the axis and usually lead to an integral in \(x\). For rotation about a vertical line, horizontal slices are perpendicular to the axis and usually lead to an integral in \(y\). The radii are distances, so they are nonnegative even when a curve or axis has a negative coordinate.
A Reliable Setup Sequence
For each problem, first locate where the boundary curves meet. Their intersection coordinates determine the endpoints of the region; do not assume the integration interval from a sketch alone. Next, decide whether vertical or horizontal slices will give one clear inner boundary and one clear outer boundary throughout that interval. Then measure both radii as distances from the rotation axis.
Use the stated boundaries and solve their intersection equations. Record the interval in the variable used for the slices.
Use slices perpendicular to the axis of rotation. Horizontal axes call for vertical slices; vertical axes call for horizontal slices.
The farther boundary gives \(R\), and the nearer boundary gives \(r\). For a disk, the nearer boundary is the axis itself.
Use \(\pi(R^2-r^2)\) for each washer, evaluate over the full interval, and verify the antiderivative and the resulting volume.
Why the Washer Integral Gives Volume
At a fixed slice position, rotating the segment of the region produces a circular cross-section. If its outer radius is \(R\) and its inner radius is \(r\), the cross-section is a large disk with a smaller disk removed. Their areas are \(\pi R^2\) and \(\pi r^2\), so subtraction gives \(\pi(R^2-r^2)\). If there is no hole, \(r=0\), which gives the disk area.
As the slice position varies over an interval, these cross-sectional areas accumulate to form the solid’s volume. Thus, for a horizontal axis and vertical slices, the setup is
For a vertical axis and horizontal slices, the corresponding setup is
These formulas require the stated radii to describe the same washer at every slice position in the integration interval. If the nearer or farther boundary changes, the interval must be split where that change occurs. A good sketch or a short boundary comparison helps establish which curve gives each radius.
Worked Example: A Line and a Parabola About the \(x\)-Axis
Worked Example: A Line and a Parabola About the \(x\)-Axis
Find the volume formed by rotating the region between \(y=x\) and \(y=x^2\) about the \(x\)-axis.
Step 1: Find the endpoints. At an intersection, \(x=x^2\), so \(x(x-1)=0\). The curves meet at \(x=0\) and \(x=1\); the integration interval is \([0,1]\). On this interval, \(x^2\leq x\), so \(y=x\) is the upper curve and \(y=x^2\) is the lower curve.
Step 2: Identify the radii. The axis is \(y=0\), and the entire region is above it. A vertical slice rotates to a washer whose outer radius is the distance from \(y=0\) to \(y=x\), namely \(R(x)=x\). Its inner radius is the distance from \(y=0\) to \(y=x^2\), namely \(r(x)=x^2\).
Step 3: Set up and evaluate.
Check. The derivative of \(x^3/3-x^5/5\) is \(x^2-x^4\), the integrand after factoring out \(\pi\). The antiderivative is zero at \(x=0\) and equals \(1/3-1/5=2/15\) at \(x=1\), giving \(V=2\pi/15\). The result is positive, as a volume must be.
Shifted Horizontal Axes: Measure from the Axis, Not the Origin
When the axis is \(y=k\) rather than \(y=0\), a vertical distance to a boundary \(y=f(x)\) is \(\lvert f(x)-k\rvert\). If the region lies entirely above the axis, the distance is \(f(x)-k\); if it lies entirely below, the distance is \(k-f(x)\). For a region on one side of the axis, compare these distances to decide which boundary is outer and which is inner.
Worked Example: A Parabola and a Line About a Shifted Axis
Find the volume formed by rotating the region between \(y=x^2\) and \(y=2\) about the line \(y=-1\).
Step 1: Find the interval and order the curves. Intersections satisfy \(x^2=2\), so \(x=-\sqrt{2}\) and \(x=\sqrt{2}\). On this interval, \(x^2\leq 2\), so the line \(y=2\) is above the parabola \(y=x^2\).
Step 2: Measure radii from \(y=-1\). The upper boundary is 3 units above the axis, giving \(R(x)=2-(-1)=3\). The lower boundary is \(x^2+1\) units above the axis, giving \(r(x)=x^2-(-1)=x^2+1\). Both radii are measured from the shifted axis.
Step 3: Integrate.
An antiderivative of the integrand without \(\pi\) is \(F(x)=8x-\frac{2x^3}{3}-\frac{x^5}{5}\), since \(F'(x)=8-2x^2-x^4\). Set \(a=\sqrt{2}\). Because \(F\) is odd, \(F(a)-F(-a)=2F(a)\). Evaluating each term gives
Check. Differentiating \(F\) gives \(8-2x^2-x^4\), as required. Also, throughout the interval \(x^2\leq 2\), so \(r(x)=x^2+1\leq 3=R(x)\); the washer area is nonnegative. The evaluated volume is therefore consistent with the geometry.
Worked Example: A Circle and a Line About a Shifted Vertical Axis
Worked Example: A Circle and a Line About a Shifted Vertical Axis
Find the volume formed by rotating the region inside \(x^2+y^2=4\) and to the right of the \(y\)-axis, \(x=0\), about \(x=-1\).
Step 1: Choose horizontal slices and find the limits. The axis of rotation is vertical, so use horizontal slices. The right half of the circle is described by \(x=\sqrt{4-y^2}\), and it meets \(x=0\) at \(y=-2\) and \(y=2\). The \(y\)-limits are therefore \([-2,2]\).
Step 2: Measure the radii horizontally from \(x=-1\). The slice begins at the \(y\)-axis, \(x=0\), and ends at \(x=\sqrt{4-y^2}\). The inner radius is \(r(y)=0-(-1)=1\). The outer radius is \(R(y)=\sqrt{4-y^2}-(-1)=1+\sqrt{4-y^2}\).
Step 3: Set up and simplify.
An antiderivative of the integrand without \(\pi\) is \(G(y)=4y-\frac{y^3}{3}+y\sqrt{4-y^2}+4\arcsin(y/2)\). Indeed, differentiating \(y\sqrt{4-y^2}\) gives \(\sqrt{4-y^2}-y^2/\sqrt{4-y^2}\), while differentiating \(4\arcsin(y/2)\) gives \(4/\sqrt{4-y^2}\). Their sum is \(2\sqrt{4-y^2}\), so \(G'(y)=4-y^2+2\sqrt{4-y^2}\) for \(-2<y<2\). Evaluate the continuous endpoint values:
Check. The differentiated antiderivative matches the simplified integrand, and the endpoint subtraction gives \(32/3+4\pi\) before multiplication by the outer factor \(\pi\). The inner radius is 1 and the outer radius is at least 1 throughout, so no cross-section has negative area.
A Common Mistake: Using Coordinates Instead of Distances
In the shifted-axis example, a tempting but incorrect setup is to use the boundary heights \(2\) and \(x^2\) directly as radii, giving \(\pi\int(2^2-(x^2)^2)\,dx\). Those are distances from \(y=0\), not from the actual axis \(y=-1\). The resulting washers would be centered at the wrong line.
The correction is to subtract the axis coordinate from each boundary coordinate. The correct distances are \(2-(-1)=3\) and \(x^2-(-1)=x^2+1\), so the integrand is \(\pi\bigl(3^2-(x^2+1)^2\bigr)\). More generally, when an axis is shifted, write down the distance formula before simplifying it; this prevents a sign error from being hidden inside a radius.
Final Checks Before Accepting a Setup
Before evaluating a volume integral, check that the endpoints come from the region’s boundaries, the slice direction is perpendicular to the rotation axis, and each radius is measured from that axis. Then check that \(R\geq r\geq 0\) throughout the interval. A negative washer area signals that the radii have been reversed or the interval needs to be split.
Finally, differentiate the antiderivative and compare it with the integrand, then check that the evaluated volume is positive. If the problem gives physical dimensions, the result should have cubic units; without specified dimensions, it is understood in cubic units of the coordinate system.
Check Your Understanding
For each question, identify the slice direction and explain how the radii or limits are determined.
- The region between \(y=3x\) and \(y=x^2\) is rotated about the \(x\)-axis. What are the intersection points, and which curve gives the outer radius between them?
- A region lies between \(y=1\) and \(y=4\) and is rotated about \(y=0\). How do you express the radii as distances from the axis?
- For rotation about \(x=2\), why are horizontal slices usually used, and how would you find the distance from a boundary \(x=g(y)\) to the axis?
- In a washer integral, what geometric condition should hold between \(R\) and \(r\) throughout the interval?
- Why is \(\pi\int (2^2-x^4)\,dx\) not the correct setup for the region between \(y=x^2\) and \(y=2\) rotated about \(y=-1\)?
