From a Region to an Integral
A free-response question about area or volume is not just an invitation to write down a familiar formula. The reasoning must connect the region in the prompt to the integral: identify its boundaries, establish the limits, decide what a representative slice looks like, and translate that slice into an expression to integrate.
A sketch helps make those choices visible. It can reveal which curve is above another, whether the boundary changes within the interval, and how distances should be measured from an axis of rotation. The sketch need not be to scale, but its labels and relative positions must be correct.
A Reliable Free-Response Workflow
Use this sequence before evaluating. It separates the geometric decisions from the algebra and makes it easier to catch an incorrect setup.
List the curves and any lines that bound the requested region. Note whether the coordinates represent lengths, such as meters or centimeters.
Solve equations for intersections that could bound the region. Include stated vertical or horizontal boundaries when they determine an endpoint.
Mark the intersections, the interval, and which boundary is above or below, or to the left or right. For volume, also draw and label the axis of rotation.
Vertical slices usually give differences of \(y\)-values and integrate with respect to \(x\). Horizontal slices give differences of \(x\)-values and integrate with respect to \(y\). For volume, use slices perpendicular to the axis of rotation for disks or washers.
Use a positive height for area and the correct cross-sectional area for volume. Split the integral if the boundary order or cross-section formula changes.
Show the antiderivative and endpoint substitution. Check the sign and size of the result, then state the answer with the appropriate squared or cubed units.
Area: Match the Integrand to the Slices
For vertical slices, the height of a rectangle is the upper \(y\)-value minus the lower \(y\)-value. For horizontal slices, its width is the right \(x\)-value minus the left \(x\)-value. If one curve is not consistently above or to the right of the other, the region must be split into intervals where the ordering is consistent.
Worked Example: Area Between Intersecting Curves
Find the area bounded by \(y=4-x^2\) and \(y=x+2\). Assume both axes are measured in meters.
Find the intersections. Set the equations equal:
Rearrange and factor:
Thus \(x=-2\) and \(x=1\). At \(x=-2\), both curves have \(y=0\); at \(x=1\), both have \(y=3\). These give the endpoints of the bounded region.
Determine which curve is on top. Subtract the line from the parabola:
For \(-2<x<1\), \(x+2>0\) and \(x-1<0\), so \(-(x+2)(x-1)>0\). Therefore the parabola is above the line throughout the interior of the interval. The area is
An antiderivative is \(2x-\frac{x^2}{2}-\frac{x^3}{3}\). Substituting the upper endpoint gives
At the lower endpoint, the value is
Therefore
Check. The difference \(2-x-x^2\) is zero at both intersection values, positive between them, and has antiderivative \(2x-\frac{x^2}{2}-\frac{x^3}{3}\), whose derivative is \(2-x-x^2\). The positive result and square units are consistent with an area.
When Horizontal Slices Simplify the Region
The slice direction is a choice, not a rule that every area problem must use \(dx\). If the boundaries are easier to express as \(x\) in terms of \(y\), horizontal slices can make the setup simpler. The limits must then be \(y\)-coordinates, and the integrand is a horizontal length.
Worked Example: Area Using Horizontal Slices
Find the area enclosed by \(x=y^2\) and \(x=2y\), with coordinates measured in meters.
Find the intersections. Equate the \(x\)-values:
The curves meet at \(y=0\) and \(y=2\). Their corresponding \(x\)-coordinates are \(0\) and \(4\). On \(0<y<2\), \(2y-y^2=y(2-y)>0\), so \(x=2y\) is the right boundary and \(x=y^2\) is the left boundary.
Set up and evaluate the area.
The upper-end value is \(2^2-\frac{2^3}{3}=4-\frac83=\frac43\). The lower-end value is \(0^2-\frac{0^3}{3}=0\). Hence
Check. The width \(2y-y^2\) is positive between the intersections and zero at both endpoints. Differentiating \(y^2-\frac{y^3}{3}\) gives \(2y-y^2\), as required.
Why the Slice Integrals Give Area
The area formula follows from adding thin rectangular strips. For vertical slices across \(a\leq x\leq b\), a strip near \(x\) has approximate width \(\Delta x\) and height \(u(x)-\ell(x)\), where \(u\) is the upper boundary and \(\ell\) is the lower boundary. Its approximate area is \([u(x)-\ell(x)]\Delta x\).
Partition the interval into subintervals and add the strip areas. As the widths of the subintervals approach zero, this sum approaches the definite integral of the height:
This reasoning explains both the integrand and the limits: each contribution is a slice height times a small width, and the slices together cover the interval. For horizontal slices, the same argument uses width \(\text{right}-\text{left}\) times a small change in \(y\).
Volume: Measure Radii from the Axis
For a solid formed by rotating a region, a cross-section perpendicular to the axis of rotation may be a disk or a washer. A disk has no hole. A washer has an outer radius \(R\) and an inner radius \(r\), with \(R\geq r\geq 0\). Its area is \(\pi R^2-\pi r^2\). The radii are distances from the axis, not simply the function values unless the axis is an appropriate coordinate axis.
The volume formula follows by adding thin cylinders of washer-shaped cross-sections. A slice of thickness \(\Delta x\) has approximate volume \(\pi(R^2-r^2)\Delta x\). In the limit as the slice thickness approaches zero, these contributions give the integral of the cross-sectional area.
Worked Example: Washer Volume About a Shifted Axis
The triangular region bounded by \(x=0\), \(y=0\), and \(y=2-x\) is rotated about \(y=-1\). Find the volume. Coordinates are measured in meters.
Identify the limits and radii. The sloping line meets \(y=0\) when \(0=2-x\), so \(x=2\). Together with \(x=0\), the \(x\)-limits are \(0\) to \(2\). A vertical slice runs from \(y=0\) to \(y=2-x\), so it is perpendicular to the horizontal axis \(y=-1\). The outer radius is the distance from \(y=-1\) to the upper curve, \(R(x)=(2-x)-(-1)=3-x\). The inner radius is the distance from the axis to \(y=0\), \(r(x)=0-(-1)=1\).
Set up the washer integral.
Expand the integrand:
Thus
At \(x=2\), the bracket equals \(\frac83-3(4)+16=\frac83-12+16=\frac83+4=\frac{20}{3}\). At \(x=0\), it equals \(0\). Therefore
Check. Differentiating \(\frac{x^3}{3}-3x^2+8x\) gives \(x^2-6x+8\), the expanded cross-sectional area factor. At \(x=0\), the radii are \(R=3\) and \(r=1\); at \(x=2\), both radii are \(1\), so the washer area decreases to zero at the endpoint. The result has cubic units, as volume should.
Common Set-Up Errors
Many lost points come from a setup that does not describe the pictured region. A correct antiderivative cannot repair incorrect limits or an incorrectly ordered difference.
Other checks help prevent related errors:
- Confirm that the proposed limits are coordinates in the variable of integration. Horizontal slices require \(y\)-limits, not \(x\)-limits.
- Do not assume that two curves keep the same order throughout the region. Test their difference or inspect a labeled sketch, and split at a crossing if needed.
- For a shifted axis, calculate each radius as a distance from the axis. Subtracting the axis value matters; a radius cannot be negative.
- Keep outer and inner radii distinct. A washer has cross-sectional area \(\pi(R^2-r^2)\), not \(\pi(R-r)^2\).
- Check units at the end: area uses square units, while volume uses cubic units.
Presenting a Complete Response
A strong written response makes the geometry and calculus traceable. State how the intersections or boundary lines determine the limits. Identify the upper and lower curves, or the right and left boundaries. For volume, name the axis and show how the radii are measured from it. Then write the definite integral, show the antiderivative and endpoint evaluation, and state the result with units.
A sketch supports this explanation, but it does not replace it. Likewise, an unevaluated integral may show that the region was recognized, but if the prompt asks for the area or volume, evaluate it and report the result. If a calculator is allowed and used for a numerical result, retain enough precision during the calculation and round only at the final step as requested.
Check Your Understanding
For each question, identify the geometric information needed before evaluating any integral.
- For a region between two curves integrated with respect to \(x\), what do the limits and integrand represent?
- If two curves intersect inside the proposed interval, what should you check before writing one area integral?
- For horizontal slices, which boundaries determine the integrand, and which coordinate determines the limits?
- A region is rotated about \(y=3\). How should you find the radius of a boundary curve \(y=f(x)\) at a given \(x\)?
- What are the correct units for an area result and for a volume result?
