Why Slice Widths Matter
A slice-based area or volume calculation depends on the size of each slice. When the boundaries are given by curves, that size is often not a fixed number: it changes with position. The first task is to express the distance between the boundaries as a function of the slice location.
For curves written as \(y=f(x)\) and \(y=g(x)\), a vertical slice at a fixed \(x\) runs from one curve to the other. Its width is the vertical distance between their \(y\)-coordinates. That distance is nonnegative, so the order of subtraction matters: subtract the lower curve from the upper curve.
The same width can serve two related purposes. It can be the height of a thin vertical strip used to find planar area, or it can be a side length or diameter used in the area formula for a solid's cross section. In either case, getting the boundary order right is essential.
Finding Which Curve Is Above
Suppose a region is bounded by two graphs on an interval \([a,b]\). The curves can exchange order only at points where they meet, so first solve \(f(x)=g(x)\) to find their intersection \(x\)-coordinates. Those points divide \([a,b]\) into subintervals. On each open subinterval, compare the function values at a test point. The curve with the greater \(y\)-value is above the other one there.
A test point identifies the ordering on its subinterval, not necessarily on the entire original interval. If there are several intersections, check each interval between consecutive intersections. At an intersection, the width is zero because the two boundary heights are equal.
Find the endpoints of the region and solve the equality of the boundary functions for any interior crossing points.
Use the intersection coordinates to form intervals on which the curve order can be checked consistently.
At a test value in each interval, evaluate both functions. The larger output is the upper boundary.
Subtract the lower function from the upper function on each interval. Use each resulting width in the relevant area formula.
Area Between Two Curves
When \(f\) and \(g\) are continuous on \([a,b]\), and \(f(x)\ge g(x)\) throughout that interval, the region between their graphs has area equal to the integral of its vertical width. If the curves cross, divide the interval at their intersections and add the areas of the resulting pieces.
The width is always upper minus lower; it is not always \(f-g\) for the same pair of functions. Equivalently, the width at each \(x\) is the absolute difference \(|f(x)-g(x)|\). When calculating the integral, splitting at crossings makes that absolute difference an ordinary, nonnegative function on each piece.
Why Integrating the Width Gives Area
Here is a justification of the area formula. Consider a partition of \([a,b]\) into short subintervals. Over one subinterval of width \(\Delta x\), a narrow strip of the region is approximately a rectangle: its horizontal length is \(\Delta x\), and its vertical height is approximately \(f(x_i^*)-g(x_i^*)\) at a sample point \(x_i^*\) in the subinterval. Its approximate area is therefore \(\bigl(f(x_i^*)-g(x_i^*)\bigr)\Delta x\).
Adding the strip areas gives a Riemann sum. Because \(f\) and \(g\) are continuous, their difference is continuous, and these sums converge to the definite integral as the partition is refined. The limit is the region's area. If the curves cross, perform the same reasoning on each subinterval of fixed order and add the results.
Worked Examples
Worked Example: Area Between a Line and a Parabola
Find the area enclosed by \(y=4-x^2\) and \(y=x+2\).
First find where the curves meet: $$ 4-x^2=x+2 \quad\Longrightarrow\quad x^2+x-2=0 \quad\Longrightarrow\quad (x+2)(x-1)=0. $$ Thus the intersection coordinates are \(x=-2\) and \(x=1\). To establish which curve is above between them, test \(x=0\). The parabola has value \(4-0^2=4\), while the line has value \(0+2=2\). The parabola is above the line on this interval, so the width is $$ w(x)=(4-x^2)-(x+2)=2-x-x^2. $$ The area is $$ A=\int_{-2}^{1}(2-x-x^2)\,dx =\left[2x-\frac{x^2}{2}-\frac{x^3}{3}\right]_{-2}^{1}. $$ At the upper endpoint, the antiderivative is \(2-\frac12-\frac13=\frac76\). At the lower endpoint, it is \(-4-2+\frac83=-\frac{10}{3}\). Therefore $$ A=\frac76-\left(-\frac{10}{3}\right) =\frac76+\frac{20}{6} =\frac{27}{6} =\frac92. $$ For a check, differentiating the antiderivative gives \(2-x-x^2\), exactly the width being integrated. Also, the width is zero at both endpoints: \(w(-2)=2+2-4=0\) and \(w(1)=2-1-1=0\), as expected where the curves meet. The enclosed area is \(\frac92\) square units.
Worked Example: Curves That Switch Order
Find the total area between \(y=x\) and \(y=x^2\) on \([-1,2]\).
The intersections satisfy \(x=x^2\), so \(x(x-1)=0\) and the curves meet at \(x=0\) and \(x=1\). These points split the interval into \([-1,0]\), \([0,1]\), and \([1,2]\). Compare the functions at one test point in each interval:
- At \(x=-\frac12\), \(x^2=\frac14\) is greater than \(x=-\frac12\), so the width is \(x^2-x\) on \([-1,0]\).
- At \(x=\frac12\), \(x=\frac12\) is greater than \(x^2=\frac14\), so the width is \(x-x^2\) on \([0,1]\).
- At \(x=\frac32\), \(x^2=\frac94\) is greater than \(x=\frac32\), so the width is \(x^2-x\) on \([1,2]\).
Integrate each nonnegative width separately: $$ \begin{aligned} A &=\int_{-1}^{0}(x^2-x)\,dx +\int_{0}^{1}(x-x^2)\,dx +\int_{1}^{2}(x^2-x)\,dx\\ &=\left[\frac{x^3}{3}-\frac{x^2}{2}\right]_{-1}^{0} +\left[\frac{x^2}{2}-\frac{x^3}{3}\right]_{0}^{1} +\left[\frac{x^3}{3}-\frac{x^2}{2}\right]_{1}^{2}. \end{aligned} $$ The first integral is \(0-(-\frac13-\frac12)=\frac56\). The second is \(\frac12-\frac13=\frac16\). For the third, the antiderivative has value \(\frac83-2=\frac23\) at \(2\), and \(\frac13-\frac12=-\frac16\) at \(1\), giving \(\frac23-(-\frac16)=\frac56\). Hence $$ A=\frac56+\frac16+\frac56=\frac{12}{6}=\frac{11}{6}. $$ To check, the derivative of \(\frac{x^3}{3}-\frac{x^2}{2}\) is \(x^2-x\), and the derivative of \(\frac{x^2}{2}-\frac{x^3}{3}\) is \(x-x^2\). Each is the width used on its interval. The widths vanish at the crossing points \(0\) and \(1\); the total area is \(\frac{11}{6}\) square units.
Using Widths for Cross-Sectional Area
A slice width can also determine the area of a cross section in a solid. Suppose a solid has cross sections perpendicular to the \(x\)-axis, and each cross section uses the vertical segment between two boundary curves as a specified side or diameter. First compute that segment's length \(w(x)\). Then substitute it into the appropriate cross-sectional area formula to obtain \(A(x)\). The volume is the integral of the cross-sectional areas over the interval, as given by the slice principle.
The width itself is not automatically the cross-sectional area. For example, when the slices are squares with side length \(w(x)\), their area is \(A(x)=w(x)^2\). The volume integral is then \(\int_a^b w(x)^2\,dx\). When the cross section has a different shape, use that shape's area formula with the width in the role specified by the geometry.
Worked Example: Square Cross Sections Between a Parabola and a Line
A solid has square cross sections perpendicular to the \(x\)-axis. Its base region lies between \(y=4\) and \(y=x^2\), for \(-2\le x\le2\). Find its volume.
On this interval, \(4\ge x^2\), so the upper curve is \(y=4\) and the lower curve is \(y=x^2\). The square's side length is the vertical width $$ w(x)=4-x^2. $$ Because the area of a square with side length \(w\) is \(w^2\), the cross-sectional area is $$ A(x)=(4-x^2)^2=16-8x^2+x^4. $$ Integrating the area across the base interval gives $$ \begin{aligned} V &=\int_{-2}^{2}(16-8x^2+x^4)\,dx\\ &=\left[16x-\frac{8x^3}{3}+\frac{x^5}{5}\right]_{-2}^{2}. \end{aligned} $$ At \(x=2\), the antiderivative is \(32-\frac{64}{3}+\frac{32}{5}\). At \(x=-2\), it is \(-32+\frac{64}{3}-\frac{32}{5}\). Subtracting gives $$ V=64-\frac{128}{3}+\frac{64}{5} =\frac{960-640+192}{15} =\frac{512}{15}. $$ For a check, differentiating the antiderivative gives \(16-8x^2+x^4=(4-x^2)^2\), the cross-sectional area function. The side length is nonnegative on the whole interval and is zero at \(x=-2\) and \(x=2\), where the curves meet. The volume is \(\frac{512}{15}\) cubic units.
A Common Mistake: Subtracting in the Wrong Order
A frequent error is to subtract the functions in the order they are named rather than in the order determined by their heights. In the first worked example, using line minus parabola would give $$ (x+2)-(4-x^2)=x^2+x-2. $$ At \(x=0\), this expression is \(-2\), which is a signed difference, not a slice width. Integrating it over \([-2,1]\) gives \(-\frac92\), a negative value that cannot represent the enclosed area.
The correction is to identify the upper curve first. Here \(4-x^2\) is above \(x+2\), so the width is \((4-x^2)-(x+2)\), which is nonnegative between the intersections. If the curves exchange order, as in the second example, do not force one subtraction order across the whole domain. Split at the crossings and use upper minus lower on every piece.
Check Your Understanding
For each question, identify the boundary order before writing an area or cross-sectional formula.
- For \(y=3-x^2\) and \(y=1\) on \([-1,1]\), which curve is upper, and what is the vertical width?
- The curves \(y=x^2\) and \(y=2x\) meet at \(x=0\) and \(x=2\). Which is above on \([0,2]\), and what width should be integrated there?
- Why must an interval be divided at an interior intersection if the curves switch order there?
- A solid has square cross sections perpendicular to the \(x\)-axis, with vertical width \(w(x)\). What expression gives the area of a cross section?
- What does a negative value of \(f(x)-g(x)\) at a test point tell you about the order of the graphs \(y=f(x)\) and \(y=g(x)\) at that point?
