Why the Complete Boundary Matters
A region described using graphs and coordinate axes is enclosed only when its boundary forms a complete loop. The axes can supply some of that boundary, but they do not automatically supply all of it. To set up an area integral correctly, first identify every boundary segment and how the segments meet.
For a curve \(y=f(x)\), intersections with the axes can help locate the boundary: points on the \(x\)-axis satisfy \(f(x)=0\), while points on the \(y\)-axis have \(x=0\). These intersections are useful landmarks, but they do not by themselves tell us which region is intended. We must also determine which component of the graph and axes forms a finite enclosed region.
Once the boundary is clear, choose slices that run across the region. With vertical slices, the area integrand is the upper boundary minus the lower boundary, and the bounds are the leftmost and rightmost \(x\)-coordinates. With horizontal slices, the integrand is the right boundary minus the left boundary, and the bounds are the lowest and highest \(y\)-coordinates. The earlier material on choosing between vertical and horizontal slices develops this comparison; here the key is to identify the actual boundaries first.
A Boundary-to-Integral Method
Work through the following checks before integrating. They help prevent a common error: using an axis as a boundary simply because it appears in the description, even though another graph forms that edge instead.
Determine where the graphs meet each other and where they meet the axes. These points provide possible corners or endpoints of the enclosed boundary.
Follow the relevant graph and axis segments in order. Check that they close around the intended region, and identify which curve or axis supplies each edge.
For vertical slices, identify top and bottom edges; for horizontal slices, identify left and right edges. Prefer the direction that describes the region with the simplest bounds and integrand.
Use the outermost \(x\)-coordinates for vertical slices, or the lowest and highest \(y\)-coordinates for horizontal slices. Do not assume a bound is zero unless the region reaches the corresponding axis.
Use upper minus lower for vertical slices or right minus left for horizontal slices, and check that the width is nonnegative throughout the interval.
Why Integrating Slice Width Gives Area
The integral setup follows from approximating the region by thin strips. For a vertical slice at \(x\), suppose the upper boundary is \(u(x)\) and the lower boundary is \(l(x)\), with \(u(x)\ge l(x)\) throughout \(a\le x\le b\). A thin strip has height approximately \(u(x)-l(x)\) and width approximately \(\Delta x\), so its area is approximately \([u(x)-l(x)]\Delta x\). Adding the strips and taking the limit gives the area.
For horizontal slices, if the right boundary is \(r(y)\) and the left boundary is \(l(y)\), the same strip argument gives horizontal width \(r(y)-l(y)\). Thus, for \(c\le y\le d\), the area is \(\int_c^d [r(y)-l(y)]\,dy\). These formulas require the stated boundaries to describe the region across the entire interval; if the boundary changes, the interval must be split. That is why tracing the complete boundary comes before writing the integral.
Worked Examples
Worked Example: A Curve and Both Coordinate Axes
Find the area enclosed by \(y=4-x^2\), the \(x\)-axis, and the \(y\)-axis in the first quadrant.
First locate the relevant endpoints. The curve meets the \(y\)-axis at \(x=0\), so \(y=4\) and the point is \((0,4)\). It meets the \(x\)-axis when \(4-x^2=0\), giving \(x^2=4\), so \(x=2\) in the first quadrant and the point is \((2,0)\). Together with the origin, these points identify a closed boundary: the curve, the \(x\)-axis segment from \((0,0)\) to \((2,0)\), and the \(y\)-axis segment from \((0,0)\) to \((0,4)\).
Vertical slices run from the \(x\)-axis up to the curve. Their bounds are \(x=0\) to \(x=2\), and their height is \((4-x^2)-0\). Therefore,
Check the antiderivative: \(\frac{d}{dx}(4x-x^3/3)=4-x^2\), the integrand. The endpoint calculation gives \(16/3\), a positive area, consistent with the curve lying above the \(x\)-axis between \(0\) and \(2\).
Worked Example: The \(y\)-Axis and Two Curves
Find the area enclosed by \(y=x^2\), \(y=2-x\), and the \(y\)-axis in the first quadrant.
The curves meet where \(x^2=2-x\), or \(x^2+x-2=0\). Factoring gives \((x+2)(x-1)=0\), so their intersection coordinates are \(x=-2\) and \(x=1\). The relevant intersection in the first quadrant is \(x=1\), where \(y=1\). At the \(y\)-axis, \(x=0\), the curves have values \(0\) and \(2\). The boundary is the \(y\)-axis segment between \((0,0)\) and \((0,2)\), then the line segment to \((1,1)\), then the parabola back to \((0,0)\). The \(x\)-axis is not an edge of this region.
On \(0\le x\le1\), the line is above the parabola: their difference is \(2-x-x^2=(1-x)(x+2)\), which is nonnegative on this interval. Vertical slices therefore have height \((2-x)-x^2\), giving
Check: the derivative of \(2x-x^2/2-x^3/3\) is \(2-x-x^2\), and the endpoint difference is \(7/6>0\). The positive result agrees with the positive slice widths.
Worked Example: A Sideways Curve and Both Axes
Find the area enclosed in the first quadrant by \(x=6-2y^2\), the \(x\)-axis, and the \(y\)-axis.
The curve meets the \(x\)-axis at \(y=0\), where \(x=6\). It meets the \(y\)-axis when \(6-2y^2=0\), so \(y^2=3\) and \(y=\sqrt{3}\) in the first quadrant. The boundary consists of the curve between \((6,0)\) and \((0,\sqrt{3})\), the \(x\)-axis from \((0,0)\) to \((6,0)\), and the \(y\)-axis from \((0,\sqrt{3})\) to \((0,0)\).
Horizontal slices describe this boundary directly. For \(0\le y\le\sqrt{3}\), a slice begins at the \(y\)-axis, \(x=0\), and ends at \(x=6-2y^2\). Its width is \(6-2y^2\), so
Check: \(\frac{d}{dy}(6y-2y^3/3)=6-2y^2\), and the endpoint difference is \(4\sqrt{3}>0\). Horizontal slices fit the given sideways curve without first solving for \(y\) as a function of \(x\).
A Common Boundary Mistake
In the second example, it would be incorrect to use the \(x\)-axis as the lower boundary just because the region lies above it. The actual lower edge is \(y=x^2\), and the \(x\)-axis is not part of the closed boundary. The mistaken setup would be
This includes extra area between the parabola and the \(x\)-axis. That extra amount is \(\int_0^1x^2\,dx=1/3\), so subtracting it gives \(3/2-1/3=7/6\), the correct area. The correction is to follow the closed boundary: at each \(x\) from \(0\) to \(1\), the slice begins at \(y=x^2\), not at \(y=0\).
Check Your Understanding
For each question, identify the boundary before choosing an integral.
- For \(y=9-x^2\) in the first quadrant, what are the intersection points with the coordinate axes, and what are the vertical-slice bounds?
- For the region bounded by \(y=x\), \(y=3-x\), and the \(y\)-axis in the first quadrant, which curve is the upper boundary between the \(y\)-axis and the curves’ intersection?
- For \(x=9-y^2\) in the first quadrant together with both coordinate axes, which slice direction gives the direct width without rewriting the curve?
- In a vertical-slice integral, what does the lower function represent geometrically?
- Why is it not valid to use the \(x\)-axis as a lower boundary in the region bounded by \(y=x^2\), \(y=2-x\), and the \(y\)-axis?
