Why Intersections Give Candidate Bounds
When a region is enclosed by two curves, the curves meet at the ends of the boundary arcs. Those shared points are natural places to look for the limits of an area integral. The algebraic task is to find the coordinates of the shared points; the coordinate used as a limit depends on whether the slices are vertical or horizontal.
For vertical slices, the integration variable is \(x\), so candidate limits are the \(x\)-coordinates of the relevant endpoints. For horizontal slices, the integration variable is \(y\), so candidate limits are their \(y\)-coordinates. The equations for the curves tell you which variable to solve for.
The word candidate matters. Solving the equality finds common points, but a particular common point is a limit for a chosen region only when it is an endpoint of that region's boundary. The algebra supplies the coordinates needed to identify possible limits.
Setting Two Equations Equal
Suppose two curves are graphs \(y=f(x)\) and \(y=g(x)\). At an intersection, the point has one \(x\)-coordinate and one \(y\)-coordinate. Since both equations give the \(y\)-coordinate of that same point, their outputs must agree:
Solving this equation gives the \(x\)-coordinates of intersections. Substitute each real solution into either original curve equation to find its \(y\)-coordinate. For curves expressed as \(x=p(y)\) and \(x=q(y)\), use the corresponding equation \(p(y)=q(y)\); its solutions give \(y\)-coordinates first.
Algebraically, it is often useful to move all terms to one side, factor if possible, and solve the resulting equation. Keep track of every real solution. A value that satisfies the equation after simplification must still be checked in the original curve equations, especially if the algebra involved division or squaring.
Why the Equality Method Works
The method follows directly from what it means for two graphs to share a point. If a point \((r,s)\) lies on \(y=f(x)\), then \(s=f(r)\). If it also lies on \(y=g(x)\), then \(s=g(r)\). Therefore \(f(r)=g(r)\). Conversely, if \(f(r)=g(r)\), the point \((r,f(r))\) lies on both graphs, because its \(y\)-coordinate equals both function values.
This proves that solving \(f(x)=g(x)\) finds exactly the common \(x\)-coordinates for the two graphs, provided both curves are defined at those values. The same reasoning applies to curves written as \(x=p(y)\) and \(x=q(y)\).
Worked Example: A Parabola and a Line
Worked Example: A Parabola and a Line
Find the intersection coordinates of \(y=x^2\) and \(y=2x\). State the candidate \(x\)-limits and the corresponding \(y\)-coordinates.
At an intersection, both expressions equal \(y\), so set them equal:
Move all terms to one side and factor:
The zero-product property gives \(x=0\) or \(x-2=0\), so \(x=0\) or \(x=2\). Find the matching \(y\)-coordinates using \(y=2x\): when \(x=0\), \(y=0\); when \(x=2\), \(y=4\). The candidate intersection points are \((0,0)\) and \((2,4)\). If the intended region is the one bounded by these curve segments, its candidate vertical limits are \(x=0\) and \(x=2\); the endpoint \(y\)-coordinates are \(0\) and \(4\).
Check both points in both equations. At \((0,0)\), \(x^2=0^2=0\) and \(2x=2(0)=0\). At \((2,4)\), \(x^2=2^2=4\) and \(2x=2(2)=4\). Each point satisfies both curve equations.
Using the Quadratic Formula
Not every intersection equation factors into integer factors. When it is quadratic, the quadratic formula gives exact solutions, including irrational ones. Keeping exact forms is useful: rounded approximations can obscure the precise location of an endpoint.
Worked Example: Exact Irrational Intersection Bounds
Find the intersection points of \(y=x^2\) and \(y=x+1\), and give the candidate \(x\)-limits for the region bounded by the two arcs.
Set the expressions for \(y\) equal and collect terms:
Here \(a=1\), \(b=-1\), and \(c=-1\). Applying the quadratic formula gives
For either solution, use \(y=x+1\). Thus the corresponding \(y\)-coordinates are \(\frac{3-\sqrt{5}}{2}\) and \(\frac{3+\sqrt{5}}{2}\). The intersection points are
Therefore the candidate vertical limits for the region bounded by these arcs are \(x=\frac{1-\sqrt{5}}{2}\) and \(x=\frac{1+\sqrt{5}}{2}\). To check the points, let \(x=\frac{1\pm\sqrt{5}}{2}\). Squaring gives \(x^2=\frac{(1\pm\sqrt{5})^2}{4}=\frac{6\pm2\sqrt{5}}{4}=\frac{3\pm\sqrt{5}}{2}\). Also \(x+1=\frac{1\pm\sqrt{5}}{2}+1=\frac{3\pm\sqrt{5}}{2}\). Thus \(x^2=x+1\) for both solutions, as required.
When the Bounds Are \(y\)-Coordinates
If the curves are naturally written as \(x\) in terms of \(y\), solve their \(x\)-expressions equal to each other. The roots are then \(y\)-coordinates, which are the candidate limits for horizontal slices. This is the same intersection principle with the roles of the coordinates exchanged.
Worked Example: Bounds for Horizontal Slices
Find the intersections of \(x=y^2\) and \(x=2-y\), and identify the candidate \(y\)-limits for the region enclosed by the two curve segments.
At a common point, the \(x\)-coordinates agree:
Factor the quadratic:
Calculate \(x\) from \(x=y^2\). If \(y=-2\), then \(x=(-2)^2=4\). If \(y=1\), then \(x=1^2=1\). The intersection points are \((4,-2)\) and \((1,1)\), and the candidate horizontal limits are \(y=-2\) and \(y=1\).
Check both equations at each point. For \((4,-2)\), \(y^2=(-2)^2=4\) and \(2-y=2-(-2)=4\). For \((1,1)\), \(y^2=1^2=1\) and \(2-y=2-1=1\). Both points satisfy both equations.
Common Mistake: Dividing Away an Intersection
A tempting shortcut is to divide an equation by a variable before solving it. If that variable could equal zero, division can remove a valid solution and therefore lose an intersection.
Worked Example: Do Not Lose the Zero Root
For \(y=x^2\) and \(y=2x\), the intersection equation is \(x^2=2x\). A mistaken solution divides both sides by \(x\) and obtains \(x=2\), then reports only the point \((2,4)\). This division assumes \(x\ne0\), so it has discarded the possibility \(x=0\).
The correction is to avoid dividing by a variable that might be zero. Instead, factor:
Substitution gives \(y=0\) when \(x=0\), and \(y=4\) when \(x=2\). Checking in the original equations confirms \(0^2=2(0)=0\) and \(2^2=2(2)=4\). Both intersections must be retained.
From Intersection Coordinates to Candidate Limits
Once you have the intersection points, select the coordinate that matches the integration variable. For vertical slices, the limits are \(x\)-values; for horizontal slices, they are \(y\)-values. The vertical-slice area formula uses \(x\)-limits, while the horizontal-slice area formula uses \(y\)-limits.
| Curve forms | Equation to solve | Candidate limits |
|---|---|---|
| \(y=f(x)\) and \(y=g(x)\) | \(f(x)=g(x)\) | The resulting \(x\)-coordinates for vertical slices |
| \(x=p(y)\) and \(x=q(y)\) | \(p(y)=q(y)\) | The resulting \(y\)-coordinates for horizontal slices |
Keep the coordinate pairs together as you work. A root of the equality is one coordinate, not an entire intersection point; the other coordinate must be calculated from the curve equations. Substitution into both equations provides a direct check and catches algebra errors before the coordinates are used as bounds.
Check Your Understanding
For each question, show the equation you solve and check any intersection coordinates in both original curve equations.
- Find the intersection points of \(y=x^2\) and \(y=3x-2\). What are the candidate \(x\)-limits for the region bounded by the corresponding arcs?
- Find the intersection points of \(y=2x+1\) and \(y=x^2+1\). State the \(x\)-coordinates first and then calculate the corresponding \(y\)-coordinates.
- The curves are \(x=y^2\) and \(x=y+2\). Find their intersection points and the candidate limits for horizontal slices.
- Explain why dividing the equation \(x^2=5x\) by \(x\) before checking whether \(x=0\) is valid can lose an intersection.
- For curves written as \(y=f(x)\) and \(y=g(x)\), which coordinate is found by solving \(f(x)=g(x)\), and which coordinate must then be calculated?
