Why Horizontal Slices Fit Sideways Parabolas
For a curve written as \(x\) in terms of \(y\), a horizontal line at a fixed height \(y\) meets the curve at a particular \(x\)-coordinate. That makes horizontal slices a natural way to measure a region between sideways parabolas: each slice runs from a left boundary to a right boundary.
A standard sideways parabola has the form \(x=a(y-k)^2+h\), where \(a\ne 0\). Its vertex is \((h,k)\). If \(a>0\), it opens to the right; if \(a<0\), it opens to the left. For area calculations, the essential information is the \(x\)-coordinate of each boundary at a given height.
The quantity \(x_R(y)-x_L(y)\) is the horizontal width of the region at height \(y\). The integration variable is \(y\), and the bounds are heights—not \(x\)-coordinates. If the curves exchange left-right order within the region, split the interval at the height where the order changes and calculate the areas separately.
Why the Horizontal-Slice Formula Works
Here is a justification for the area formula. Divide the height interval \([c,d]\) into small subintervals. Over a thin subinterval of height \(\Delta y\), the region is approximately a rectangle whose width is the horizontal distance between its boundaries. Its approximate area is therefore width times height.
As the subintervals become thinner, the sum of these approximate rectangle areas approaches the area of the region. This is the Riemann-sum definition of the definite integral. Since the boundary functions are continuous, their horizontal width is also continuous, so the sums converge to the integral of that width.
This also explains why the order of subtraction matters: a width is right coordinate minus left coordinate. Reversing the subtraction gives a negative signed integral, not a geometric area.
Worked Example: A Region Between Oppositely Opening Parabolas
Worked Example: A Region Between Oppositely Opening Parabolas
Find the area enclosed by \(x=y^2\) and \(x=2y-y^2\).
Step 1: Find the intersection heights. At an intersection, the \(x\)-coordinates agree: $$ y^2=2y-y^2. $$ Thus \(2y^2-2y=0\), so \(2y(y-1)=0\). The intersection heights are \(y=0\) and \(y=1\).
Step 2: Identify left and right. The difference between the second curve and the first is $$ (2y-y^2)-y^2=2y-2y^2=2y(1-y). $$ For \(0\le y\le 1\), this difference is nonnegative. Therefore \(x_L(y)=y^2\) and \(x_R(y)=2y-y^2\) on the enclosed interval.
Step 3: Integrate the horizontal width. $$ A=\int_0^1\bigl((2y-y^2)-y^2\bigr)\,dy =\int_0^1(2y-2y^2)\,dy. $$ An antiderivative is \(y^2-\frac{2}{3}y^3\), so $$ A=\left[y^2-\frac{2}{3}y^3\right]_0^1 =\left(1-\frac{2}{3}\right)-0 =\frac{1}{3}. $$
Check: Differentiating \(y^2-\frac{2}{3}y^3\) gives \(2y-2y^2\), the width being integrated. The width is zero at both intersection heights and positive between them, as expected. The enclosed area is \(\frac{1}{3}\) square unit.
Finding Intersection Heights and Boundary Order
For two sideways parabolas, find intersections by setting their \(x\)-expressions equal and solving for \(y\). These solutions are the heights at which the curves meet. The region’s area bounds come from those heights.
Do not assume which curve is on the right just from the way the parabolas open. Instead, subtract the two \(x\)-expressions. On an interval between consecutive intersection heights, the sign of the difference tells you their order. A positive value of \(x_2(y)-x_1(y)\) means curve 2 is to the right of curve 1 at that height.
Worked Example: Shifted Sideways Parabolas
Worked Example: Shifted Sideways Parabolas
Find the area enclosed by \(x=(y-1)^2-1\) and \(x=3-(y+1)^2\).
Step 1: Rewrite and equate the expressions. $$ (y-1)^2-1=y^2-2y, \qquad 3-(y+1)^2=2-y^2-2y. $$ At an intersection, $$ y^2-2y=2-y^2-2y. $$ Adding \(y^2+2y\) to both sides gives \(2y^2=2\), so \(y^2=1\). The intersection heights are \(y=-1\) and \(y=1\).
Step 2: Determine the left and right boundaries. Subtract the first \(x\)-expression from the second: $$ (2-y^2-2y)-(y^2-2y)=2-2y^2. $$ For \(-1\le y\le 1\), \(2-2y^2\ge 0\). Thus \(x_L(y)=y^2-2y\), and \(x_R(y)=2-y^2-2y\).
Step 3: Integrate. $$ A=\int_{-1}^{1}\bigl((2-y^2-2y)-(y^2-2y)\bigr)\,dy =\int_{-1}^{1}(2-2y^2)\,dy. $$ An antiderivative is \(2y-\frac{2}{3}y^3\). Therefore, $$ A=\left[2y-\frac{2}{3}y^3\right]_{-1}^{1} =\left(2-\frac{2}{3}\right)-\left(-2+\frac{2}{3}\right) =\frac{4}{3}-\left(-\frac{4}{3}\right) =\frac{8}{3}. $$
Check: The derivative of \(2y-\frac{2}{3}y^3\) is \(2-2y^2\), the horizontal width. That width is zero at \(y=-1\) and \(y=1\), and positive between them. The area is \(\frac{8}{3}\) square units.
A Common Mistake: Reversing the Boundaries
Suppose the curves are \(x=y^2\) and \(x=2-y^2\). Their intersections satisfy \(y^2=2-y^2\), so \(2y^2=2\) and \(y=-1,1\). On this interval, the difference between the second curve and the first is \(2-2y^2\ge 0\). Hence the right boundary is \(2-y^2\), and the left boundary is \(y^2\).
A common error is to subtract right from left: $$ \int_{-1}^{1}\bigl(y^2-(2-y^2)\bigr)\,dy =\int_{-1}^{1}(2y^2-2)\,dy =-\frac{8}{3}. $$ The negative result is a warning: the subtraction gives a signed quantity, not geometric area. Correct the order to right minus left: $$ A=\int_{-1}^{1}\bigl((2-y^2)-y^2\bigr)\,dy =\int_{-1}^{1}(2-2y^2)\,dy =\left[2y-\frac{2}{3}y^3\right]_{-1}^{1} =\frac{8}{3}. $$ The width is nonnegative throughout the interval, and the area is positive.
Worked Example: Parabolas with Different Widths
Worked Example: Parabolas with Different Widths
Find the area enclosed by \(x=2y^2-2\) and \(x=4-y^2\).
Step 1: Find the intersection heights. $$ 2y^2-2=4-y^2 \quad\Longrightarrow\quad 3y^2=6 \quad\Longrightarrow\quad y^2=2. $$ Thus \(y=-\sqrt{2}\) and \(y=\sqrt{2}\).
Step 2: Identify the boundaries. Subtract the first curve from the second: $$ (4-y^2)-(2y^2-2)=6-3y^2. $$ For \(-\sqrt{2}\le y\le\sqrt{2}\), \(y^2\le 2\), so \(6-3y^2\ge 0\). Therefore \(x_L(y)=2y^2-2\) and \(x_R(y)=4-y^2\).
Step 3: Evaluate the area. $$ A=\int_{-\sqrt{2}}^{\sqrt{2}}(6-3y^2)\,dy =\left[6y-y^3\right]_{-\sqrt{2}}^{\sqrt{2}}. $$ At \(y=\sqrt{2}\), the antiderivative is \(6\sqrt{2}-(\sqrt{2})^3=6\sqrt{2}-2\sqrt{2}=4\sqrt{2}\). At \(y=-\sqrt{2}\), it is \(-6\sqrt{2}-(-\sqrt{2})^3=-6\sqrt{2}+2\sqrt{2}=-4\sqrt{2}\). Hence $$ A=4\sqrt{2}-(-4\sqrt{2})=8\sqrt{2}. $$
Check: Differentiating \(6y-y^3\) gives \(6-3y^2\), the horizontal width. That width is zero at both intersection heights and positive between them. The enclosed area is \(8\sqrt{2}\) square units.
A Reliable Setup
When both boundaries are sideways parabolas, use this sequence to keep the setup clear:
For example, a sideways parabola may be written as \(x=a(y-k)^2+h\).
Solve for \(y\); the solutions give the intersection heights.
Subtract one \(x\)-expression from the other on the interval between intersections.
Use the intersection heights as the bounds and \(dy\) as the integration variable.
Confirm the width is nonnegative, verify the antiderivative by differentiation, and check that the final area is positive.
These checks serve different purposes. The sign of the width confirms the boundary order; differentiating checks the integration; and a positive final result is necessary for geometric area. Together they help catch errors before treating the answer as complete.
Check Your Understanding
For each question, identify the horizontal width before evaluating or describing the area.
- Find the intersection heights of \(x=y^2\) and \(x=3-y^2\).
- For \(-1\le y\le 1\), which curve is farther right: \(x=y^2-1\) or \(x=2-y^2\)? Show the difference that establishes your answer.
- Set up, but do not evaluate, the area integral between \(x=(y-1)^2\) and \(x=4-(y-1)^2\), using their intersection heights.
- Why does integrating left boundary minus right boundary give the wrong sign for area?
- What two checks can confirm that a proposed area integral has been set up and evaluated consistently?
