A Workshop on Horizontal-Slice Area Problems

Why Slice Horizontally?

A bounded region can be described using thin vertical slices or thin horizontal slices. The useful choice depends on the shape of the region and how simply its boundaries can be written. When a horizontal slice runs from one boundary to another, its length is a difference of $x$-coordinates, and its thickness is a change in $y$.

This is especially convenient when a boundary is a sideways curve, such as $x=y^2$, or when the region's left edge changes from one curve to another. The essential task is to describe each horizontal slice correctly before integrating.

Key formula: If, for every $y$ in $c\le y\le d$, a region runs horizontally from $x=L(y)$ to $x=R(y)$, where $L(y)\le R(y)$, its area is $$ A=\int_c^d \bigl(R(y)-L(y)\bigr)\,dy. $$ Here $L(y)$ is the left boundary and $R(y)$ is the right boundary.

How to Set Up a Horizontal-Slice Integral

The bounds $c$ and $d$ are the lowest and highest $y$-values in the region. At a fixed height $y$, the slice's width is the right $x$-coordinate minus the left $x$-coordinate. If either boundary changes over the region, split the $y$-interval at each change and integrate the appropriate width on each part.

Find the vertical extent.
Determine the lowest and highest $y$-values in the bounded region, usually by finding relevant intersections.

Write the boundaries as $x$-values.
For each height $y$, identify which curves give the region's left and right edges.

Check the order and look for changes.
Confirm that the proposed right boundary lies to the right of the left boundary. Find any heights where the identity of a boundary changes.

Integrate the width.
Use right minus left, with respect to $y$, over each interval where the boundaries are fixed.

Check the result.
Differentiate each antiderivative and verify the endpoint substitutions and the total area.

Why the Formula Works

Suppose the region is bounded between $y=c$ and $y=d$, with continuous left and right boundaries $x=L(y)$ and $x=R(y)$, and $L(y)\le R(y)$ throughout. Divide the vertical interval into many thin subintervals. In a subinterval of height $\Delta y$, a horizontal strip has approximate width $R(y)-L(y)$, so its approximate area is $(R(y)-L(y))\Delta y$.

Adding these strip areas gives a Riemann sum. As the heights of the strips approach zero, the sums approach the exact area. By the definition of the definite integral, their limit is the integral of the width:

A=\int_c^d \bigl(R(y)-L(y)\bigr)\,dy.

If the boundaries change at a height $y=a$, apply the same argument separately below and above $a$. The resulting areas add, so the total is the sum of the corresponding integrals. The condition $L(y)\le R(y)$ matters: subtracting in the wrong order would give negative widths rather than areas.

Worked Example: A Sideways Parabola and a Vertical Line

Find the area enclosed by $x=y^2$ and $x=4$.

First find their intersections. Set the two $x$-coordinates equal:

y^2=4 \quad\Longrightarrow\quad y=-2\ \text{or}\ y=2.

Thus the region extends from $y=-2$ to $y=2$. At each height in this interval, $y^2\le4$, so $x=y^2$ is the left boundary and $x=4$ is the right boundary. The area is

\begin{aligned} A &=\int_{-2}^{2}(4-y^2)\,dy\\ &=\left[4y-\frac{y^3}{3}\right]_{-2}^{2}\\ &=\left(8-\frac{8}{3}\right) -\left(-8+\frac{8}{3}\right)\\ &=\frac{16}{3}-\left(-\frac{16}{3}\right)\\ &=\frac{32}{3}. \end{aligned}

Check: the derivative of $4y-y^3/3$ is $4-y^2$, the integrand. At $y=-2$ and $y=2$, the two boundaries meet, so the slice width is zero at both ends. The computed area is positive, as required.

Worked Example: A Radical Boundary

Find the area enclosed by $x=\sqrt{y+1}$ and $x=y+1$.

The radical requires $y+1\ge0$. To find intersections, solve $\sqrt{y+1}=y+1$. Let $t=y+1$, so $t\ge0$. Then $\sqrt{t}=t$, which gives $t=0$ or $t=1$. These correspond to $y=-1$ and $y=0$.

For $-1<y<0$, the value $t=y+1$ lies between $0$ and $1$, and $\sqrt{t}>t$. Therefore $x=y+1$ is the left boundary and $x=\sqrt{y+1}$ is the right boundary. The area is

\begin{aligned} A &=\int_{-1}^{0}\bigl(\sqrt{y+1}-(y+1)\bigr)\,dy\\ &=\left[\frac{2}{3}(y+1)^{3/2} -\frac{(y+1)^2}{2}\right]_{-1}^{0}\\ &=\left(\frac{2}{3}-\frac{1}{2}\right)-0\\ &=\frac{4}{6}-\frac{3}{6}\\ &=\frac{1}{6}. \end{aligned}

Check: differentiating the antiderivative gives $\sqrt{y+1}-(y+1)$. At both intersection heights, the boundaries coincide; the antiderivative values are $0$ at $y=-1$ and $1/6$ at $y=0$. Their difference is positive and equals the stated area.

When the Left Boundary Changes

A region may have a different curve on its left edge at different heights. The change occurs where the candidate boundary curves meet and exchange order. Find those heights, then use a separate width formula on each interval. Do not choose one curve as the left boundary for the whole region unless its position is actually consistent throughout.

Worked Example: A Left Boundary That Changes

Consider the region bounded by $x=4$, $x=y^2$, and $x=2-y^2$. Find its area.

The right-hand boundary is $x=4$. The two curves that may form the left boundary meet where

y^2=2-y^2 \quad\Longrightarrow\quad 2y^2=2 \quad\Longrightarrow\quad y=-1\ \text{or}\ y=1.

The parabola $x=y^2$ meets $x=4$ at $y=-2$ and $y=2$. Thus the region extends from $-2$ to $2$. For $|y|\ge1$, $y^2\ge2-y^2$, so $x=y^2$ is farther right and forms the left edge of the region. For $|y|\le1$, $2-y^2\ge y^2$, so $x=2-y^2$ forms that edge. The width changes at $y=-1$ and $y=1$, giving three integrals:

\begin{aligned} A &=\int_{-2}^{-1}(4-y^2)\,dy +\int_{-1}^{1}\bigl(4-(2-y^2)\bigr)\,dy +\int_{1}^{2}(4-y^2)\,dy\\ &=\int_{-2}^{-1}(4-y^2)\,dy +\int_{-1}^{1}(2+y^2)\,dy +\int_{1}^{2}(4-y^2)\,dy. \end{aligned}

Evaluate the first integral:

\begin{aligned} \int_{-2}^{-1}(4-y^2)\,dy &=\left[4y-\frac{y^3}{3}\right]_{-2}^{-1}\\ &=\left(-4+\frac{1}{3}\right) -\left(-8+\frac{8}{3}\right)\\ &=-\frac{11}{3}-\left(-\frac{16}{3}\right) =\frac{5}{3}. \end{aligned}

For the middle integral:

\begin{aligned} \int_{-1}^{1}(2+y^2)\,dy &=\left[2y+\frac{y^3}{3}\right]_{-1}^{1}\\ &=\left(2+\frac{1}{3}\right) -\left(-2-\frac{1}{3}\right) =\frac{14}{3}. \end{aligned}

For the last integral:

\begin{aligned} \int_{1}^{2}(4-y^2)\,dy &=\left[4y-\frac{y^3}{3}\right]_{1}^{2}\\ &=\left(8-\frac{8}{3}\right) -\left(4-\frac{1}{3}\right)\\ &=\frac{16}{3}-\frac{11}{3} =\frac{5}{3}. \end{aligned}

Adding the parts gives $A=5/3+14/3+5/3=24/3=8$. Check: the derivatives of the three antiderivatives are their respective widths. At $y=-2$ and $y=2$, the outer width is zero because $y^2=4$. At $y=-1$ and $y=1$, the two candidate left boundaries meet, so the width formulas agree across each split. The total area is positive.

Common Pitfall: Ignoring a Boundary Switch

In the last example, a tempting but incorrect setup is to use $y^2$ as the left boundary throughout:

\int_{-2}^{2}(4-y^2)\,dy=\frac{32}{3}.

This calculation is wrong for the stated region. When $-1<y<1$, the curve $x=2-y^2$ lies to the right of $x=y^2$, so it forms the region's left boundary. Using $y^2$ there counts extra area to the left of the actual boundary. The correction is to split at $y=-1$ and $y=1$, use $4-(2-y^2)$ in the middle, and obtain the area $8$.

Takeaway: For a horizontal slice, identify the leftmost and rightmost points that bound the region at that height. If either point is supplied by different curves at different heights, split the integral where the boundary changes.

Check Your Understanding

For each question, identify the horizontal bounds and explain how you know the left and right boundaries are in the correct order.

For the region bounded by $x=y^2$ and $x=9$, what are the $y$-bounds and the horizontal width?
For $x=\sqrt{y}$ and $x=y$, which curve is on the left for $0<y<1$?
What equation should be solved to find the heights where two candidate left boundaries exchange order?
Why must an area integral be split when its left boundary changes?
What does a negative value of $R(y)-L(y)$ indicate about a proposed boundary ordering?

A Workshop on Horizontal-Slice Area Problems

What You'll Learn

Why Slice Horizontally?

How to Set Up a Horizontal-Slice Integral

Why the Formula Works

Worked Example: A Sideways Parabola and a Vertical Line

Worked Example: A Sideways Parabola and a Vertical Line

Worked Example: A Radical Boundary

Worked Example: A Radical Boundary

When the Left Boundary Changes

Worked Example: A Left Boundary That Changes

Common Pitfall: Ignoring a Boundary Switch

Check Your Understanding

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A Workshop on Horizontal-Slice Area Problems

What You'll Learn

Why Slice Horizontally?

How to Set Up a Horizontal-Slice Integral

Why the Formula Works

Worked Example: A Sideways Parabola and a Vertical Line

Worked Example: A Sideways Parabola and a Vertical Line

Worked Example: A Radical Boundary

Worked Example: A Radical Boundary

When the Left Boundary Changes

Worked Example: A Left Boundary That Changes

Common Pitfall: Ignoring a Boundary Switch

Check Your Understanding

Test yourself: Calculus I final exam practice

Want all 1,114 practice problems?

Get the Calc Pass App

Vol. 1: 1,114 Questions

Vol. 2: Solutions Manual

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