Single-Variable-Calculus · Unit 17 · Video 4 · Interactive Practice

Vertical vs Horizontal Slices: One Region, Two Integrals

IKey Formulas

Slicing	Area	Hypotheses
Vertical strips, width $dx$	$\displaystyle A = \int_a^b \big(y_{\text{top}}(x) - y_{\text{bottom}}(x)\big)\,dx$	$y_{\text{top}} \ge y_{\text{bottom}}$ on $[a, b]$ , each given by one formula there; $a$ , $b$ are the leftmost and rightmost $x$ of the region
Horizontal strips, thickness $dy$	$\displaystyle A = \int_c^d \big(x_{\text{right}}(y) - x_{\text{left}}(y)\big)\,dy$	$x_{\text{right}} \ge x_{\text{left}}$ on $[c, d]$ , each given by one formula there; $c$ , $d$ are the lowest and highest $y$ of the region
Splitting	$\displaystyle \int_a^b = \int_a^s + \int_s^b$	cut at every $s$ where a boundary changes formula; each cut adds one integral
Rewriting a boundary	$y = x - 2 \iff x = y + 2$ , $\quad x = y^2 \iff y = \pm\sqrt{x}$	solve each curve for the variable that is not the variable of integration

Key Insight: Both directions give the same area, $\frac{9}{2}$ for $x = y^2$ and $y = x - 2$ . The direction with one formula on each side of the strip needs one integral; pick it from the picture before integrating.

IIVisualization 1 — One Region Swept Two Ways

Vertical strips change bottom formula at $x = 1$ ; horizontal strips always run from $x = y^2$ to $x = y + 2$ .

Slices Strip at x 0.5

IIIVisualization 2 — Which Direction Needs Fewer Integrals?

Every switch of a boundary formula along the sweep costs one more integral, and the region decides which direction switches less.

Region Slices

IVQuiz Questions

Problem 1 · Set Up Horizontal Slices

Given: the region bounded by $x = 3 - y^2$ and $x = 2y$ — find the integral in $y$ that gives its area.

✅ Correct! The strip runs from the line

x = 2y

on the left to the parabola

x = 3 - y^2

on the right, for

y

between the crossings

-3

and

1

; the area is

\frac{32}{3}

❌ Left minus right. At

y = 0

the line is at

x = 0

and the parabola at

x = 3

, so the parabola is on the right. This order gives

-\frac{32}{3}

, which cannot be an area.

❌ Those are $x$ -values. The crossings are

(-6, -3)

and

(2, 1)

; an integral in

dy

needs their

y

-coordinates,

-3

and

1

❌ Check the crossing equation.

3 - y^2 = 2y

gives

y^2 + 2y - 3 = (y + 3)(y - 1) = 0

, so

y = -3

and

y = 1

, not

-1

and

3

❌ Not quite. Find where the curves cross, then subtract left from right.

Show solution

Crossings: $3 - y^2 = 2y \iff y^2 + 2y - 3 = 0 \iff (y + 3)(y - 1) = 0$ , so $y = -3$ (point $(-6, -3)$ ) and $y = 1$ (point $(2, 1)$ ).

Right and left: at $y = 0$ , $x = 3 - 0 = 3$ on the parabola and $x = 2 \cdot 0 = 0$ on the line, so $x_{\text{right}} = 3 - y^2$ and $x_{\text{left}} = 2y$ for every $y$ in $[-3, 1]$ .

A = \int_{-3}^{1} \big[(3 - y^2) - 2y\big]\,dy = \int_{-3}^{1} (3 - 2y - y^2)\,dy = \Big[3y - y^2 - \tfrac{y^3}{3}\Big]_{-3}^{1}

= \Big(3 - 1 - \tfrac{1}{3}\Big) - \big(-9 - 9 + 9\big) = \tfrac{5}{3} + 9 = \tfrac{32}{3}

Problem 2 · Right Minus Left Against the $y$ -Axis

Given: the region between the curve $x = y^2 - 2y$ and the $y$ -axis — find its area.

✅ Correct! For

0 \le y \le 2

the curve lies left of the

y

-axis, so the length is

0 - (y^2 - 2y) = 2y - y^2

and the area is

\frac{4}{3}

❌ That is left minus right.

y^2 - 2y = y(y - 2) \le 0

[0, 2]

, so the curve is the LEFT boundary and the

y

-axis

x = 0

the right one. Subtracting in the other order makes the answer negative.

❌ Only half the region.

y = 1

is where the curve is furthest left, not where the region ends. The curve meets the

y

-axis at

y = 0

and

y = 2

, so integrate over

[0, 2]

❌ The parentheses were dropped.

0 - (y^2 - 2y) = -y^2 + 2y

; writing

0 - y^2 - 2y

flips the sign of the

2y

term.

❌ Not quite. Find where the curve meets

x = 0

, then integrate right minus left.

Show solution

Limits: $y^2 - 2y = 0 \iff y(y - 2) = 0$ , so the curve meets the $y$ -axis at $y = 0$ and $y = 2$ .

Right and left: at $y = 1$ the curve is at $x = 1 - 2 = -1$ , left of $x = 0$ . So $x_{\text{right}} = 0$ and $x_{\text{left}} = y^2 - 2y$ on all of $[0, 2]$ — one formula on each side, one integral.

A = \int_0^2 \big[0 - (y^2 - 2y)\big]\,dy = \int_0^2 (2y - y^2)\,dy = \Big[y^2 - \tfrac{y^3}{3}\Big]_0^2 = 4 - \tfrac{8}{3} = \tfrac{4}{3}

Vertical strips would need $y = 1 \pm \sqrt{1 + x}$ for $-1 \le x \le 0$ ; the horizontal setup avoids solving the quadratic.

Problem 3 · When Right and Left Trade Places

Given: the region enclosed by $x = y^3$ and $x = 4y$ , sliced horizontally — find how many integrals the setup needs and the total area.

How many integrals in $dy$ ?

What is the total area?

✅ Correct! The region has two lobes of area

4

each; right and left trade places at

y = 0

, so the setup needs two integrals.

❌ The sides swap. At

y = 1

4y = 4

is right of

y^3 = 1

; at

y = -1

y^3 = -1

is right of

4y = -4

. One formula for right minus left cannot hold on all of

[-2, 2]

❌ Cut only where a side changes formula. The curves cross at

y = -2, 0, 2

; between consecutive crossings the right and left curves stay the same, so

y = \pm 1

are not cut points.

❌ Signed areas cancelled.

\int_{-2}^{2} (4y - y^3)\,dy = 0

because on

[-2, 0]

that integrand is right-minus-left reversed and negative. Use

y^3 - 4y

there.

❌ That is one lobe.

\int_0^2 (4y - y^3)\,dy = 4

covers only

0 \le y \le 2

; the lobe on

[-2, 0]

adds another

4

❌ The left curve is missing.

2\int_0^2 4y\,dy = 16

measures from

x = 0

, not from

x = y^3

; subtract the left boundary.

❌ Not quite. Compare the two curves at a

y

above

0

and at a

y

below

0

❌ Not quite. Add the areas of the two lobes, each computed as right minus left.

Show solution

Crossings: $y^3 = 4y \iff y(y - 2)(y + 2) = 0$ , so $y = -2, 0, 2$ , at the points $(-8, -2)$ , $(0, 0)$ , $(8, 2)$ .

Right and left: on $[0, 2]$ (test $y = 1$ : $4 \gt 1$ ) the right side is $x = 4y$ ; on $[-2, 0]$ (test $y = -1$ : $-1 \gt -4$ ) the right side is $x = y^3$ . The sides trade places at $y = 0$ , so there are two integrals.

A = \int_{-2}^{0} (y^3 - 4y)\,dy + \int_{0}^{2} (4y - y^3)\,dy = \Big[\tfrac{y^4}{4} - 2y^2\Big]_{-2}^{0} + \Big[2y^2 - \tfrac{y^4}{4}\Big]_{0}^{2}

= \big(0 - (4 - 8)\big) + (8 - 4) = 4 + 4 = 8

Vertical strips need two integrals as well (top and bottom swap at $x = 0$ ), so here the choice of direction does not save an integral.

Problem 4 · The Same Region, Sliced Vertically

Given: the region of Problem 1, bounded by $x = 3 - y^2$ and $x = 2y$ with corners $(-6, -3)$ and $(2, 1)$ — find the vertical-slice setup that gives its area.

✅ Correct! The top changes formula at

x = 2

, so vertical slices need two integrals; they still total

\frac{32}{3}

, the value Problem 1 reached with one.

❌ The line is not the bottom. The region lies to the RIGHT of

x = 2y

, i.e. below

y = \frac{x}{2}

; so

\sqrt{3 - x} - \frac{x}{2}

is not the height of a strip of this region; the integral gives

\frac{99}{4}

❌ The first piece is outside the region. For

-6 \le x \le 2

the points of the region satisfy

y \le \frac{x}{2}

, so the line is the TOP there and the lower branch

-\sqrt{3 - x}

the bottom. This sum gives

\frac{80}{3}

❌ Those are $y$ -limits. An integral in

dx

runs from the leftmost

x = -6

to the rightmost

x = 3

(the vertex of the parabola), and the top changes at

x = 2

❌ Not quite. Solve each curve for

y

and find where the top or bottom changes formula.

Show solution

Solve for $y$ : $x = 3 - y^2 \iff y = \pm\sqrt{3 - x}$ , and $x = 2y \iff y = \frac{x}{2}$ . The region is $2y \le x \le 3 - y^2$ , i.e. $y \le \frac{x}{2}$ and $-\sqrt{3 - x} \le y \le \sqrt{3 - x}$ .

Top and bottom: the bottom is always the lower branch $y = -\sqrt{3 - x}$ . The top is the line $y = \frac{x}{2}$ while it is below the upper branch, i.e. for $-6 \le x \le 2$ , and the upper branch $y = \sqrt{3 - x}$ for $2 \le x \le 3$ (the leftmost point is $(-6, -3)$ , the rightmost the vertex $(3, 0)$ ).

A = \int_{-6}^{2} \big[\tfrac{x}{2} + \sqrt{3 - x}\big]\,dx + \int_{2}^{3} 2\sqrt{3 - x}\,dx

= \Big[\tfrac{x^2}{4} - \tfrac{2}{3}(3 - x)^{3/2}\Big]_{-6}^{2} + \Big[-\tfrac{4}{3}(3 - x)^{3/2}\Big]_{2}^{3} = \Big(\big(1 - \tfrac{2}{3}\big) - (9 - 18)\Big) + \tfrac{4}{3} = \tfrac{28}{3} + \tfrac{4}{3} = \tfrac{32}{3}

Two integrals and a square root, against one polynomial integral in Problem 1: for this region, horizontal slices are the better choice.

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