Application of Integrals · Lesson 1 of 3

Finding Area with Strips and Symmetry

“Turn curved regions into definite integrals by choosing useful strips, limits, and symmetries.”

Learning Objectives

• Explain how thin strips connect geometric area with a definite integral. • Choose vertical or horizontal strips and identify their dimensions and limits. • Derive the areas of a circle and an ellipse by integration in either direction. • Use quadrant symmetry without counting a region twice. • Set up and evaluate areas bounded by polynomial curves, axes, and straight lines.

From familiar shapes to curved boundaries

A rectangle has a constant height, so multiplying its height by its width gives its area. Under a curved graph, the height usually changes from one position to the next. We can still use rectangles: make each one narrow enough that the graph changes very little across it, then add their areas. Integration makes this limiting addition precise.

The definite integral was introduced as a limit of sums. This chapter uses that idea as a practical geometric method. We first identify the region, then translate its shape into an integral, and finally use an antiderivative to evaluate it. Knowing how to integrate an expression is only one part of the task; choosing the correct expression and limits comes first.

Begin with a continuous graph $y = f (x)$ lying on or above the horizontal axis between $x = a$ and $x = b$ , where $a < b$ . The region is bounded above by the graph, below by the axis, and at its ends by the two vertical lines. These four boundaries tell us both the height of a strip and where the strips begin and end.

Definition

Elementary strip

A very thin part of a region, used to build its total area. For a vertical strip, its height is measured vertically and its small width is measured along the horizontal axis. The notation $d A$ describes its elementary area in the integral model.

Building the area with vertical strips

At a chosen horizontal position, a strip stretches from the axis up to the curve. Its height is the function value at that position. Moving the strip from the left boundary to the right boundary sweeps through the entire region exactly once.

A finite rectangle of width $Δ x$ gives an approximation because the actual top boundary may curve. In the limit as the widths shrink and the rectangles fill the region, the sum becomes an integral. The notation $d x$ records this limiting width; it is not an extra constant that we substitute after integrating.

Area from vertical stripsLaTeX

d A = y d x = f (x) d x, A = \int_{a}^{b} f (x) d x

Here

A

is the total area,

x

is the strip's horizontal position,

f (x)

is its nonnegative height, and

a

and

b

are the left and right boundaries. This formula gives geometric area directly when

f (x) \geq 0

throughout the interval.

To evaluate the integral, find an antiderivative $F$ of $f$ . The Fundamental Theorem of Calculus then gives $F (b) - F (a)$ . The subtraction belongs to evaluating the accumulated area from the starting position to the finishing position. If the coordinates are measured in centimetres, strip height times strip width has units of square centimetres.

Evaluating the area integralLaTeX

F' (x) = f (x) ⟹ A = F (b) - F (a)

The prime indicates differentiation. The endpoints supplied to

F

must be the same endpoints used in the area integral.

Example — A straight boundary as a check

Problem
Find the area below $y = 3 x + 1$ , above the horizontal axis, and between $x = 0$ and $x = 2$ .

1.On $[0, 2]$ , the graph is positive. A vertical strip has height $3 x + 1$ and width $d x$ . Thus $A = \int_{0}^{2} (3 x + 1) d x$ .
2.An antiderivative is $3 x^{2} /2 + x$ . Therefore $A = [3 x^{2} /2 + x]_{0}^{2} = 6 + 2 = 8$ square units.
3.The endpoint heights are $1$ and $7$ , separated by width $2$ . The trapezium formula gives $(1 + 7) \times 2/2 = 8$ , confirming that the integral measures the same region.

Example — Area beneath a parabola

Problem
Find the area below $y = x^{2}$ , above the horizontal axis, and between $x = 1$ and $x = 2$ .

1.The function is nonnegative on this interval. The strip height is $x^{2}$ , so $A = \int_{1}^{2} x^{2} d x$ .
2.An antiderivative is $x^{3} /3$ . Subtract the value at the lower limit from the value at the upper limit: $A = [x^{3} /3]_{1}^{2} = 8/3 - 1/3 = 7/3$ square units.
3.The limits are positions along the horizontal axis, not the endpoint heights $1$ and $4$ . Choosing heights as limits would describe a different calculation.

Example — A higher power uses the same geometric idea

Problem
Find the area under $y = x^{4}$ from $x = 1$ to $x = 5$ , bounded below by the horizontal axis.

1.The height remains nonnegative, so the setup is $A = \int_{1}^{5} x^{4} d x$ . Only the height formula has changed from the previous example.
2.Since an antiderivative is $x^{5} /5$ , the area is $[x^{5} /5]_{1}^{5} = (3125 - 1) /5 = 3124/5$ square units.
3.A large answer is reasonable: near the right end the curve reaches height $625$ . The graph should guide our interpretation of the number, as well as the setup.

Turning the strips sideways

Sometimes the region is easier to describe by starting at the vertical axis and moving horizontally to a curve. We then use the vertical coordinate as the variable. Nothing about area has changed; we are simply adding the pieces in a different direction.

Suppose the boundary can be written as $x = g (y)$ and lies on or to the right of the vertical axis for $c \leq y \leq d$ . A horizontal strip has length $g (y)$ and small vertical thickness $d y$ . Its position moves upward from the lower line $y = c$ to the upper line $y = d$ .

Area from horizontal stripsLaTeX

d A = x d y = g (y) d y, A = \int_{c}^{d} g (y) d y

The bounds

c

and

d

are vertical coordinates. The quantity

g (y)

is a horizontal length, not a height. We assume

g (y) \geq 0

here.

Two ways to sweep through a region— Notice which length varies and which direction the strip moves.

Choose the direction that gives a simple strip length and simple limits. A vertical strip requires a height expressed in terms of $x$ ; a horizontal strip requires a horizontal length expressed in terms of $y$ . Either choice can work, but a useful choice can avoid a square root or an unnecessary split into regions.

Strip direction	Length of each strip	Variable and bounds
Vertical	$f (x)$ above the horizontal axis	$d x$ ; left to right, $a$ to $b$
Horizontal	$g (y)$ to the right of the vertical axis	$d y$ ; bottom to top, $c$ to $d$

Match the limits to the variable

An integral with $d y$ needs vertical-coordinate limits. An integral with $d x$ needs horizontal-coordinate limits. Changing the letter while keeping the old bounds does not preserve the region unless those bounds happen to agree for a geometric reason.

Recovering the area of a circle

A familiar area formula is a useful test of the strip method. A circle centred at the origin is symmetric about both axes, so its four quadrants have equal area. We can calculate one quadrant and multiply by four instead of trying to describe the upper and lower halves at the same time.

Let the radius be $a > 0$ , so the boundary is $x^{2} + y^{2} = a^{2}$ . Solving for the vertical coordinate gives $y = \pm a^{2} - x^{2}$ . In the first quadrant the coordinate is nonnegative, so the top boundary is the positive square root. A vertical strip runs from the horizontal axis up to that branch, with $0 \leq x \leq a$ .

Circle area using a quadrantLaTeX

A = 4 \int_{0}^{a} a^{2} - x^{2} d x

The factor four comes from four equal quadrant regions. The square root supplies the height of just the first-quadrant strip.

Use the standard square-root antiderivative from integration. It contains an algebraic term and an inverse-sine term. At the right boundary the algebraic term becomes zero because the square root becomes zero, leaving a particularly simple endpoint evaluation.

Square-root antiderivativeLaTeX

\int a^{2} - x^{2} d x = \frac{x}{2} a^{2} - x^{2} + \frac{a ^{2}}{2} sin^{- 1} (\frac{x}{a}) + C

Here

a > 0

, the square root is real for

- a \leq x \leq a

, and inverse sine takes its principal value in radians. We use the antiderivative at the definite-integral bounds.

Example — Deriving the circle formula

Problem
Evaluate the quadrant integral and hence find the full area inside $x^{2} + y^{2} = a^{2}$ .

1.At $x = a$ , the antiderivative has value $0 + (a^{2} /2) sin^{- 1} (1) = (a^{2} /2) (π /2) = π a^{2} /4$ .
2.At $x = 0$ , both terms are zero because $sin^{- 1} (0) = 0$ . The first-quadrant area is therefore $π a^{2} /4$ .
3.Multiply by four: $A = 4 (π a^{2} /4) = π a^{2}$ . The strip calculation recovers the familiar formula, rather than assuming it at the start.
4.With horizontal strips, solve instead for $x = a^{2} - y^{2}$ in the first quadrant. Then $A = 4 \int_{0}^{a} a^{2} - y^{2} d y$ . Its endpoint evaluation is the same, with the integration variable renamed, so it again gives $π a^{2}$ .

Circle area in either direction— Each picture shades one quadrant. Multiply its area by four only once.

Example — Recognising the requested fraction of a circle

Problem
Find the first-quadrant area bounded by $x^{2} + y^{2} = 4$ , the horizontal axis, and the lines $x = 0$ and $x = 2$ .

1.The radius is $2$ . Because only the first quadrant is requested, we use just the upper branch and the interval $[0, 2]$ .
2.The area is $\int_{0}^{2} 4 - x^{2} d x$ . This is one quarter of the full area $π (2)^{2} = 4 π$ .
3.The requested area is $π$ square units. Multiplying this integral by four would answer a different question: the area of the entire circle.

Extending the calculation to an ellipse

An ellipse has two axis scales instead of one radius. Its symmetry still gives four equal quadrants, and its equation supplies the strip height. The main new reasoning is to keep the horizontal and vertical semiaxes separate throughout the calculation.

For $a > 0$ and $b > 0$ , consider $x^{2} / a^{2} + y^{2} / b^{2} = 1$ . The boundary meets the horizontal axis at $x = \pm a$ and the vertical axis at $y = \pm b$ . Thus its full horizontal and vertical spans are $2 a$ and $2 b$ . Either semiaxis may be the longer one; the area argument does not depend on which is larger.

Definition

Semiaxis

Half of an ellipse's span along one of its symmetry axes. In the standard equation, $a$ is the horizontal semiaxis and $b$ is the vertical semiaxis.

In the first quadrant, rearranging the equation gives $y = (b / a) a^{2} - x^{2}$ . The positive sign is required because the strip lies above the horizontal axis. Compared with the circle of radius $a$ , the strip heights have all been multiplied by $b / a$ .

Ellipse area from vertical stripsLaTeX

A = 4 \int_{0}^{a} \frac{b}{a} a^{2} - x^{2} d x = \frac{4 b}{a} \int_{0}^{a} a^{2} - x^{2} d x

The constant

b / a

scales every height. The limits are still

0

and

a

because the strips move horizontally.

Example — Deriving the ellipse formula in two directions

Problem
Use integration to find the area inside $x^{2} / a^{2} + y^{2} / b^{2} = 1$ .

1.From the circle calculation, $\int_{0}^{a} a^{2} - x^{2} d x = π a^{2} /4$ . Multiplying by $4 b / a$ gives $A = (4 b / a) (π a^{2} /4) = π ab$ .
2.For a separate horizontal-strip setup, solve the first-quadrant boundary for $x = (a / b) b^{2} - y^{2}$ . The strip positions run from $y = 0$ to $y = b$ .
3.The whole area is $4 (a / b) \int_{0}^{b} b^{2} - y^{2} d y$ . Using the same antiderivative with parameter $b$ , this becomes $(4 a / b) (π b^{2} /4) = π ab$ .
4.If $a = b$ , the ellipse is a circle and the expression becomes $π a^{2}$ . This special case is a useful consistency check.

Ellipse area in either direction— Horizontal and vertical strips have different formulas and bounds but cover the same quadrant.

Area inside an ellipseLaTeX

A = π ab

Both

a

and

b

are semiaxes. Using full axis lengths here would make the result four times too large.

Example — Reading the semiaxes from the equation

Problem
Find the full areas inside $x^{2} /16 + y^{2} /9 = 1$ and $x^{2} /4 + y^{2} /9 = 1$ .

1.In the first equation, $a^{2} = 16$ and $b^{2} = 9$ , so $a = 4$ and $b = 3$ . Its area is $π \times 4 \times 3 = 12 π$ square units.
2.In the second equation, $a = 2$ and $b = 3$ , so its area is $6 π$ square units. Its longer axis is vertical, which does not alter the formula.
3.The denominators are squared semiaxes, not the semiaxes themselves. Reading $16$ and $9$ directly as lengths would give a much larger, incorrect region.

Choosing the easier strip direction

A sideways-opening curve is a good opportunity to choose horizontal strips. Read the boundaries before manipulating the equation: the axis supplies one end of each strip, while the curve supplies the other. Then the line that cuts off the region supplies a limit.

Example — A parabola bounded by a horizontal line

Problem
Find the area bounded by $y^{2} = 4 x$ , the vertical axis, and the line $y = 3$ .

1.The enclosed region lies between $y = 0$ and $y = 3$ , to the right of the vertical axis and to the left of the parabola. Write the right boundary as $x = y^{2} /4$ .
2.A horizontal strip has length $y^{2} /4$ and thickness $d y$ . Hence $A = \int_{0}^{3} (y^{2} /4) d y$ .
3.Integrating gives $A = [y^{3} /12]_{0}^{3} = 27/12 = 9/4$ square units.
4.A vertical-strip check is possible, but the height is $3 - 2 x$ , not $2 x$ . The interval is $0 \leq x \leq 9/4$ , so $\int_{0}^{9/4} (3 - 2 x) d x = 27/4 - 9/2 = 9/4$ . Both setups describe the same region.

A horizontal cutoff on a sideways parabola— The horizontal strip ends at the curve, while the top boundary fixes the upper limit.

Use the region, not an isolated equation

For the parabola problem, the region is above the upper branch and below the horizontal cutoff when viewed with vertical strips. Integrating the upper branch alone would measure the area beneath that branch, which is a different region. A sketch prevents this boundary error.

A sign condition to keep in view

The direct strip formulas in this lesson use nonnegative strip lengths. A signed function value below an axis cannot be used as a negative geometric length. The next lesson explains how to keep the geometric area positive when a graph lies below an axis or crosses it.

Check your understanding

For each question, identify the region and the direction of the strips before choosing a formula. Some wrong choices use the correct integration rule on the wrong geometric region.

Quick check

For a nonnegative continuous function $f$ on $[a, b]$ , what does a vertical strip contribute?

Quick check

For a first-quadrant region described by $x = g (y)$ , which bounds belong with horizontal strips?

Quick check

The ellipse $x^{2} /25 + y^{2} /4 = 1$ has area:

Quick check

What area is $\int_{0}^{3} 9 - x^{2} d x$ ?

Quick check

For $y^{2} = 4 x$ between the vertical axis and $y = 3$ , the horizontal-strip setup is:

Quick check

The area under $y = x^{2}$ from $x = 1$ to $x = 2$ is:

Quick check

A full ellipse has horizontal span $10$ cm and vertical span $6$ cm. Its area is:

Quick check

Which reason justifies multiplying a first-quadrant integral by four for a full origin-centred circle?

Practice Problems

Sketch the region under $y = 2 x + 2$ on $[0, 3]$ . Find its area by integration and check with a trapezium.
Find the area under $y = x^{3}$ on $[0, 2]$ . Explain why no sign correction is needed.
Write both vertical-strip and horizontal-strip integrals for the whole circle $x^{2} + y^{2} = 25$ . Evaluate one of them.
Find the first-quadrant area inside $x^{2} + y^{2} = 36$ . State why your answer is not the full circle area.
An ellipse is $x^{2} /36 + y^{2} /16 = 1$ . Identify both semiaxes, write a quadrant integral, and find the whole area.
For $x^{2} /9 + y^{2} /49 = 1$ , write a horizontal-strip integral for the whole ellipse. Explain the choice of upper limit.
Find the area bounded by $y^{2} = 8 x$ , the vertical axis, and $y = 4$ . Use horizontal strips.
A student obtains $144 π$ for the area of $x^{2} /16 + y^{2} /9 = 1$ . Identify the mistaken lengths and correct the answer.
Find the area under $y = x^{4}$ on $[0, 2]$ . Compare the size of the answer with the rectangle of height $16$ and width $2$ .
The equation $x^{2} / a^{2} + y^{2} / b^{2} = 1$ has positive semiaxes. Explain why doubling both semiaxes multiplies the area by four.

In order, the numerical results are $15$ , $4$ , $25 π$ , $9 π$ , $24 π$ , $21 π$ , $8/3$ , $12 π$ , and $32/5$ square units; the final scaling factor is four. For the horizontal ellipse setup use $4 \int_{0}^{7} (3/7) 49 - y^{2} d y$ . For the sideways parabola use $\int_{0}^{4} y^{2} /8 d y$ .

Key Takeaways

• Area comes from adding elementary strip areas; identify the region before choosing an integrand. • Use $d x$ with horizontal-coordinate bounds and $d y$ with vertical-coordinate bounds. • For nonnegative strip lengths, vertical strips give $\int f (x) d x$ and horizontal strips give $\int g (y) d y$ . • The full circle area is $π a^{2}$ ; the full ellipse area is $π ab$ for positive semiaxes. • Use symmetry only for equal regions, and multiply a quadrant area by four only when the whole region is requested.

Start of chapter

Next · Lesson 2

Areas When a Curve Crosses an Axis

From familiar shapes to curved boundaries

Building the area with vertical strips

Turning the strips sideways

Recovering the area of a circle

Extending the calculation to an ellipse

Choosing the easier strip direction

Check your understanding

Quiz

Practice Problems

Key Takeaways