8.7 - Volumes with Cross Sections: Squares and Rectangles

Introduction to Cross-Sections

We’ve been finding the area between curves. Now we use those regions as the base of a three-dimensional solid.

Picture looking straight down at a solid from above. All you see is the outline of its base, the bounded region. To get the volume, we need to know about cross-sections.

A cross-section is what you see when you slice through a solid perpendicular to an axis. Think of slicing a loaf of bread. Each slice shows you a cross-section.

8.7 - Volumes with Cross Sections: Squares and Rectangles

Two things matter every time: - Perpendicular to which axis? This determines the slicing direction and your integration variable - What shape is the cross-section? Square, rectangle, triangle, and so on, which determines the area formula

As you move along the base, each cross-section changes size based on the distance between the boundary curves at that location.

Volume Formula for Cross-Sections

\boxed{V = \int_{a}^{b} A(x)\,dx}

where: -

A(x)

is the area of one cross-section at position

x

[a, b]

is the interval the base spans - The integral sums up all the infinitely many cross-sections

Perpendicular to the

y

-axis instead:

V = \int_{c}^{d} A(y)\,dy

Square Cross-Sections

For a square:

A = s^2

where

s

is the side length, which equals the distance between the boundary curves:

s = f(x) - g(x)

with

f

the upper curve and

g

the lower one. So:

\boxed{V = \int_{a}^{b} [f(x) - g(x)]^2\,dx}

Example 1: Squares Perpendicular to the x-axis

Find the volume of the solid whose base is bounded by

y = \sqrt{x}

and

y = x^2

, with square cross-sections perpendicular to the

x

-axis.

Step 1: find the intersection points

\sqrt{x} = x^2 \quad \Rightarrow \quad x = x^4 \quad \Rightarrow \quad x^4 - x = 0 \quad \Rightarrow \quad x(x^3 - 1) = 0

x = 0 \text{ or } x = 1

Step 2: which function is on top? At

x = 0.5

\sqrt{0.5} \approx 0.707

and

(0.5)^2 = 0.25

. So

\sqrt{x}

is on top.

Step 3: side length

s = \sqrt{x} - x^2

Step 4: set up

V = \int_{0}^{1} (\sqrt{x} - x^2)^2\,dx

Step 5: expand and integrate

(\sqrt{x} - x^2)^2 = x - 2x^{5/2} + x^4

V = \int_{0}^{1} (x - 2x^{5/2} + x^4)\,dx = \left[\frac{x^2}{2} - \frac{4x^{7/2}}{7} + \frac{x^5}{5}\right]_{0}^{1}

= \frac{1}{2} - \frac{4}{7} + \frac{1}{5} = \frac{35 - 40 + 14}{70} = \frac{9}{70}

Volume:

\frac{9}{70}

cubic units

Example 2: Squares Perpendicular to the y-axis

Same base region, but with square cross-sections perpendicular to the

y

-axis.

Step 1: convert to $x = f(y)$ form

y = \sqrt{x} \;\Rightarrow\; x = y^2 \qquad\qquad y = x^2 \;\Rightarrow\; x = \sqrt{y}

(taking the positive root)

Step 2: boundaries. The curves meet at

(0, 0)

and

(1, 1)

, so

y

runs from 0 to 1.

Step 3: which is on the right? At

y = 0.5

x = y^2 = 0.25

(left) and

x = \sqrt{y} \approx 0.707

(right). So

\sqrt{y}

is on the right.

Step 4: set up

V = \int_{0}^{1} (\sqrt{y} - y^2)^2\,dy

Notice this is structurally identical to Example 1, just with

y

in place of

x

, so it evaluates to the same

\frac{9}{70}

. That happens here because the base region is symmetric about the line

y = x

Rectangular Cross-Sections

For a rectangle:

A = \text{width} \times \text{height}

The big difference from squares: the height is extra information the problem has to give you. It isn’t automatically equal to the width.

\boxed{V = \int_{a}^{b} [\text{width}(x)] \cdot [\text{height}(x)]\,dx}

Common phrasings: - “Height is twice the width” →

h = 2w

- “Height is constant at 5 units” →

h = 5

- “Height equals

3x

” →

h = 3x

Example 3: Rectangles with Height Proportional to Width

The base is bounded by

y = x^3

, the

x

-axis, and

x = 2

. Cross-sections perpendicular to the

x

-axis are rectangles whose height is 3 times the width.

Step 1: width. The distance from the

x

-axis to the curve:

w = x^3 - 0 = x^3

Step 2: height. Three times the width:

h = 3w = 3x^3

Step 3: set up

V = \int_{0}^{2} (x^3)(3x^3)\,dx = \int_{0}^{2} 3x^6\,dx = 3\int_{0}^{2} x^6\,dx

Step 4: evaluate

= 3\left[\frac{x^7}{7}\right]_{0}^{2} = 3 \cdot \frac{128}{7} = \frac{384}{7}

Volume:

\frac{384}{7}

cubic units

Example 4: Rectangles Perpendicular to the y-axis

Same base as Example 3, but rectangles perpendicular to the

y

-axis with a constant height of 4.

Step 1: convert. From

y = x^3

x = \sqrt[3]{y}

Step 2: boundaries. When

x = 2

y = 8

. So

y

runs from 0 to 8.

Step 3: width and height

Width: from the left edge to the right edge. - Left:

x = \sqrt[3]{y}

- Right:

x = 2

- So

w = 2 - \sqrt[3]{y}

Height: given as constant,

h = 4

Step 4: set up and evaluate

V = \int_{0}^{8} (2 - \sqrt[3]{y}) \cdot 4\,dy = 4\int_{0}^{8} (2 - y^{1/3})\,dy

= 4\left[2y - \frac{3y^{4/3}}{4}\right]_{0}^{8} = 4\left[16 - \frac{3(16)}{4}\right] = 4[16 - 12] = 16

Volume:

16

cubic units

Example 5: No Graph Given

Find the volume of the solid whose base is bounded by

y = 3x - x^2

and

y = x^2 - 5x

, with square cross-sections perpendicular to the

x

-axis.

Step 1: find the intersections

3x - x^2 = x^2 - 5x \quad \Rightarrow \quad -2x^2 + 8x = 0 \quad \Rightarrow \quad -2x(x - 4) = 0

x = 0 \text{ or } x = 4

Step 2: which is on top? Test

x = 2

: -

3(2) - 2^2 = 2

(top) -

2^2 - 5(2) = -6

(bottom)

3x - x^2

is on top.

Step 3: side length

s = (3x - x^2) - (x^2 - 5x) = -2x^2 + 8x

Step 4: set up

V = \int_{0}^{4} (-2x^2 + 8x)^2\,dx

Step 5: expand and evaluate

(-2x^2 + 8x)^2 = 4x^4 - 32x^3 + 64x^2

V = \int_{0}^{4} (4x^4 - 32x^3 + 64x^2)\,dx = \left[\frac{4x^5}{5} - 8x^4 + \frac{64x^3}{3}\right]_{0}^{4}

= \frac{4096}{5} - 2048 + \frac{4096}{3} = \frac{12288 - 30720 + 20480}{15} = \frac{2048}{15}

Volume:

\frac{2048}{15} \approx 136.533

cubic units

Common Patterns and Tips

Width determination

Orientation	Width Formula
Perpendicular to $x$ -axis	$w = f(x) - g(x)$ (top minus bottom)
Perpendicular to $y$ -axis	$w = f(y) - g(y)$ (right minus left)

Height given as…

Description	Formula
“Height equals width”	$h = w$ (that makes it a square)
“Height is twice the width”	$h = 2w$
“Height is constant at $k$ ”	$h = k$
“Height equals $3x + 1$ ”	$h = 3x + 1$

Setup checklist

Before you integrate, check: - Perpendicular to which axis? (determines your integration variable) - What shape? (determines the area formula) - Did you find the intersection points? (they’re your limits) - Which curve is on top or right? (test a point if you’re unsure) - Is the height specified correctly? (for rectangles)

Units

If the base is measured in cm, the volume comes out in cm³. Area involves length × length; volume involves length × length × length.

Calculator: set up the integral, then use MATH → 9:fnInt( to evaluate numerically. This is especially handy when expanding a square creates a messy integral.

Problem-Solving Strategy

Step 1: identify the base region. Sketch it, and find the intersection points if they aren’t given.

Step 2: determine orientation. Perpendicular to

x

-axis means

dx

. Perpendicular to

y

-axis means

dy

, and convert the equations.

Step 3: identify the cross-section shape. - Square:

A = s^2

, where

s

is the distance between curves - Rectangle:

A = w \times h

, where

w

is the distance and

h

is given

Step 4: set up the integral.

V = \int_{a}^{b} A(x)\,dx \quad \text{or} \quad V = \int_{c}^{d} A(y)\,dy

Step 5: evaluate. Expand if needed, integrate term by term, or use a calculator.

Common Mistakes to Avoid

Forgetting to square for square cross-sections. It’s $\int [f(x) - g(x)]^2\,dx$ , not $\int [f(x) - g(x)]\,dx$

Wrong curve order. Always subtract bottom from top (or left from right)

Confusing width with height for rectangles. Width is the distance between curves; height is extra given information

Wrong integration variable. Perpendicular to $x$ -axis means $dx$ ; perpendicular to $y$ -axis means $dy$

Not converting equations for $dy$ integrals. Always solve for $x = f(y)$

Key Takeaways

Cross-sections: slices through a solid perpendicular to an axis

Volume formula: $V = \int_{a}^{b} A(x)\,dx$ , where $A(x)$ is the area of one cross-section

Square cross-sections: $V = \int_{a}^{b} [f(x) - g(x)]^2\,dx$

Rectangular cross-sections: $V = \int_{a}^{b} [\text{width}] \cdot [\text{height}]\,dx$

Width always equals the distance between the boundary curves

Height (for rectangles) has to be given in the problem

Perpendicular to $x$ -axis: integrate with $dx$

Perpendicular to $y$ -axis: integrate with $dy$ , converting equations to $x = f(y)$

Most common on the AP: squares perpendicular to the $x$ -axis

Calculator friendly: set up the integral, then use fnInt(

Units: volume is always in cubic units

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