8.4.1 Using cross sections

Achievable AP Calculus AB

8. Applications of integrals

8.4. Volume

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Using cross sections

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What you’ll learn

Volume by slicing: Set up definite integrals that accumulate cross-sectional areas to find 3D volumes.
Known cross sections: Apply area templates for squares, rectangles, triangles, and circles based on base length $s$ .

Previously, we found the areas of 2D regions by summing the areas of infinitely many thin rectangles. Similarly, the volume of a 3D solid can be found by summing the areas of infinitely many thin cross-sectional slices.

For these solids, the 2D region serves as the base, and the cross sections extend vertically out of the page. Because the distance between the bounding functions varies across the interval, the base side length, $s$ , of each cross section changes. the area of each cross section changes. This variable side length determines the cross-sectional area function, $A (x)$ or $A (y)$ . Integrating this area function over the interval accumulates the slices to give the total volume of the solid.

To see what solids with certain cross sections look like, visit this GeoGebra application.

Slices perpendicular to the $x$ -axis ( $d x$ ):

$V = \int_{a}^{b} A (x) d x where s = Top - Bottom$

Perpendicular to the $y$ -axis ( $d y$ ):

$V = \int_{c}^{d} A (y) d y where s = Right - Left$

Geometric area formulas

For all shapes, first find the base length $s$ (the distance between the boundary curves) and then plug it into the matching area formula:

Cross-Sectional Shape	Area Formula $A (s)$	Notes
Square	$s^{2}$	Base $s$ is the entire side.
Rectangle	$s \cdot h$	Height $h$ is given as a constant or a function of $s$ .
Equilateral triangle	$\frac{3}{4} s^{2}$	Do not confuse with standard triangles.
Isosceles right triangle (leg on base)	$\frac{1}{2} s^{2}$	The base $s$ is one of the perpendicular legs.
Isosceles right triangle (hypotenuse on base)	$\frac{1}{4} s^{2}$	The base $s$ is the hypotenuse ( $\frac{1}{2} (\frac{s}{2})^{2}$ ).
Semicircle	$\frac{π}{8} s^{2}$	The base $s$ is the diameter, so radius $r = \frac{s}{2}$ .
Quarter circle	$\frac{π}{4} s^{2}$	The base $s$ is the full radius ( $r = s$ ).

Example 1 Squares $(s^{2})$

The region bounded by $y = x, y = - x$ , and $x = 4$ forms the base of a solid. Find the volume if cross sections are squares perpendicular to:

a) The $x$ -axis.

b) The $y$ -axis.

Solutions

a) Perpendicular to $x$ -axis ( $d x$ )

(spoiler)

Sketch the region and draw vertical slice from the upper function to the lower function. The length of this slice is the base, $s$ .

Note: The square cross sections extend directly out of the page toward you, using $s$ as their bottom edge.

s is the side length of the square cross sections perpendicular to the x-axis

1. Bounds:

Integrating over the $x$ -bounds: $[0, 4]$

2. Find side length:

$s = Top - Bottom = x - (- x) = 2 x$

3. Area of each cross section (formula):

$A (x) = s^{2} = (2 x)^{2} = 4 x$

4. Volume integral:

$V = \int_{0}^{4} A (x) d x = \int_{0}^{4} 4 x d x = 32$

b) Perpendicular to $y$ -axis ( $d y$ )

(spoiler)

Shown below is a horizontal slice perpendicular to the $y$ -axis that spans from the left to the right boundary

s is the side length of the square cross sections perpendicular to the y-axis

For slices perpendicular to the $y$ -axis, the base length $s$ should be expressed as a function of $y$ , representing the horizontal distance between the left and right boundary curves at any $y$ .

1. Rewrite equations in terms of $y$ :

$y = \pm x ⟹ x = y^{2}$

2. Bounds:

Find where $x = y^{2}$ intersects $x = 4$ :

$y$ -bounds: $[- 2, 2]$

3. Find side length:

$s = Right - Left = 4 - y^{2}$

4. Area formula:

$A (y) = s^{2} = (4 - y^{2})^{2}$

5. Volume integral:

$V = \int_{- 2}^{2} (4 - y^{2})^{2} d y \approx 34.133$

Example 2: Rectangles $(s \cdot h)$

The region bounded by $y = x$ and $y = x^{2}$ on $[0, 1]$ forms the base of a solid. Find the volume if cross sections perpendicular to the $x$ -axis have a height that is $2$ times its width.

Solution

(spoiler)

1. Bounds:

$x$ -bounds: $[0, 1]$

2. Find base length:

$s = Top - Bottom = x - x^{2}$

3. Area formula: $A (x) = base \cdot height$

“Height is $2$ times its width” $⟹ h = 2 s$ .
So $A (x) = s \cdot 2 s = 2 (x - x^{2})^{2}$

4. Volume integral:

$V = \int_{0}^{1} 2 (x - x^{2})^{2} d x = 0.067$

Example 3: Equilateral triangles $(\frac{3}{4} s^{2})$

The region bounded by $x = \frac{1}{y}$ , $x = - 1$ , $y = 0.5$ , and $y = 2$ forms the base of a solid. Cross sections perpendicular to the $y$ -axis are equilateral triangles. Find the volume.

Solution

(spoiler)

Perpendicular to $y ⟹$ integrate using $d y$

1. Bounds:

$y$ -bounds: $[0.5, 2]$

2. Find base length:

$s = Right - Left = \frac{1}{y} - (- 1) = \frac{1}{y} + 1$

3. Area formula:

$A (y) = \frac{3}{4} s^{2} = \frac{3}{4} (\frac{1}{y} + 1)^{2}$

4. Volume integral:

$V = \frac{3}{4} \int_{0.5}^{2} (\frac{1}{y} + 1)^{2} d y \approx 2.500$

Example 4: Isosceles right triangles (leg vs. hypotenuse)

The region bounded by $y = x - 1$ , the $x$ -axis, and the $y$ -axis forms a triangular base in the fourth quadrant. Find the volume if cross sections are isosceles right triangles with:

a) The hypotenuse lying along the base, sliced perpendicular to the $y$ -axis.

b) One leg lying along the base, sliced perpendicular to the $x$ -axis.

Solutions

a) Hypotenuse on base

(spoiler)

Perpendicular to the $y$ -axis $⟹$ integrate using $d y$

1. Rewrite equation in terms of $y$ :

$y = x - 1 ⟹ x = y + 1$

2. Bounds:

Find where $x = y + 1$ meets both axes:

$y$ -bounds: $[- 1, 0]$

3. Find side length:

$s = Right - Left = (y + 1) - 0 = y + 1$

4. Area formula:

Area formula: $A (y) = \frac{1}{4} s^{2} = \frac{1}{4} (y + 1)^{2}$

5. Volume integral:

$V = \frac{1}{4} \int_{- 1}^{0} (y + 1)^{2} d y \approx 0.083$

b) Leg on base

(spoiler)

1. Bounds:

$x$ -bounds: $[0, 1]$

2. Find base length:

$s = Top - Bottom = 0 - (x - 1) = 1 - x$

3. Area formula:

$A (x) = \frac{1}{2} s^{2} = \frac{1}{2} (1 - x)^{2}$

4. Volume integral:

$V = \frac{1}{2} \int_{0}^{1} (1 - x)^{2} d x \approx 0.167$

Example 5: Semicircles $(\frac{π}{8} s^{2})$

The region bounded by $y = e^{x}$ , $y = - x$ , and the $y$ -axis forms the base of a solid. Find the volume if cross sections perpendicular to the $x$ -axis are semicircles. (Calculator active)

Solution

(spoiler)

Find intersection point with calculator: $e^{x} = - x ⟹ x \approx - 0.567$ .
Bounds: Integrate over $[- 0.567, 0]$ .
Find diameter length: $s = Top - Bottom = e^{x} - (- x) = e^{x} + x$
Area formula: $A (x) = \frac{π}{8} s^{2} = \frac{π}{8} (e^{x} + x)^{2}$
Volume:

$V = \frac{π}{8} \int_{- 0.567}^{0} (e^{x} + x)^{2} d x \approx 0.070$

Example 6: Quarter circles $(\frac{π}{4} s^{2})$ - with a split base

The region bounded by $y = e^{x}$ , $y = - x$ , and the $y$ -axis forms the base of a solid. Find the volume if cross sections perpendicular to the $y$ -axis are quarter circles. (Calculator active)

Solution

(spoiler)

Perpendicular to the $y$ -axis $⟹$ integrate using $d y$

1. Rewrite equations in terms of $y$ :

$y = e^{x} ⟹ x = ln (y)$
$y = - x ⟹ x = - y$

2. Bounds (Watch out!) + Base length:

The “left” function changes at the intersection level $y = 0.567$ . Split the integral:

Bottom section ( $[0, 0.567]$ ): Bounded by $x = - y$ and $x = 0 ⟹ s = 0 - (- y) = y$
Top section ( $[0.567, 1]$ ): Bounded by $x = ln (y)$ and $x = 0 ⟹ s = 0 - ln (y) = - ln (y)$

3. Area template:

$A (y) = \frac{π}{4} s^{2}$

4. Volume:

$V = \frac{π}{4} \int_{0}^{0.567} (y)^{2} d y + \frac{π}{4} \int_{0.567}^{1} (- ln (y))^{2} d y \approx 0.048 + 0.031 = 0.079$

Always identify:

The direction of slicing (perpendicular to $x$ or $y$ -axis) to determine the expression for the side length $s$ .
The area formula $A (x)$ or $A (y)$ depending on the shape of the cross section
The limits of integration

Common area formulas, where $s$ is the distance between the curves and/or axes that defines the size of the shape:

Shape	Area
Square	$A = s^{2}$
Rectangle	$A = s \times h$
Equilateral triangle	$A = \frac{3}{4} s^{2}$
Isosceles right triangle, hypotenuse as base	$A = \frac{1}{4} s^{2}$
Isosceles right triangle, leg as base	$A = \frac{1}{2} s^{2}$
Semicircle	$A = \frac{π}{8} s^{2}$
Quarter circle	$A = \frac{π}{4} s^{2}$

Using cross sections

What you’ll learn

Volume by slicing: Set up definite integrals that accumulate cross-sectional areas to find 3D volumes.
Known cross sections: Apply area templates for squares, rectangles, triangles, and circles based on base length $s$ .

To see what solids with certain cross sections look like, visit this GeoGebra application.

Slices perpendicular to the $x$ -axis ( $d x$ ):

$V = \int_{a}^{b} A (x) d x where s = Top - Bottom$

Perpendicular to the $y$ -axis ( $d y$ ):

$V = \int_{c}^{d} A (y) d y where s = Right - Left$

Geometric area formulas

For all shapes, first find the base length $s$ (the distance between the boundary curves) and then plug it into the matching area formula:

Cross-Sectional Shape	Area Formula $A (s)$	Notes
Square	$s^{2}$	Base $s$ is the entire side.
Rectangle	$s \cdot h$	Height $h$ is given as a constant or a function of $s$ .
Equilateral triangle	$\frac{3}{4} s^{2}$	Do not confuse with standard triangles.
Isosceles right triangle (leg on base)	$\frac{1}{2} s^{2}$	The base $s$ is one of the perpendicular legs.
Isosceles right triangle (hypotenuse on base)	$\frac{1}{4} s^{2}$	The base $s$ is the hypotenuse ( $\frac{1}{2} (\frac{s}{2})^{2}$ ).
Semicircle	$\frac{π}{8} s^{2}$	The base $s$ is the diameter, so radius $r = \frac{s}{2}$ .
Quarter circle	$\frac{π}{4} s^{2}$	The base $s$ is the full radius ( $r = s$ ).

Example 1 Squares $(s^{2})$

The region bounded by $y = x, y = - x$ , and $x = 4$ forms the base of a solid. Find the volume if cross sections are squares perpendicular to:

a) The $x$ -axis.

b) The $y$ -axis.

Solutions

a) Perpendicular to $x$ -axis ( $d x$ )

(spoiler)

Sketch the region and draw vertical slice from the upper function to the lower function. The length of this slice is the base, $s$ .

Note: The square cross sections extend directly out of the page toward you, using $s$ as their bottom edge.

1. Bounds:

Integrating over the $x$ -bounds: $[0, 4]$

2. Find side length:

$s = Top - Bottom = x - (- x) = 2 x$

3. Area of each cross section (formula):

$A (x) = s^{2} = (2 x)^{2} = 4 x$

4. Volume integral:

$V = \int_{0}^{4} A (x) d x = \int_{0}^{4} 4 x d x = 32$

b) Perpendicular to $y$ -axis ( $d y$ )

(spoiler)

Shown below is a horizontal slice perpendicular to the $y$ -axis that spans from the left to the right boundary

For slices perpendicular to the $y$ -axis, the base length $s$ should be expressed as a function of $y$ , representing the horizontal distance between the left and right boundary curves at any $y$ .

1. Rewrite equations in terms of $y$ :

$y = \pm x ⟹ x = y^{2}$

2. Bounds:

Find where $x = y^{2}$ intersects $x = 4$ :

$y$ -bounds: $[- 2, 2]$

3. Find side length:

$s = Right - Left = 4 - y^{2}$

4. Area formula:

$A (y) = s^{2} = (4 - y^{2})^{2}$

5. Volume integral:

$V = \int_{- 2}^{2} (4 - y^{2})^{2} d y \approx 34.133$

Example 2: Rectangles $(s \cdot h)$

The region bounded by $y = x$ and $y = x^{2}$ on $[0, 1]$ forms the base of a solid. Find the volume if cross sections perpendicular to the $x$ -axis have a height that is $2$ times its width.

Solution

(spoiler)

1. Bounds:

$x$ -bounds: $[0, 1]$

2. Find base length:

$s = Top - Bottom = x - x^{2}$

3. Area formula: $A (x) = base \cdot height$

“Height is $2$ times its width” $⟹ h = 2 s$ .
So $A (x) = s \cdot 2 s = 2 (x - x^{2})^{2}$

4. Volume integral:

$V = \int_{0}^{1} 2 (x - x^{2})^{2} d x = 0.067$

Example 3: Equilateral triangles $(\frac{3}{4} s^{2})$

The region bounded by $x = \frac{1}{y}$ , $x = - 1$ , $y = 0.5$ , and $y = 2$ forms the base of a solid. Cross sections perpendicular to the $y$ -axis are equilateral triangles. Find the volume.

Solution

(spoiler)

Perpendicular to $y ⟹$ integrate using $d y$

1. Bounds:

$y$ -bounds: $[0.5, 2]$

2. Find base length:

$s = Right - Left = \frac{1}{y} - (- 1) = \frac{1}{y} + 1$

3. Area formula:

$A (y) = \frac{3}{4} s^{2} = \frac{3}{4} (\frac{1}{y} + 1)^{2}$

4. Volume integral:

$V = \frac{3}{4} \int_{0.5}^{2} (\frac{1}{y} + 1)^{2} d y \approx 2.500$

Example 4: Isosceles right triangles (leg vs. hypotenuse)

The region bounded by $y = x - 1$ , the $x$ -axis, and the $y$ -axis forms a triangular base in the fourth quadrant. Find the volume if cross sections are isosceles right triangles with:

a) The hypotenuse lying along the base, sliced perpendicular to the $y$ -axis.

b) One leg lying along the base, sliced perpendicular to the $x$ -axis.

Solutions

a) Hypotenuse on base

(spoiler)

Perpendicular to the $y$ -axis $⟹$ integrate using $d y$

1. Rewrite equation in terms of $y$ :

$y = x - 1 ⟹ x = y + 1$

2. Bounds:

Find where $x = y + 1$ meets both axes:

$y$ -bounds: $[- 1, 0]$

3. Find side length:

$s = Right - Left = (y + 1) - 0 = y + 1$

4. Area formula:

Area formula: $A (y) = \frac{1}{4} s^{2} = \frac{1}{4} (y + 1)^{2}$

5. Volume integral:

$V = \frac{1}{4} \int_{- 1}^{0} (y + 1)^{2} d y \approx 0.083$

b) Leg on base

(spoiler)

1. Bounds:

$x$ -bounds: $[0, 1]$

2. Find base length:

$s = Top - Bottom = 0 - (x - 1) = 1 - x$

3. Area formula:

$A (x) = \frac{1}{2} s^{2} = \frac{1}{2} (1 - x)^{2}$

4. Volume integral:

$V = \frac{1}{2} \int_{0}^{1} (1 - x)^{2} d x \approx 0.167$

Example 5: Semicircles $(\frac{π}{8} s^{2})$

The region bounded by $y = e^{x}$ , $y = - x$ , and the $y$ -axis forms the base of a solid. Find the volume if cross sections perpendicular to the $x$ -axis are semicircles. (Calculator active)

Solution

(spoiler)

Find intersection point with calculator: $e^{x} = - x ⟹ x \approx - 0.567$ .
Bounds: Integrate over $[- 0.567, 0]$ .
Find diameter length: $s = Top - Bottom = e^{x} - (- x) = e^{x} + x$
Area formula: $A (x) = \frac{π}{8} s^{2} = \frac{π}{8} (e^{x} + x)^{2}$
Volume:

$V = \frac{π}{8} \int_{- 0.567}^{0} (e^{x} + x)^{2} d x \approx 0.070$

Example 6: Quarter circles $(\frac{π}{4} s^{2})$ - with a split base

The region bounded by $y = e^{x}$ , $y = - x$ , and the $y$ -axis forms the base of a solid. Find the volume if cross sections perpendicular to the $y$ -axis are quarter circles. (Calculator active)

Solution

(spoiler)

Perpendicular to the $y$ -axis $⟹$ integrate using $d y$

1. Rewrite equations in terms of $y$ :

$y = e^{x} ⟹ x = ln (y)$
$y = - x ⟹ x = - y$

2. Bounds (Watch out!) + Base length:

The “left” function changes at the intersection level $y = 0.567$ . Split the integral:

Bottom section ( $[0, 0.567]$ ): Bounded by $x = - y$ and $x = 0 ⟹ s = 0 - (- y) = y$
Top section ( $[0.567, 1]$ ): Bounded by $x = ln (y)$ and $x = 0 ⟹ s = 0 - ln (y) = - ln (y)$

3. Area template:

$A (y) = \frac{π}{4} s^{2}$

4. Volume:

$V = \frac{π}{4} \int_{0}^{0.567} (y)^{2} d y + \frac{π}{4} \int_{0.567}^{1} (- ln (y))^{2} d y \approx 0.048 + 0.031 = 0.079$

Achievable AP Calculus AB

8. Applications of integrals

8.4. Volume

Our AP Calculus AB course is now in "early access" - get 50% off for a limited time.

Using cross sections

9 min read

Font

Discuss

Feedback

What you’ll learn

Volume by slicing: Set up definite integrals that accumulate cross-sectional areas to find 3D volumes.
Known cross sections: Apply area templates for squares, rectangles, triangles, and circles based on base length $s$ .

To see what solids with certain cross sections look like, visit this GeoGebra application.

Slices perpendicular to the $x$ -axis ( $d x$ ):

$V = \int_{a}^{b} A (x) d x where s = Top - Bottom$

Perpendicular to the $y$ -axis ( $d y$ ):

$V = \int_{c}^{d} A (y) d y where s = Right - Left$

Geometric area formulas

For all shapes, first find the base length $s$ (the distance between the boundary curves) and then plug it into the matching area formula:

Cross-Sectional Shape	Area Formula $A (s)$	Notes
Square	$s^{2}$	Base $s$ is the entire side.
Rectangle	$s \cdot h$	Height $h$ is given as a constant or a function of $s$ .
Equilateral triangle	$\frac{3}{4} s^{2}$	Do not confuse with standard triangles.
Isosceles right triangle (leg on base)	$\frac{1}{2} s^{2}$	The base $s$ is one of the perpendicular legs.
Isosceles right triangle (hypotenuse on base)	$\frac{1}{4} s^{2}$	The base $s$ is the hypotenuse ( $\frac{1}{2} (\frac{s}{2})^{2}$ ).
Semicircle	$\frac{π}{8} s^{2}$	The base $s$ is the diameter, so radius $r = \frac{s}{2}$ .
Quarter circle	$\frac{π}{4} s^{2}$	The base $s$ is the full radius ( $r = s$ ).

Example 1 Squares $(s^{2})$

The region bounded by $y = x, y = - x$ , and $x = 4$ forms the base of a solid. Find the volume if cross sections are squares perpendicular to:

a) The $x$ -axis.

b) The $y$ -axis.

Solutions

a) Perpendicular to $x$ -axis ( $d x$ )

(spoiler)

Sketch the region and draw vertical slice from the upper function to the lower function. The length of this slice is the base, $s$ .

Note: The square cross sections extend directly out of the page toward you, using $s$ as their bottom edge.

1. Bounds:

Integrating over the $x$ -bounds: $[0, 4]$

2. Find side length:

$s = Top - Bottom = x - (- x) = 2 x$

3. Area of each cross section (formula):

$A (x) = s^{2} = (2 x)^{2} = 4 x$

4. Volume integral:

$V = \int_{0}^{4} A (x) d x = \int_{0}^{4} 4 x d x = 32$

b) Perpendicular to $y$ -axis ( $d y$ )

(spoiler)

Shown below is a horizontal slice perpendicular to the $y$ -axis that spans from the left to the right boundary

For slices perpendicular to the $y$ -axis, the base length $s$ should be expressed as a function of $y$ , representing the horizontal distance between the left and right boundary curves at any $y$ .

1. Rewrite equations in terms of $y$ :

$y = \pm x ⟹ x = y^{2}$

2. Bounds:

Find where $x = y^{2}$ intersects $x = 4$ :

$y$ -bounds: $[- 2, 2]$

3. Find side length:

$s = Right - Left = 4 - y^{2}$

4. Area formula:

$A (y) = s^{2} = (4 - y^{2})^{2}$

5. Volume integral:

$V = \int_{- 2}^{2} (4 - y^{2})^{2} d y \approx 34.133$

Example 2: Rectangles $(s \cdot h)$

The region bounded by $y = x$ and $y = x^{2}$ on $[0, 1]$ forms the base of a solid. Find the volume if cross sections perpendicular to the $x$ -axis have a height that is $2$ times its width.

Solution

(spoiler)

1. Bounds:

$x$ -bounds: $[0, 1]$

2. Find base length:

$s = Top - Bottom = x - x^{2}$

3. Area formula: $A (x) = base \cdot height$

“Height is $2$ times its width” $⟹ h = 2 s$ .
So $A (x) = s \cdot 2 s = 2 (x - x^{2})^{2}$

4. Volume integral:

$V = \int_{0}^{1} 2 (x - x^{2})^{2} d x = 0.067$

Example 3: Equilateral triangles $(\frac{3}{4} s^{2})$

The region bounded by $x = \frac{1}{y}$ , $x = - 1$ , $y = 0.5$ , and $y = 2$ forms the base of a solid. Cross sections perpendicular to the $y$ -axis are equilateral triangles. Find the volume.

Solution

(spoiler)

Perpendicular to $y ⟹$ integrate using $d y$

1. Bounds:

$y$ -bounds: $[0.5, 2]$

2. Find base length:

$s = Right - Left = \frac{1}{y} - (- 1) = \frac{1}{y} + 1$

3. Area formula:

$A (y) = \frac{3}{4} s^{2} = \frac{3}{4} (\frac{1}{y} + 1)^{2}$

4. Volume integral:

$V = \frac{3}{4} \int_{0.5}^{2} (\frac{1}{y} + 1)^{2} d y \approx 2.500$

Example 4: Isosceles right triangles (leg vs. hypotenuse)

The region bounded by $y = x - 1$ , the $x$ -axis, and the $y$ -axis forms a triangular base in the fourth quadrant. Find the volume if cross sections are isosceles right triangles with:

a) The hypotenuse lying along the base, sliced perpendicular to the $y$ -axis.

b) One leg lying along the base, sliced perpendicular to the $x$ -axis.

Solutions

a) Hypotenuse on base

(spoiler)

Perpendicular to the $y$ -axis $⟹$ integrate using $d y$

1. Rewrite equation in terms of $y$ :

$y = x - 1 ⟹ x = y + 1$

2. Bounds:

Find where $x = y + 1$ meets both axes:

$y$ -bounds: $[- 1, 0]$

3. Find side length:

$s = Right - Left = (y + 1) - 0 = y + 1$

4. Area formula:

Area formula: $A (y) = \frac{1}{4} s^{2} = \frac{1}{4} (y + 1)^{2}$

5. Volume integral:

$V = \frac{1}{4} \int_{- 1}^{0} (y + 1)^{2} d y \approx 0.083$

b) Leg on base

(spoiler)

1. Bounds:

$x$ -bounds: $[0, 1]$

2. Find base length:

$s = Top - Bottom = 0 - (x - 1) = 1 - x$

3. Area formula:

$A (x) = \frac{1}{2} s^{2} = \frac{1}{2} (1 - x)^{2}$

4. Volume integral:

$V = \frac{1}{2} \int_{0}^{1} (1 - x)^{2} d x \approx 0.167$

Example 5: Semicircles $(\frac{π}{8} s^{2})$

The region bounded by $y = e^{x}$ , $y = - x$ , and the $y$ -axis forms the base of a solid. Find the volume if cross sections perpendicular to the $x$ -axis are semicircles. (Calculator active)

Solution

(spoiler)

Find intersection point with calculator: $e^{x} = - x ⟹ x \approx - 0.567$ .
Bounds: Integrate over $[- 0.567, 0]$ .
Find diameter length: $s = Top - Bottom = e^{x} - (- x) = e^{x} + x$
Area formula: $A (x) = \frac{π}{8} s^{2} = \frac{π}{8} (e^{x} + x)^{2}$
Volume:

$V = \frac{π}{8} \int_{- 0.567}^{0} (e^{x} + x)^{2} d x \approx 0.070$

Example 6: Quarter circles $(\frac{π}{4} s^{2})$ - with a split base

The region bounded by $y = e^{x}$ , $y = - x$ , and the $y$ -axis forms the base of a solid. Find the volume if cross sections perpendicular to the $y$ -axis are quarter circles. (Calculator active)

Solution

(spoiler)

Perpendicular to the $y$ -axis $⟹$ integrate using $d y$

1. Rewrite equations in terms of $y$ :

$y = e^{x} ⟹ x = ln (y)$
$y = - x ⟹ x = - y$

2. Bounds (Watch out!) + Base length:

The “left” function changes at the intersection level $y = 0.567$ . Split the integral:

Bottom section ( $[0, 0.567]$ ): Bounded by $x = - y$ and $x = 0 ⟹ s = 0 - (- y) = y$
Top section ( $[0.567, 1]$ ): Bounded by $x = ln (y)$ and $x = 0 ⟹ s = 0 - ln (y) = - ln (y)$

3. Area template:

$A (y) = \frac{π}{4} s^{2}$

4. Volume:

$V = \frac{π}{4} \int_{0}^{0.567} (y)^{2} d y + \frac{π}{4} \int_{0.567}^{1} (- ln (y))^{2} d y \approx 0.048 + 0.031 = 0.079$

Always identify:

The direction of slicing (perpendicular to $x$ or $y$ -axis) to determine the expression for the side length $s$ .
The area formula $A (x)$ or $A (y)$ depending on the shape of the cross section
The limits of integration

Common area formulas, where $s$ is the distance between the curves and/or axes that defines the size of the shape:

Shape	Area
Square	$A = s^{2}$
Rectangle	$A = s \times h$
Equilateral triangle	$A = \frac{3}{4} s^{2}$
Isosceles right triangle, hypotenuse as base	$A = \frac{1}{4} s^{2}$
Isosceles right triangle, leg as base	$A = \frac{1}{2} s^{2}$
Semicircle	$A = \frac{π}{8} s^{2}$
Quarter circle	$A = \frac{π}{4} s^{2}$

Using cross sections

What you’ll learn

Volume by slicing: Set up definite integrals that accumulate cross-sectional areas to find 3D volumes.
Known cross sections: Apply area templates for squares, rectangles, triangles, and circles based on base length $s$ .

To see what solids with certain cross sections look like, visit this GeoGebra application.

Slices perpendicular to the $x$ -axis ( $d x$ ):

$V = \int_{a}^{b} A (x) d x where s = Top - Bottom$

Perpendicular to the $y$ -axis ( $d y$ ):

$V = \int_{c}^{d} A (y) d y where s = Right - Left$

Geometric area formulas

For all shapes, first find the base length $s$ (the distance between the boundary curves) and then plug it into the matching area formula:

Cross-Sectional Shape	Area Formula $A (s)$	Notes
Square	$s^{2}$	Base $s$ is the entire side.
Rectangle	$s \cdot h$	Height $h$ is given as a constant or a function of $s$ .
Equilateral triangle	$\frac{3}{4} s^{2}$	Do not confuse with standard triangles.
Isosceles right triangle (leg on base)	$\frac{1}{2} s^{2}$	The base $s$ is one of the perpendicular legs.
Isosceles right triangle (hypotenuse on base)	$\frac{1}{4} s^{2}$	The base $s$ is the hypotenuse ( $\frac{1}{2} (\frac{s}{2})^{2}$ ).
Semicircle	$\frac{π}{8} s^{2}$	The base $s$ is the diameter, so radius $r = \frac{s}{2}$ .
Quarter circle	$\frac{π}{4} s^{2}$	The base $s$ is the full radius ( $r = s$ ).

Example 1 Squares $(s^{2})$

The region bounded by $y = x, y = - x$ , and $x = 4$ forms the base of a solid. Find the volume if cross sections are squares perpendicular to:

a) The $x$ -axis.

b) The $y$ -axis.

Solutions

a) Perpendicular to $x$ -axis ( $d x$ )

(spoiler)

Sketch the region and draw vertical slice from the upper function to the lower function. The length of this slice is the base, $s$ .

Note: The square cross sections extend directly out of the page toward you, using $s$ as their bottom edge.

1. Bounds:

Integrating over the $x$ -bounds: $[0, 4]$

2. Find side length:

$s = Top - Bottom = x - (- x) = 2 x$

3. Area of each cross section (formula):

$A (x) = s^{2} = (2 x)^{2} = 4 x$

4. Volume integral:

$V = \int_{0}^{4} A (x) d x = \int_{0}^{4} 4 x d x = 32$

b) Perpendicular to $y$ -axis ( $d y$ )

(spoiler)

Shown below is a horizontal slice perpendicular to the $y$ -axis that spans from the left to the right boundary

For slices perpendicular to the $y$ -axis, the base length $s$ should be expressed as a function of $y$ , representing the horizontal distance between the left and right boundary curves at any $y$ .

1. Rewrite equations in terms of $y$ :

$y = \pm x ⟹ x = y^{2}$

2. Bounds:

Find where $x = y^{2}$ intersects $x = 4$ :

$y$ -bounds: $[- 2, 2]$

3. Find side length:

$s = Right - Left = 4 - y^{2}$

4. Area formula:

$A (y) = s^{2} = (4 - y^{2})^{2}$

5. Volume integral:

$V = \int_{- 2}^{2} (4 - y^{2})^{2} d y \approx 34.133$

Example 2: Rectangles $(s \cdot h)$

The region bounded by $y = x$ and $y = x^{2}$ on $[0, 1]$ forms the base of a solid. Find the volume if cross sections perpendicular to the $x$ -axis have a height that is $2$ times its width.

Solution

(spoiler)

1. Bounds:

$x$ -bounds: $[0, 1]$

2. Find base length:

$s = Top - Bottom = x - x^{2}$

3. Area formula: $A (x) = base \cdot height$

“Height is $2$ times its width” $⟹ h = 2 s$ .
So $A (x) = s \cdot 2 s = 2 (x - x^{2})^{2}$

4. Volume integral:

$V = \int_{0}^{1} 2 (x - x^{2})^{2} d x = 0.067$

Example 3: Equilateral triangles $(\frac{3}{4} s^{2})$

The region bounded by $x = \frac{1}{y}$ , $x = - 1$ , $y = 0.5$ , and $y = 2$ forms the base of a solid. Cross sections perpendicular to the $y$ -axis are equilateral triangles. Find the volume.

Solution

(spoiler)

Perpendicular to $y ⟹$ integrate using $d y$

1. Bounds:

$y$ -bounds: $[0.5, 2]$

2. Find base length:

$s = Right - Left = \frac{1}{y} - (- 1) = \frac{1}{y} + 1$

3. Area formula:

$A (y) = \frac{3}{4} s^{2} = \frac{3}{4} (\frac{1}{y} + 1)^{2}$

4. Volume integral:

$V = \frac{3}{4} \int_{0.5}^{2} (\frac{1}{y} + 1)^{2} d y \approx 2.500$

Example 4: Isosceles right triangles (leg vs. hypotenuse)

The region bounded by $y = x - 1$ , the $x$ -axis, and the $y$ -axis forms a triangular base in the fourth quadrant. Find the volume if cross sections are isosceles right triangles with:

a) The hypotenuse lying along the base, sliced perpendicular to the $y$ -axis.

b) One leg lying along the base, sliced perpendicular to the $x$ -axis.

Solutions

a) Hypotenuse on base

(spoiler)

Perpendicular to the $y$ -axis $⟹$ integrate using $d y$

1. Rewrite equation in terms of $y$ :

$y = x - 1 ⟹ x = y + 1$

2. Bounds:

Find where $x = y + 1$ meets both axes:

$y$ -bounds: $[- 1, 0]$

3. Find side length:

$s = Right - Left = (y + 1) - 0 = y + 1$

4. Area formula:

Area formula: $A (y) = \frac{1}{4} s^{2} = \frac{1}{4} (y + 1)^{2}$

5. Volume integral:

$V = \frac{1}{4} \int_{- 1}^{0} (y + 1)^{2} d y \approx 0.083$

b) Leg on base

(spoiler)

1. Bounds:

$x$ -bounds: $[0, 1]$

2. Find base length:

$s = Top - Bottom = 0 - (x - 1) = 1 - x$

3. Area formula:

$A (x) = \frac{1}{2} s^{2} = \frac{1}{2} (1 - x)^{2}$

4. Volume integral:

$V = \frac{1}{2} \int_{0}^{1} (1 - x)^{2} d x \approx 0.167$

Example 5: Semicircles $(\frac{π}{8} s^{2})$

The region bounded by $y = e^{x}$ , $y = - x$ , and the $y$ -axis forms the base of a solid. Find the volume if cross sections perpendicular to the $x$ -axis are semicircles. (Calculator active)

Solution

(spoiler)

Find intersection point with calculator: $e^{x} = - x ⟹ x \approx - 0.567$ .
Bounds: Integrate over $[- 0.567, 0]$ .
Find diameter length: $s = Top - Bottom = e^{x} - (- x) = e^{x} + x$
Area formula: $A (x) = \frac{π}{8} s^{2} = \frac{π}{8} (e^{x} + x)^{2}$
Volume:

$V = \frac{π}{8} \int_{- 0.567}^{0} (e^{x} + x)^{2} d x \approx 0.070$

Example 6: Quarter circles $(\frac{π}{4} s^{2})$ - with a split base

The region bounded by $y = e^{x}$ , $y = - x$ , and the $y$ -axis forms the base of a solid. Find the volume if cross sections perpendicular to the $y$ -axis are quarter circles. (Calculator active)

Solution

(spoiler)

Perpendicular to the $y$ -axis $⟹$ integrate using $d y$

1. Rewrite equations in terms of $y$ :

$y = e^{x} ⟹ x = ln (y)$
$y = - x ⟹ x = - y$

2. Bounds (Watch out!) + Base length:

The “left” function changes at the intersection level $y = 0.567$ . Split the integral:

Bottom section ( $[0, 0.567]$ ): Bounded by $x = - y$ and $x = 0 ⟹ s = 0 - (- y) = y$
Top section ( $[0.567, 1]$ ): Bounded by $x = ln (y)$ and $x = 0 ⟹ s = 0 - ln (y) = - ln (y)$

3. Area template:

$A (y) = \frac{π}{4} s^{2}$

4. Volume:

$V = \frac{π}{4} \int_{0}^{0.567} (y)^{2} d y + \frac{π}{4} \int_{0.567}^{1} (- ln (y))^{2} d y \approx 0.048 + 0.031 = 0.079$

Key points

Always identify:

The direction of slicing (perpendicular to $x$ or $y$ -axis) to determine the expression for the side length $s$ .
The area formula $A (x)$ or $A (y)$ depending on the shape of the cross section
The limits of integration

Common area formulas, where $s$ is the distance between the curves and/or axes that defines the size of the shape:

Shape	Area
Square	$A = s^{2}$
Rectangle	$A = s \times h$
Equilateral triangle	$A = \frac{3}{4} s^{2}$
Isosceles right triangle, hypotenuse as base	$A = \frac{1}{4} s^{2}$
Isosceles right triangle, leg as base	$A = \frac{1}{2} s^{2}$
Semicircle	$A = \frac{π}{8} s^{2}$
Quarter circle	$A = \frac{π}{4} s^{2}$

Using cross sections

What you’ll learn

Geometric area formulas

Example 1 Squares (s2)

Solutions

Example 2: Rectangles (s⋅h)

Solution

Example 3: Equilateral triangles (43​​s2)

Solution

Example 4: Isosceles right triangles (leg vs. hypotenuse)

Solutions

Example 5: Semicircles (8π​s2)

Solution

Example 6: Quarter circles (4π​s2) - with a split base

Solution

Using cross sections

What you’ll learn

Geometric area formulas

Example 1 Squares (s2)

Solutions

Example 2: Rectangles (s⋅h)

Solution

Example 3: Equilateral triangles (43​​s2)

Solution

Example 4: Isosceles right triangles (leg vs. hypotenuse)

Solutions

Example 5: Semicircles (8π​s2)

Solution

Example 6: Quarter circles (4π​s2) - with a split base

Solution

Using cross sections

What you’ll learn

Geometric area formulas

Example 1 Squares (s2)

Solutions

Example 2: Rectangles (s⋅h)

Solution

Example 3: Equilateral triangles (43​​s2)

Solution

Example 4: Isosceles right triangles (leg vs. hypotenuse)

Solutions

Example 5: Semicircles (8π​s2)

Solution

Example 6: Quarter circles (4π​s2) - with a split base

Solution

Using cross sections

What you’ll learn

Geometric area formulas

Example 1 Squares (s2)

Solutions

Example 2: Rectangles (s⋅h)

Solution

Example 3: Equilateral triangles (43​​s2)

Solution

Example 4: Isosceles right triangles (leg vs. hypotenuse)

Solutions

Example 5: Semicircles (8π​s2)

Solution

Example 6: Quarter circles (4π​s2) - with a split base

Solution

Example 1 Squares $(s^{2})$

Example 2: Rectangles $(s \cdot h)$

Example 3: Equilateral triangles $(\frac{3}{4} s^{2})$

Example 5: Semicircles $(\frac{π}{8} s^{2})$

Example 6: Quarter circles $(\frac{π}{4} s^{2})$ - with a split base

Example 1 Squares $(s^{2})$

Example 2: Rectangles $(s \cdot h)$

Example 3: Equilateral triangles $(\frac{3}{4} s^{2})$

Example 5: Semicircles $(\frac{π}{8} s^{2})$

Example 6: Quarter circles $(\frac{π}{4} s^{2})$ - with a split base

Example 1 Squares $(s^{2})$

Example 2: Rectangles $(s \cdot h)$

Example 3: Equilateral triangles $(\frac{3}{4} s^{2})$

Example 5: Semicircles $(\frac{π}{8} s^{2})$

Example 6: Quarter circles $(\frac{π}{4} s^{2})$ - with a split base

Example 1 Squares $(s^{2})$

Example 2: Rectangles $(s \cdot h)$

Example 3: Equilateral triangles $(\frac{3}{4} s^{2})$

Example 5: Semicircles $(\frac{π}{8} s^{2})$

Example 6: Quarter circles $(\frac{π}{4} s^{2})$ - with a split base