Topic 8.7 Notes – Volumes with Cross Sections: Squares and Rectangles

Volume from Known Cross Sections

If you know the area of each cross section, volume comes from accumulation:

$$V = \int_a^b A(x)\,dx \quad \text{or} \quad V = \int_a^b A(y)\,dy$$

• $A(x)$ or $A(y)$ is the area of one cross section
• $dx$ or $dy$ is the slice thickness
• $a$ and $b$ are bounds of the base region

Think back to area between curves. There, you integrated a height.
Here, you integrate an area formula.

So the entire problem boils down to this:

What is the area of one slice?

The Base Region and Slice Direction

Everything starts with the base region.

Consider the region bounded by $y = 4 - x^2$ and $y = x$ on an $x$-$y$ coordinate plane, shaded between their intersection points. A vertical slice inside the region represents one cross section.

Two decisions matter:

1. Which variable are you integrating with respect to?

• Perpendicular to the x-axis → integrate $dx$
  → subtract top − bottom
• Perpendicular to the y-axis → integrate $dy$→ subtract right − left

That tells you how to find the side length or width.

Square Cross Sections

If each slice is a square, then

$$A = s^2$$

where $s$ is the distance across the base region.

If slices are vertical:

$$s = \text{top function} - \text{bottom function}$$

So

$$A(x) = (\text{top} - \text{bottom})^2$$

Example structure:

If region is between $y=f(x)$ and $y=g(x)$, then

$$V = \int_a^b (f(x) - g(x))^2 dx$$

That squared is where students lose points. The difference gives the side length. The square gives the area.

No π. No radii. Just geometry.

Rectangular Cross Sections

For rectangles:

$$A = \text{width} \cdot \text{height}$$

The width usually comes from the base region.
The height is often given in words.

Common situations:

• Height is a constant (like 5)
• Height is a multiple of the base (like “twice the width”)
• Height is another function

Example reasoning:

If slices are perpendicular to the y-axis and height is 3, then

• Width = right − left
• Area = $3(\text{right} - \text{left})$

Then:

$$V = \int_a^b 3(\text{right} - \text{left})\,dy$$

Be careful: if integrating with respect to $y$, everything must be written in terms of $y$. That may require solving an equation for $x$.

A Quick Setup Walkthrough

Suppose the base is bounded by:

• $y = x^2$
• $y = 6 - x$

Cross sections perpendicular to the x-axis are squares.

1. Find intersections
   Solve $x^2 = 6 - x$

2. Identify top and bottom
   Top is $6 - x$, bottom is $x^2$

3. Side length
   $s = (6 - x) - x^2$

4. Area
   $A(x) = [(6 - x) - x^2]^2$

5. Volume
   $$V = \int_a^b [(6 - x) - x^2]^2 dx$$

On an FRQ, if your setup is correct, you earn most of the credit even before integrating.

Common Mistakes That Cost Points

• Forgetting to square for square cross sections
• Using top − bottom when you needed right − left
• Mixing variables, like integrating $dy$ but leaving x’s in the integrand
• Using washer/disk formulas when the problem never mentioned rotation
• Wrong bounds because intersections weren’t solved carefully

What This Topic Is Really Testing

They’re checking whether you can:

• Translate geometry into an integral
• Connect area between curves to volume
• Handle algebra cleanly inside a setup

This is usually no-calculator on the AP exam. The algebra is manageable if your setup is clean.

The hard part is not integration.
It’s writing the correct area expression.