Course Content
AP Calculus BC

Volumes by Cross Sections


What Are Cross-Sectional Volumes?

Suppose a region lies in the xy-plane, and perpendicular to the x-axis (or y-axis) we place cross-sections with a known geometric shape.

Each slice has:

  • Base along the region

  • Area determined by the shape (square, semicircle, triangle, etc.)

  • Thickness dxdx (for vertical cross-sections) or dydy (for horizontal ones)

The total volume is the sum (integral) of all the areas:

where A(x) is the area of the cross-section at x.


Typical Shapes

Cross-sections are usually:

Shape Area Formula
Square A=s^2
Semicircle A=\frac{1}{2}\pi{r^2}
Equilateral triangle A=\frac{\sqrt3}{4}s^2
Isosceles right triangle A=\frac{1}{2}s^2
Rectangle (constant height k) A=ks

Where ss is the length of the base (distance between curves).


Step-by-Step

To find a volume by cross-sections:

Step 1 — Identify the base region

Usually bounded by curves:

$$y=f(x) , y=g(x) , x=a , x=b$$

Step 2 — Determine slice direction
  • If cross-sections are perpendicular to x-axis → integrate in dx

  • If perpendicular to y-axis → integrate in dy

Step 3 — Find the base length

This becomes the side or diameter of the cross-section:

$$s(x) = f(x) – g(x)$$

Step 4 — Write the area formula for that shape

For example, if a square:

A(x)=$$[s(x)]^2$$

If a semicircle (radius = half the base):

A(x)=$$\frac{1}{2}\pi\left(\frac{s(x)}{2}\right)^2$$

Step 5 — Integrate

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


Cross Sections Are Squares

The base region is bounded by:

$$y=0 , y=\sqrt{x} , x=0 , x=4$$

Cross-sections ⟂ to the x-axis are squares.

Step 1: Base length

Step 2: Area of cross-section (square)

A(x)=$$(\sqrt{x})^2=x$$

Step 3: Integrate

V=$$\int_{0}^{4} x dx = \left[\frac{x^2}{2}\right]_{0}^{4}=8$$

So the volume = 8 cubic units.


Cross Sections Are Semicircles

Region between

$$y=4-x^2 , y=0$$

Cross-sections perpendicular to x-axis are semicircles.

Step 1: Base (diameter of semicircle)

$$s(x) = 4-x^2$$

Radius:

Step 2: Area of semicircle

A(x)=$$\frac{1}{2}\pi{r^2}$$

A(x)=$$\frac{1}{2}\pi\left(\frac{4-x^2}{2}\right)^2=\frac{\pi}{8}(4-x^2)^2$$

Step 3: Integrate

Bounds where the curves intersect:

$$4-x^2=0 \longrightarrow x=-2 , 2$$

V=$$\int_{-2}^{2}\frac{\pi}{8}(4-x^2)^2dx$$


Equilateral Triangles

Region bounded by:

$$y=x , y=2$$

Cross-sections ⟂ to y-axis → integrate with respect to y.

Solve for x:

$$x=y$$

The base is from:

$$x=0 , x=y$$

So the side length is:

$$s(y)=y$$

Area of equilateral triangle:

A(y)=$$\frac{\sqrt3}{4}s^2=\frac{\sqrt3}{4}y^2$$

Volume:

V=$$\int_{0}^{2}\frac{\sqrt3}{4}y^2dy$$


Cross Sections Are Rectangles

The base is between two curves:

$$y=3x , y=x^2$$

Cross-sections perpendicular to x-axis have height 5.

Base length:

$$s(x)=3x-x^2$$

Area:

A(x)=$$5(3x-x^2)$$

Volume:

the curves intersect at x(3−x)=0 → x = 0,3

V=$$\int_{0}^{3}5(3x-x^2)dx$$