Squares as Cross Sections (College Board AP® Calculus AB): Revision Note

Volumes with cross sections as squares

How can I find the volume of a solid with a square cross section?

• Use the basic concept

  • If the area of the cross section of a solid is given by $A\left(x\right)$

    • and $A\left(x\right)$ is continuous on $\left[a,b\right]$

  • Then the volume of the corresponding solid from $x=a$ to $x=b$ is

    • $\mathrm{Volume}={\int }_a^{b}A\left(x\right)dx$

• You may need to create the cross-sectional area function $A\left(x\right)$

  • E.g. to calculate the volume of a square-based right pyramid of height $h$ and base side length $2a$

    • Consider a side view of the pyramid laid out along the $x$-axis:

• The line in red from $a$ to $h$ has the equation $y=a-\frac{a}{h}x$

• At each value of $x$ between 0 and $h$

  • the cross section of the pyramid is a square with side length $2\left(a-\frac{a}{h}x\right)$

    • and area $A\left(x\right)={\left(2\left(a-\frac{a}{h}x\right)\right)}^2=4a^2{\left(1-\frac{1}{h}x\right)}^2$

  • Therefore

$\begin{array}{rcl}\mathrm{Volume}& =& {\int }_0^{h}4a^2{\left(1-\frac{1}{h}x\right)}^{2}dx\\ & =& 4a^2{\int }_0^h\left(1-\frac{2}{h}x+\frac{1}{h^2}{x}^2\right)dx\\ & =& 4a^2{\left[x-\frac{1}{h}{x}^2+\frac{1}{3h^2}{x}^3\right]}_0^h\\ & =& 4a^2\left(\left(h-\frac{1}{h}{\left(h\right)}^2+\frac{1}{3h^2}{\left(h\right)}^3\right)-0\right)\\ & =& 4a^2·\frac{1}{3}h\\ & =& \frac{4}{3}{a}^{2}h\end{array}$

• Alternatively, $A\left(x\right)$ may depend on the values of another function given to you in the question

  • See the Worked Example