Rectangles as Cross Sections (College Board AP® Calculus AB)

Revision Note

Author

Roger B

Expertise

Maths

Volumes with cross sections as rectangles

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

Use the basic concept
- If the area of the cross section of a solid is given by $A (x)$
  - and $A (x)$ is continuous on $[a, b]$
- Then the volume of the corresponding solid from $x = a$ to $x = b$ is
  - $Volume = \int_{a}^{b} A (x) d x$
You may need to create the cross sectional area function $A (x)$ based on information provided
- For example $A (x)$ may depend on the values of another function (or functions) given to you in the question

Worked Example

Let $R$ be the region enclosed by the graph of $f (x) = 54 - 2 x^{3}$ , the $x$ - and $y$ -axes, and the vertical line $x = 2$ , as shown in the figure below.

Graph shows a shaded region R enclosed by the x-axis, y-axis, the line x = 2, and the curve y=54+2x^3.

Region $R$ is the base of a solid. For the solid, at each $x$ the cross section perpendicular to the $x$ -axis is a rectangle with height $h (x) = 1 + x$ . Find the volume of the solid.

Answer:

Use $Volume = \int_{a}^{b} A (x) d x$

At each $x$ , the cross-sectional area is $A (x) = f (x) \cdot h (x) = (54 - 2 x^{3}) (1 + x)$

$\begin{array}{rcl} Volume & = & \int_{0}^{2} (54 - 2 x^{3}) (1 + x) d x \\ = & \int_{0}^{2} (54 + 54 x - 2 x^{3} - 2 x^{4}) d x \\ = & {[54 x + 27 x^{2} - \frac{1}{2} x^{4} - \frac{2}{5} x^{5}]}_{0}^{2} \\ = & (54 (2) + 27 {(2)}^{2} - \frac{1}{2} {(2)}^{4} - \frac{2}{5} {(2)}^{5}) - (0) \\ = & 108 + 108 - 8 - \frac{64}{5} \\ = & \frac{976}{5} \\ = & 195.2 \end{array}$

The question doesn't specify units, so the units of volume will be ${units}^{3}$

195.2 units³

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