Volumes with Cross Sections: Squares and Rectangles

AP Calculus · Applications of Integration

Volumes with Cross Sections: Squares & Rectangles

The base supplies one number per slice. The named shape turns it into an area.

key idea$V=\int_a^b A(x)\,dx\qquad A=s^{2}\ \text{(square)}\quad A=sh\ \text{(rectangle)}$

Picture the region

The solid is not a shape you have to visualise in three dimensions. It is a stack of flat pieces, each standing on a slice of a flat base region. Find the side length $s$ from the base, square it, and integrate.

the base region supplies $s$

square cross section: $A=s^{2}$

rectangle of height $h$: $A=sh$

The base region contributes only one number to each slice: the length $s$ of the segment it cuts off. A square cross section built on it has area $s^{2}$; a rectangle of height $h$ has area $sh$. Everything else about the picture is decoration.

[Figure: Not a solid of revolution. The base region lies flat and each cross section stands perpendicular to it — here a square whose side is the vertical extent s(x).]

Not a solid of revolution. The base region lies flat and each cross section stands perpendicular to it — here a square whose side is the vertical extent $s(x)$.

Setting up the integral

From base region to volume

Read the direction of the slices. ‘Perpendicular to the $x$-axis’ means vertical slices and a $dx$ integral; perpendicular to the $y$-axis means $dy$.
Write $s$ as a function of that variable. If the base sits between two curves, $s=\text{top}-\text{bottom}$; if it sits between a curve and the axis, $s=f(x)$.
Substitute $s$ into the area formula for the named shape: $s^{2}$ for a square, $sh$ for a rectangle.
Set the limits from the base region — usually where the boundary curves meet.
Integrate $A$ over those limits. The answer is a volume, in cubic units.

Base description	$s$ equals	Square gives	Rectangle of height $h$
under $y=f(x)$, above the axis	$f(x)$	$\displaystyle\int f^{2}dx$	$\displaystyle\int f\,h\,dx$
between $y=f$ and $y=g$	$f-g$	$\displaystyle\int(f-g)^{2}dx$	$\displaystyle\int(f-g)h\,dx$
left of $x=g(y)$, right of the $y$-axis	$g(y)$	$\displaystyle\int g^{2}dy$	$\displaystyle\int g\,h\,dy$
a circle of radius $r$, slices $\perp$ the $x$-axis	$2\sqrt{r^{2}-x^{2}}$	$\displaystyle\int 4\left(r^{2}-x^{2}\right)dx$	—

Row 4 is the one to memorise: a full chord of a circle is twice the semicircle's height. Half the students who miss this problem use $\sqrt{r^{2}-x^{2}}$ and land on exactly a quarter of the right volume.

Worked examples

EX 1

Square cross sections

base under a parabola

The base is the region under $y=4-x^{2}$ from $x=0$ to $x=2$. Cross sections perpendicular to the $x$-axis are squares. Find the volume.

Identify $s$

The slice runs from the axis up to the curve, so $s=4-x^{2}$.

Write the area

$A(x)=s^{2}=\left(4-x^{2}\right)^{2}=16-8x^{2}+x^{4}$.

Set up the integral

$V=\displaystyle\int_{0}^{2}\left(16-8x^{2}+x^{4}\right)dx$.

Antidifferentiate

$\left[16x-\dfrac{8x^{3}}{3}+\dfrac{x^{5}}{5}\right]_{0}^{2}$.

Evaluate

$32-\dfrac{64}{3}+\dfrac{32}{5}=\dfrac{480-320+96}{15}=\dfrac{256}{15}$.

EX 2

Rectangles of fixed height

$A=sh$

The base is bounded by $y=\sqrt{x}$, the $x$-axis and $x=4$. Cross sections perpendicular to the $x$-axis are rectangles of height $3$. Find the volume.

Identify $s$

$s=\sqrt{x}$, the height of the base region at that $x$.

Write the area

$A(x)=s\cdot h=3\sqrt{x}$.

Set up

$V=\displaystyle\int_{0}^{4}3\sqrt{x}\,dx=3\int_{0}^{4}x^{1/2}dx$.

Integrate

$3\left[\dfrac{2}{3}x^{3/2}\right]_{0}^{4}=2\left[x^{3/2}\right]_{0}^{4}$.

Evaluate

$2(8)=16$. Note that $s$ came from the base and $h$ was given — two different roles.

EX 3

Height proportional to the base

$h=\tfrac12 s$

The base is bounded by $y=x$ and $y=x^{2}$. Cross sections perpendicular to the $x$-axis are rectangles whose height is half the base. Find the volume.

Find the limits

$x=x^{2}$ gives $x=0$ and $x=1$.

Identify $s$

On $(0,1)$, $x\gt x^{2}$, so $s=x-x^{2}$.

Write the area

$h=\dfrac{s}{2}$, so $A=s\cdot\dfrac{s}{2}=\dfrac{s^{2}}{2}=\dfrac{\left(x-x^{2}\right)^{2}}{2}$.

Expand and integrate

$\dfrac12\displaystyle\int_{0}^{1}\left(x^{2}-2x^{3}+x^{4}\right)dx=\dfrac12\left[\dfrac{x^{3}}{3}-\dfrac{x^{4}}{2}+\dfrac{x^{5}}{5}\right]_{0}^{1}$.

Evaluate

$\dfrac12\left(\dfrac13-\dfrac12+\dfrac15\right)=\dfrac12\cdot\dfrac{1}{30}=\dfrac{1}{60}$.

EX 4

A circular base

the full chord

The base is the circle $x^{2}+y^{2}=9$. Cross sections perpendicular to the $x$-axis are squares. Find the volume.

Find the chord length

At a given $x$ the circle runs from $y=-\sqrt{9-x^{2}}$ to $y=+\sqrt{9-x^{2}}$.

Write $s$

$s=2\sqrt{9-x^{2}}$ — the full chord, top minus bottom.

Square it

$A=s^{2}=4\left(9-x^{2}\right)$.

Integrate over the whole circle

$V=\displaystyle\int_{-3}^{3}4\left(9-x^{2}\right)dx=4\left[9x-\dfrac{x^{3}}{3}\right]_{-3}^{3}$.

Evaluate

$4\left(18-(-18)\right)=144$. Using only the half-chord would have given $36$ — exactly a quarter, because the error gets squared.

Watch out

$s$comes from the base region

$A(x)$comes from the named shape

Full chord$\text{top}-\text{bottom}$, not just the top

$s$ is the whole slice, not half of it

For a base symmetric about the $x$-axis the slice runs from $-f$ to $+f$, giving $s=2f$. Squaring turns a factor of $2$ into a factor of $4$.

No $\pi$ appears anywhere

These are squares and rectangles, not discs. A $\pi$ in the answer means the cross-section formula was the wrong one.

Slices perpendicular to the $y$-axis need $dy$

Rewrite the boundaries as functions of $y$ and use $y$-limits. Mixing the slice direction with the wrong differential is a set-up error, not an arithmetic one.

The height of a rectangle is given separately

$s$ comes from the base; $h$ comes from the problem statement. If $h$ is described in terms of $s$, substitute it before integrating.

Course map. Topic 8.7 · LO CHA-5.B — calculate volumes of solids with known cross sections using definite integrals.