Volumes With Known Cross Sections The Slice Method

The slice method computes a volume by cutting the solid with a family of parallel planes and adding up the volumes of the resulting slabs. If a plane perpendicular to the $x$ -axis meets the solid in a region of area $A (x)$ , then a slab of thickness $d x$ has volume $d V = A (x) d x$ , and the total volume is $\int_{a}^{b} A (x) d x$ . The cross section does not have to be a disk: squares, triangles, semicircles, and trapezoids are all fair game.

Item	Statement
Element of volume	$d V = A (x) d x$
Slicing $⟂$ to the $x$ -axis	$V = \int_{a}^{b} A (x) d x$
Slicing $⟂$ to the $y$ -axis	$V = \int_{c}^{d} A (y) d y$
Disk method as a special case	$A (x) = π {[R (x)]}^{2}$
Square on base $b$	$A = b^{2}$
Equilateral triangle on base $b$	$A = \frac{\sqrt{3}}{4} b^{2}$
Semicircle on diameter $b$	$A = \frac{π}{8} b^{2}$
Isosceles right triangle, legs $b$	$A = \frac{1}{2} b^{2}$
Isosceles right triangle, hypotenuse $b$	$A = \frac{1}{4} b^{2}$
Chord of the circle $x^{2} + y^{2} = a^{2}$	length $2 \sqrt{a^{2} - x^{2}}$
Sphere of radius $R$	$V = \frac{4 π}{3} R^{3}$
Two perpendicular cylinders of radius $a$	$V = \frac{16 a^{3}}{3}$
$45^{\circ}$ wedge from a cylinder of radius $a$	$V = \frac{2 a^{3}}{3}$
Strategy	Draw a typical slice, express $A$ in terms of the slicing variable, integrate over the range of that variable

The Slice Method

The volumes of many solids can be obtained by the application of the slice method. In Section 8.2 we sliced the solid into infinitely many thin disks. However, the element of volume does not have to be a disk (or a washer).

In the general case, imagine the solid is sliced into infinitesimally thin slices of thickness $d x$ by a family of planes perpendicular to the $x$ -axis, as in the figure below. Suppose you know a formula $A (x)$ for the area of the cross section cut by such a plane. Some common cross sections are triangles, squares, rectangles, trapezoids, and semicircles. Then the volume of a slice, the element of volume, is this area multiplied by the thickness of the slice:

d V = A (x) d x .

The total volume of the solid is the sum of the volumes of these slices. If the solid is bounded by two parallel planes perpendicular to the $x$ -axis at $x = a$ and $x = b$ , then

\bbox[8px, #E6F0FA, border: 3px solid #0066CC]{V=\int_{a}^{b}dV=\int_{a}^{b}A(x)\,dx.}

A solid cut by a family of parallel planes perpendicular to the x-axis, between x = a and x = b, with one thin slice of thickness dx whose cross-sectional area is A of x. — Slicing a solid by planes perpendicular to the $x$ -axis. The slice at $x$ has cross-sectional area $A (x)$ and thickness $d x$ , so $d V = A (x) d x$ .

Similarly, suppose the solid is cut by planes perpendicular to the $y$ -axis and $A (y)$ is the area of an arbitrary cross section, as in the next figure. Then

\bbox[8px, #E6F0FA, border: 3px solid #0066CC]{V=\int_{c}^{d}A(y)\,dy}

if the solid is bounded between the planes $y = c$ and $y = d$ (with $c < d$ ).

The same solid cut by a family of parallel planes perpendicular to the y-axis, between y = c and y = d, with one thin slice of thickness dy whose cross-sectional area is A of y. — Slicing by planes perpendicular to the $y$ -axis. Now the element of volume is $d V = A (y) d y$ .

The disk method of Section 8.2 is the special case $A (x) = π {[R (x)]}^{2}$ , and the washer method is the special case $A (x) = π {[R_{out} (x)]}^{2} - π {[R_{in} (x)]}^{2}$ . Nothing new is being assumed here; the slice method simply drops the requirement that the cross section be round.
The slicing variable and the thickness must agree: slices perpendicular to the $x$ -axis have thickness $d x$ and their area must be written as a function of $x$ alone. A leftover $y$ in the integrand is the most common mistake in these problems.

Areas of Common Cross Sections

In most problems the solid sits on a plane region called the base, and each cross section is erected on a chord of that base. The chord length $b$ therefore controls everything, so it pays to know the area of the standard shapes in terms of $b$ .

Cross section erected on a chord of length $b$	Area
Square with side $b$	$b^{2}$
Rectangle of height $h$	$b h$
Equilateral triangle with side $b$	$\frac{\sqrt{3}}{4} b^{2}$
Semicircle with diameter $b$	$\frac{1}{2} π {(\frac{b}{2})}^{2} = \frac{π}{8} b^{2}$
Isosceles right triangle with legs $b$ (one leg on the base)	$\frac{1}{2} b^{2}$
Isosceles right triangle with hypotenuse $b$ (hypotenuse on the base)	$\frac{1}{4} b^{2}$
Isosceles triangle of height $h$	$\frac{1}{2} b h$

Two special cases are worth memorizing, because they account for most textbook problems. If the base is the disk $x^{2} + y^{2} \leq a^{2}$ and the slices are perpendicular to the $x$ -axis, the chord runs from $y = - \sqrt{a^{2} - x^{2}}$ to $y = \sqrt{a^{2} - x^{2}}$ , so

b = 2 \sqrt{a^{2} - x^{2}} .

If the base is the region between $y = f (x)$ and $y = g (x)$ with $f \geq g$ , the chord is vertical and

b = f (x) - g (x) .

For the equilateral triangle, the height is $\frac{\sqrt{3}}{2} b$ by the Pythagorean theorem, so its area is $\frac{1}{2} \cdot b \cdot \frac{\sqrt{3}}{2} b = \frac{\sqrt{3}}{4} b^{2}$ . For the isosceles right triangle with hypotenuse $b$ , each leg is $b / \sqrt{2}$ , so the area is $\frac{1}{2} \cdot \frac{b}{\sqrt{2}} \cdot \frac{b}{\sqrt{2}} = \frac{b^{2}}{4}$ , exactly one quarter of the area of the square on the same chord.

Volume of a Sphere by Slicing

Compute the volume of a sphere with radius $R$ , knowing that the area of a circle with radius $r$ is $π r^{2}$ .

Solution

Let us cut the sphere by planes perpendicular to the $x$ -axis into thin slices with circular cross section of area $A (x)$ , as shown below.

A sphere of radius R sliced by planes perpendicular to the x-axis; the slice at distance x from the centre is a circle whose radius is the square root of R squared minus x squared. — The sphere sliced into thin circular disks. The cross section at $x$ is a circle of radius $\sqrt{R^{2} - x^{2}}$ .

If the distance of the plane from the origin is $x$ , then the radius of the cross section is $\sqrt{R^{2} - x^{2}}$ , because that radius, the distance $x$ , and a radius $R$ of the sphere form a right triangle. Therefore, the area of the cross section is

A (x) = π {(\sqrt{R^{2} - x^{2}})}^{2} = π (R^{2} - x^{2})

and the volume of the slice is

d V = A (x) d x = π (R^{2} - x^{2}) d x .

Because $x$ varies between $- R$ and $R$ , and the integrand is an even function of $x$ , the volume of the sphere is

\begin{aligned} V &= \int_{-R}^{R}\pi\left(R^{2}-x^{2}\right)dx\\ &= 2\int_{0}^{R}\pi\left(R^{2}-x^{2}\right)dx\\ &= 2\pi\left[R^{2}x-\frac{x^{3}}{3}\right]_{0}^{R}\\ &= 2\pi\left(R^{3}-\frac{R^{3}}{3}\right)\\ &= \frac{4\pi}{3}R^{3}. \end{aligned}

Two Intersecting Cylinders

Consider two circular cylinders of radius $a$ , which intersect at right angles as shown in the figure below. Compute the volume of the region common to both cylinders. (This region is called the Steinmetz solid.)

Two circular cylinders of equal radius a whose axes cross at right angles, with the solid region common to both of them at the centre. — Two circular cylinders of radius $a$ intersecting at right angles.

Solution

Take the axes of the cylinders to be the $y$ - and $z$ -axes, so that the two cylinders are $x^{2} + z^{2} \leq a^{2}$ and $x^{2} + y^{2} \leq a^{2}$ . By symmetry it is enough to work with one eighth of the common region, the piece lying in the octant $x, y, z \geq 0$ with its corner at the origin $O$ , shown in the figure below. Let us cut this piece by planes perpendicular to the $x$ -axis into thin slices. The cross section $EFGH$ is a square, so its area is

A (x) = E F \times E H = E F^{2},

since $E F = E H$ by the symmetry between the two cylinders.

One eighth of the region common to the two cylinders, with its corner at the origin O, cut by a plane perpendicular to the x-axis in the square cross section E F G H. — One eighth of the common region, sliced by a plane perpendicular to the $x$ -axis at a distance $O F = x$ from the origin.

The square cross section E F G H seen face on, with the side E F equal to the side E H. — The cross section $EFGH$ is a square, because $E F = E H$ .

If $O F = x$ , then because $O E = a$ is a radius of the cylinder, the right triangle $O F E$ gives

E F = \sqrt{a^{2} - x^{2}} .

Therefore,

A (x) = a^{2} - x^{2}

and the volume of this slice is

d V = A (x) d x = (a^{2} - x^{2}) d x .

The total volume of the common region is eight times the volume of the piece we sliced:

\begin{aligned} V &= 8\int_{0}^{a}dV\\ &= 8\int_{0}^{a}\left(a^{2}-x^{2}\right)dx\\ &= 8\left[a^{2}x-\frac{x^{3}}{3}\right]_{0}^{a}\\ &= 8\times\frac{2a^{3}}{3}\\ &= \frac{16a^{3}}{3}. \end{aligned}

Check. We may also slice the whole solid at once. At the station $x$ , the full cross section is the square $| y | \leq \sqrt{a^{2} - x^{2}}$ , $| z | \leq \sqrt{a^{2} - x^{2}}$ , whose side is $2 \sqrt{a^{2} - x^{2}}$ and whose area is $4 (a^{2} - x^{2})$ . Then

V = \int_{- a}^{a} 4 (a^{2} - x^{2}) d x = 8 \int_{0}^{a} (a^{2} - x^{2}) d x = \frac{16 a^{3}}{3},

the same answer. Notice that no $π$ appears, even though the solid is bounded entirely by round surfaces.

A Wedge Cut From a Cylinder

Consider a right circular cylinder of radius $a$ . Find the volume of the wedge cut from this cylinder by a plane through a diameter of the base of the cylinder making a $45^{\circ}$ angle with the base.

Solution

Method (a): triangular slices. Take the $y$ -axis along the diameter through which the cutting plane passes, and divide the wedge into thin slices of thickness $d y$ by planes perpendicular to the $y$ -axis. The figure below shows one of these slices. As we can see, the sections cut by these planes are right triangles.

A wedge cut from a circular cylinder by a plane through a diameter of the base at 45 degrees, sliced by planes perpendicular to the y-axis so that each cross section is a right triangle. — Method (a): slices perpendicular to the $y$ -axis are right triangles with base $b = \sqrt{a^{2} - y^{2}}$ and height $h = b \tan 45^{\circ} = b$ .

If the slice is located at a distance $y$ from the origin, then the base of the triangle is the half-chord

b = \sqrt{a^{2} - y^{2}},

and its height is

h = b \tan 45^{\circ} = \sqrt{a^{2} - y^{2}} .

Thus

A (y) = \frac{1}{2} b h = \frac{1}{2} (a^{2} - y^{2})

and the volume of the infinitely thin slice is

d V = A (y) d y = \frac{1}{2} (a^{2} - y^{2}) d y .

Because $y$ varies between $- a$ and $a$ , and the integrand is even, the total volume is

\begin{aligned} V &= \int_{-a}^{a}\frac{1}{2}\left(a^{2}-y^{2}\right)dy\\ &= 2\int_{0}^{a}\frac{1}{2}\left(a^{2}-y^{2}\right)dy\\ &= \int_{0}^{a}\left(a^{2}-y^{2}\right)dy\\ &= \left[a^{2}y-\frac{y^{3}}{3}\right]_{0}^{a}\\ &= a^{3}-\frac{a^{3}}{3}=\frac{2a^{3}}{3}. \end{aligned}

The factor $2$ coming from the symmetry cancels the $\frac{1}{2}$ in the area, which is why the bracket carries no coefficient in front of it.

Method (b): rectangular slices. We can instead cut the wedge into thin slices of rectangular cross section with planes parallel to the sharp edge of the wedge, that is, parallel to the diameter. Measure $x$ perpendicular to that diameter, as in the figure below.

The same wedge sliced by planes parallel to its sharp edge, so that each cross section is a rectangle whose width is the chord of the base circle and whose height is the distance from the diameter. — Method (b): slices parallel to the edge of the wedge are rectangles of width $2 \sqrt{a^{2} - x^{2}}$ and height $x$ .

The width of the rectangle is the full chord $2 \sqrt{a^{2} - x^{2}}$ , and its height is $x \tan 45^{\circ} = x$ , because the cutting plane rises one unit for every unit of horizontal distance from the diameter. Therefore, the area of the rectangle is

A (x) = 2 x \sqrt{a^{2} - x^{2}},

and the volume of the slice is

d V = A (x) d x = 2 x \sqrt{a^{2} - x^{2}} d x .

Because $x$ varies between $0$ and $a$ , the total volume of the wedge is

V = \int_{0}^{a} d V = \int_{0}^{a} 2 x \sqrt{a^{2} - x^{2}} d x .

To evaluate the above integral, let $u = a^{2} - x^{2}$ . Then

d u = - 2 x d x,

x = 0 \Leftrightarrow u = a^{2}, x = a \Leftrightarrow u = 0,

and

\begin{aligned} V &= \int_{0}^{a}\underbrace{\sqrt{a^{2}-x^{2}}}_{\sqrt{u}}\ \underbrace{2x\,dx}_{-du}\\ &= -\int_{a^{2}}^{0}\sqrt{u}\,du\\ &= \int_{0}^{a^{2}}\sqrt{u}\,du\\ &= \left.\frac{2}{3}u^{3/2}\right|_{0}^{a^{2}}\\ &= \frac{2}{3}\left(a^{2}\right)^{3/2}=\frac{2a^{3}}{3}, \end{aligned}

as before. The two methods slice the same solid in perpendicular directions and, as they must, they agree.

Setting Up a Slice Problem

Sketch the solid, or at least its base, and draw one typical slice.
Decide which axis the slices are perpendicular to. That choice fixes the variable of integration and the thickness, 
  d
  x
 or 
  d
  y
.
Find the length of the chord of the base on which the cross section stands, in terms of the slicing variable alone.
Convert that chord length into the area 
  A
 using the shape of the cross section (square, triangle, semicircle, and so on).
Integrate 
  A
 between the two extreme values of the slicing variable, and use symmetry whenever the integrand is even.

Exercises

A solid lies between the planes $x = 0$ and $x = 3$ . Every cross section perpendicular to the $x$ -axis is a square of side $x + 1$ . Find the volume.

$TikZ figure$

Solution

The area of the cross section is $A (x) = (x + 1)^{2}$ , so

\begin{aligned} V &= \int_{0}^{3}(x+1)^{2}\,dx\\ &= \left[\frac{(x+1)^{3}}{3}\right]_{0}^{3}\\ &= \frac{64}{3}-\frac{1}{3}=21. \end{aligned}

The base of a solid is the disk $x^{2} + y^{2} \leq 4$ . Cross sections perpendicular to the $x$ -axis are squares. Find the volume.

Solution

The chord of the base at the station $x$ runs from $y = - \sqrt{4 - x^{2}}$ to $y = \sqrt{4 - x^{2}}$ , so the side of the square is $s = 2 \sqrt{4 - x^{2}}$ and

A (x) = s^{2} = 4 (4 - x^{2}) .

Since the integrand is even,

\begin{aligned} V &= \int_{-2}^{2}4\left(4-x^{2}\right)dx\\ &= 8\int_{0}^{2}\left(4-x^{2}\right)dx\\ &= 8\left[4x-\frac{x^{3}}{3}\right]_{0}^{2}\\ &= 8\left(8-\frac{8}{3}\right)=8\cdot\frac{16}{3}=\frac{128}{3}. \end{aligned}

This is the Steinmetz formula $16 a^{3} / 3$ with $a = 2$ , as it must be: the solid built from squares on a disk of radius $a$ is exactly the region common to two perpendicular cylinders of radius $a$ .

The base of a solid is the disk $x^{2} + y^{2} \leq 9$ . Cross sections perpendicular to the $x$ -axis are semicircles whose diameters lie in the base. Find the volume.

Solution

The diameter of the semicircle is the chord $b = 2 \sqrt{9 - x^{2}}$ , so its radius is $r = \sqrt{9 - x^{2}}$ and

A (x) = \frac{1}{2} π r^{2} = \frac{π}{2} (9 - x^{2}) .

Therefore

\begin{aligned} V &= \frac{\pi}{2}\int_{-3}^{3}\left(9-x^{2}\right)dx\\ &= \pi\int_{0}^{3}\left(9-x^{2}\right)dx\\ &= \pi\left[9x-\frac{x^{3}}{3}\right]_{0}^{3}\\ &= \pi\left(27-9\right)=18\pi. \end{aligned}

The base of a solid is the disk $x^{2} + y^{2} \leq 1$ . Cross sections perpendicular to the $x$ -axis are equilateral triangles. Find the volume.

Solution

The side of the triangle is the chord $b = 2 \sqrt{1 - x^{2}}$ , so

A (x) = \frac{\sqrt{3}}{4} b^{2} = \frac{\sqrt{3}}{4} \cdot 4 (1 - x^{2}) = \sqrt{3} (1 - x^{2}) .

Hence

\begin{aligned} V &= \sqrt{3}\int_{-1}^{1}\left(1-x^{2}\right)dx\\ &= 2\sqrt{3}\int_{0}^{1}\left(1-x^{2}\right)dx\\ &= 2\sqrt{3}\left[x-\frac{x^{3}}{3}\right]_{0}^{1}\\ &= 2\sqrt{3}\cdot\frac{2}{3}=\frac{4\sqrt{3}}{3}. \end{aligned}

The base of a solid is the region bounded by the parabola $y = x^{2}$ and the line $y = 4$ . Cross sections perpendicular to the $y$ -axis are squares. Find the volume.

Solution

Because the slices are perpendicular to the $y$ -axis, everything must be written in terms of $y$ . At height $y$ the region stretches from $x = - \sqrt{y}$ to $x = \sqrt{y}$ , so the side of the square is $s = 2 \sqrt{y}$ and

A (y) = {(2 \sqrt{y})}^{2} = 4 y .

The solid lies between $y = 0$ and $y = 4$ , so

V = \int_{0}^{4} 4 y d y = {[2 y^{2}]}_{0}^{4} = 32.

The base of a solid is the region bounded by $y = 1 - x^{2}$ and the $x$ -axis. Cross sections perpendicular to the $x$ -axis are isosceles right triangles with one leg in the base. Find the volume.

Solution

The leg lying in the base has length $b = 1 - x^{2}$ , and for an isosceles right triangle with legs $b$ the area is $\frac{1}{2} b^{2}$ . So

A (x) = \frac{1}{2} {(1 - x^{2})}^{2} = \frac{1}{2} (1 - 2 x^{2} + x^{4}) .

The integrand is even, so

\begin{aligned} V &= \frac{1}{2}\int_{-1}^{1}\left(1-2x^{2}+x^{4}\right)dx\\ &= \int_{0}^{1}\left(1-2x^{2}+x^{4}\right)dx\\ &= \left[x-\frac{2x^{3}}{3}+\frac{x^{5}}{5}\right]_{0}^{1}\\ &= 1-\frac{2}{3}+\frac{1}{5}=\frac{15-10+3}{15}=\frac{8}{15}. \end{aligned}

The base of a solid is the region bounded by $y = \sqrt{x}$ , the $x$ -axis, and the line $x = 4$ . Cross sections perpendicular to the $x$ -axis are semicircles with diameter in the base. Find the volume.

Solution

Here the chord runs from $y = 0$ up to $y = \sqrt{x}$ , so the diameter is $b = \sqrt{x}$ and the radius is $r = \frac{1}{2} \sqrt{x}$ . Then

A (x) = \frac{1}{2} π r^{2} = \frac{π}{2} \cdot \frac{x}{4} = \frac{π x}{8} .

Therefore

V = \int_{0}^{4} \frac{π x}{8} d x = \frac{π}{8} {[\frac{x^{2}}{2}]}_{0}^{4} = \frac{π}{8} \cdot 8 = π .

The base of a solid is the region bounded by $y = x$ and $y = x^{2}$ . Cross sections perpendicular to the $x$ -axis are squares. Find the volume.

Solution

The curves meet where $x = x^{2}$ , that is at $x = 0$ and $x = 1$ , and on that interval $x \geq x^{2}$ . The side of the square is $s = x - x^{2}$ , so

A (x) = {(x - x^{2})}^{2} = x^{2} - 2 x^{3} + x^{4} .

Hence

\begin{aligned} V &= \int_{0}^{1}\left(x^{2}-2x^{3}+x^{4}\right)dx\\ &= \left[\frac{x^{3}}{3}-\frac{x^{4}}{2}+\frac{x^{5}}{5}\right]_{0}^{1}\\ &= \frac{1}{3}-\frac{1}{2}+\frac{1}{5}=\frac{10-15+6}{30}=\frac{1}{30}. \end{aligned}

Use the slice method to show that a pyramid with a square base of side $b$ and height $h$ has volume $\frac{1}{3} b^{2} h$ . Then adapt the argument to a right circular cone of radius $r$ and height $h$ .

Solution

Measure $y$ downward from the apex and slice with planes parallel to the base. The slice at distance $y$ from the apex is a square whose side $s$ satisfies the similar-triangle relation

\frac{s}{b} = \frac{y}{h} ⟹ s = \frac{b}{h} y .

Therefore $A (y) = s^{2} = \frac{b^{2}}{h^{2}} y^{2}$ , and

\begin{aligned} V &= \int_{0}^{h}\frac{b^{2}}{h^{2}}y^{2}\,dy\\ &= \frac{b^{2}}{h^{2}}\left[\frac{y^{3}}{3}\right]_{0}^{h}\\ &= \frac{b^{2}}{h^{2}}\cdot\frac{h^{3}}{3}=\frac{1}{3}b^{2}h. \end{aligned}

For the cone, the same similar triangles give a circular slice of radius $ρ = \frac{r}{h} y$ , so $A (y) = π \frac{r^{2}}{h^{2}} y^{2}$ and

V = \frac{π r^{2}}{h^{2}} \int_{0}^{h} y^{2} d y = \frac{1}{3} π r^{2} h .

In both cases the volume is one third of the base area times the height, which is what the factor $\int_{0}^{h} y^{2} d y = h^{3} / 3$ is really saying.

The base of a solid is the region between $y = \sin x$ and the $x$ -axis for $0 \leq x \leq π$ . Cross sections perpendicular to the $x$ -axis are equilateral triangles. Find the volume.

Solution

The side of the triangle is the vertical chord $b = \sin x$ , so

A (x) = \frac{\sqrt{3}}{4} \sin^{2} x .

Using the half-angle identity $\sin^{2} x = \frac{1 - \cos 2 x}{2}$ ,

\begin{aligned} V &= \frac{\sqrt{3}}{4}\int_{0}^{\pi}\sin^{2}x\,dx\\ &= \frac{\sqrt{3}}{8}\int_{0}^{\pi}\left(1-\cos 2x\right)dx\\ &= \frac{\sqrt{3}}{8}\left[x-\frac{\sin 2x}{2}\right]_{0}^{\pi}\\ &= \frac{\sqrt{3}}{8}\left(\pi-0\right)=\frac{\sqrt{3}\,\pi}{8}. \end{aligned}

Note that $\sin 2 π = \sin 0 = 0$ , so only the $x$ term survives, and $\int_{0}^{π} \sin^{2} x d x = π / 2$ , as one expects from the fact that $\sin^{2}$ averages $\frac{1}{2}$ over a half period.

A plane through a diameter of the base of a right circular cylinder of radius $a$ makes an angle $θ$ with the base, where $0 < θ < \frac{π}{2}$ . Find the volume of the wedge it cuts off, using both triangular and rectangular slices, and check that $θ = 45^{\circ}$ recovers the value found in this section.

Solution

Triangular slices. Put the $y$ -axis along the diameter. At the station $y$ the triangle has base $b = \sqrt{a^{2} - y^{2}}$ and height $h = b \tan θ$ , so

A (y) = \frac{1}{2} b^{2} \tan θ = \frac{1}{2} (a^{2} - y^{2}) \tan θ .

Hence

\begin{aligned} V &= \frac{\tan\theta}{2}\int_{-a}^{a}\left(a^{2}-y^{2}\right)dy\\ &= \tan\theta\int_{0}^{a}\left(a^{2}-y^{2}\right)dy\\ &= \tan\theta\left[a^{2}y-\frac{y^{3}}{3}\right]_{0}^{a}=\frac{2a^{3}}{3}\tan\theta. \end{aligned}

Rectangular slices. Measuring $x$ perpendicular to the diameter, the rectangle has width $2 \sqrt{a^{2} - x^{2}}$ and height $x \tan θ$ , so $A (x) = 2 x \tan θ \sqrt{a^{2} - x^{2}}$ and, with $u = a^{2} - x^{2}$ and $d u = - 2 x d x$ ,

\begin{aligned} V &= \tan\theta\int_{0}^{a}2x\sqrt{a^{2}-x^{2}}\,dx\\ &= \tan\theta\int_{0}^{a^{2}}\sqrt{u}\,du\\ &= \tan\theta\cdot\frac{2}{3}\left(a^{2}\right)^{3/2}=\frac{2a^{3}}{3}\tan\theta. \end{aligned}

Both methods agree. When $θ = 45^{\circ}$ we have $\tan θ = 1$ and $V = \frac{2 a^{3}}{3}$ , the value obtained in the worked example. As $θ \to \frac{π}{2}$ the volume grows without bound, which is correct: the cutting plane becomes vertical and the wedge climbs an infinitely tall cylinder.

The base of a solid is the ellipse $\frac{x^{2}}{a^{2}} + \frac{y^{2}}{b^{2}} = 1$ . Cross sections perpendicular to the $x$ -axis are squares. Find the volume, and explain what happens when $b = a$ .

Solution

Solving the equation of the ellipse for $y$ gives $y = \pm b \sqrt{1 - \frac{x^{2}}{a^{2}}}$ , so the chord at the station $x$ has length

s = 2 b \sqrt{1 - \frac{x^{2}}{a^{2}}}

and the square erected on it has area

A (x) = s^{2} = 4 b^{2} (1 - \frac{x^{2}}{a^{2}}) .

Since the integrand is even and $x$ runs from $- a$ to $a$ ,

\begin{aligned} V &= 4b^{2}\int_{-a}^{a}\left(1-\frac{x^{2}}{a^{2}}\right)dx\\ &= 8b^{2}\int_{0}^{a}\left(1-\frac{x^{2}}{a^{2}}\right)dx\\ &= 8b^{2}\left[x-\frac{x^{3}}{3a^{2}}\right]_{0}^{a}\\ &= 8b^{2}\left(a-\frac{a}{3}\right)=\frac{16ab^{2}}{3}. \end{aligned}

If $b = a$ the base is a disk of radius $a$ and the formula becomes $\frac{16 a^{3}}{3}$ , the volume of the Steinmetz solid computed earlier in this section. That is no coincidence: erecting squares on the chords of a disk of radius $a$ produces exactly the region common to two perpendicular cylinders of radius $a$ .

The Slice Method

Areas of Common Cross Sections

Volume of a Sphere by Slicing

Two Intersecting Cylinders

A Wedge Cut From a Cylinder

Setting Up a Slice Problem

Exercises

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