Calculus 1

Volumes with known cross sections

A solid is not a prism, a pyramid or a ball, but every slice of it is a shape whose area you can work out. Is that enough to find its volume?

You need

The definite integral as a limit of Riemann sumsCalculus 1 Cross sections of solidsMath 7

Where you are 10 in Applications of integrals

Calculus8

Applications of integrals10

The shell method

Volumes with known cross sections

Work as an integral Accumulation and net change3 Area between curves ▾ Show all 10 (5 more)

When you reach for it

Geometry gives the volume of a prism, a cylinder, a pyramid, a cone and a ball. Many solids are none of these: a block whose top follows a curve, a stone that tapers unevenly. What such a solid often does have is cross sections you can describe. A cross section is the flat shape a plane leaves when it cuts a solid. If every cut perpendicular to one line is a square, a semicircle or a triangle whose size is known at each position, the volume is an integral.

You reach for this method when the area of each slice is known as a function of position, either directly, from a table of measurements, or because each slice is a known shape standing on a base region drawn in the plane.

The procedure

Day 1: volume as the integral of slice area

Slicing. Place an x-axis so that the solid lies between the planes x = a and x = b, and let A(x) be the area of the cross section in the plane perpendicular to the axis at x. Cut the solid at a = x_0 < x_1 < \dots < x_n = b. The slab between x_{i-1} and x_i has thickness \Delta x_i = x_i - x_{i-1}. If it is thin, its cross sections hardly change across it, so it is nearly a prism with base area A(x_i^*), for a sample point x_i^* in the piece, and height \Delta x_i. A prism’s volume is base area times height, so

V \approx \sum_{i=1}^{n} A(x_i^*)\,\Delta x_i.

This is a Riemann sum for A on [a, b]. When A is continuous, these sums approach the integral of A as the pieces get thinner (The definite integral as a limit of Riemann sums), and that limit is taken as the definition of the volume:

V = \int_a^b A(x)\,dx.

The slab picture is the reason for the definition, not a proof of it. It agrees with Geometry where Geometry has an answer: a prism has the same cross section all along, A(x) = B, and \int_a^b B\,dx = B(b - a), base area times height.

Cavalieri’s principle. If two solids lie between the same two planes and their cross sections at every x have equal areas, they have the same A(x) and so the same volume. Geometry uses this as Cavalieri's principle to Why the volume formulas hold: dissection, slicing and Cavalieri; here it is a one-line consequence of the formula.

Checking the formula on a pyramid or cone. Take a pyramid or cone with base area B and height h, measure x from the apex along a line perpendicular to the base, so the base lies in the plane x = h. Each point of the base is joined to the apex by a segment, and the plane at x cuts every such segment a fraction \displaystyle \frac{x}{h} of the way from the apex. So the slice at x is the base shrunk by the factor \displaystyle \frac{x}{h}. Lengths scale by \displaystyle \frac{x}{h}, areas by its square, and \displaystyle A(x) = \frac{B}{h^2}x^2. Then

\begin{aligned} V &= \int_0^h \frac{B}{h^2}x^2\,dx \\ &= \frac{B}{h^2}\cdot\frac{h^3}{3} = \frac{1}{3}Bh, \end{aligned}

the formula from Geometry, for every base shape at once, and for a leaning pyramid as well as an upright one.

Squares on a base region. Often a solid is described by its base, a region in the xy-plane, together with the shape of its slices: for instance, every cross section perpendicular to the x-axis is a square with one side lying across the region. If the region runs from x = a to x = b between a lower curve y = g(x) and an upper curve y = f(x), the plane at x crosses it in a segment of length

s(x) = f(x) - g(x),

and the square on that segment has area s(x)^2. So

V = \int_a^b \big(f(x) - g(x)\big)^2\,dx.

Under one curve and above the x-axis, g(x) = 0 and s(x) = f(x). The thinking is in reading the side off the base; after that the volume is an ordinary definite integral (The second fundamental theorem of calculus).

Figure 1The base region of Example 2 with one square cross section standing on it. The side of the square is the width of the region at that x.

Day 2: other shapes, other directions, and tables

Other shapes. A slice can be any shape built on the segment of length s:

Slice on a side of length s	Area
square	s^2
equilateral triangle	\displaystyle \frac{\sqrt{3}}{4}s^2
semicircle, diameter s	\displaystyle \frac{\pi}{8}s^2
isosceles right triangle, leg on the base	\displaystyle \frac{1}{2}s^2
isosceles right triangle, hypotenuse on the base	\displaystyle \frac{1}{4}s^2
rectangle of height h	hs

An equilateral triangle of side s has height \displaystyle \frac{\sqrt{3}}{2}s, by the Pythagorean theorem on half of it, so its area is \displaystyle \frac{1}{2}\cdot s\cdot\frac{\sqrt{3}}{2}s. A semicircle of diameter s is half a circle of radius \displaystyle \frac{s}{2}, so its area is \displaystyle \frac{1}{2}\pi\frac{s^2}{4}. An isosceles right triangle with a leg s on the base has its other leg s too. With the hypotenuse s on the base, the altitude to the hypotenuse splits it into two smaller isosceles right triangles, so the altitude is \displaystyle \frac{s}{2} and the area is \displaystyle \frac{1}{2}\cdot s\cdot\frac{s}{2}. For this shape a problem must say which side lies on the base: the two readings differ by a factor of 2.

Figure 2Cross sections built on a side of length s, with their areas.

Every shape in the table except the rectangle is one fixed shape scaled to size s, so its area is ks^2 for a constant k, and

V = k\int_a^b s(x)^2\,dx.

One integral then serves for every such shape. A rectangle of fixed height h is not scaled with s, and its area grows like s, not s^2.

Slices perpendicular to the y-axis. If the slices are perpendicular to the y-axis, the slice at height y crosses the base in a horizontal segment from a left boundary x = p(y) to a right boundary x = q(y), so s(y) = q(y) - p(y) and

V = \int_c^d A(y)\,dy,

where the base runs from y = c to y = d. A boundary given as y = \dots must be solved for x. Changing the direction changes the solid, not just the arithmetic, since the slices stand on different segments.

A table of slice areas. When A is known only at measured positions, the integral can only be estimated, by a Riemann sum built from the table: a left sum, a right sum, the average of the two, or a midpoint sum when measurements sit at the midpoints. If A only decreases (or only increases) along the solid, the volume of each slab, \int A(x)\,dx over its piece, lies between its two end areas times \Delta x, so the left and right sums bracket the true volume.

Solids of revolution have circular slices, A(x) = \pi r(x)^2; that case is the The disk method, and with a hole the The washer method.

Worked examples

Example 1

Day 1: Mosese checks the formula on a pyramid

The base of a solid is the triangle in the xy-plane bounded by the x-axis, the y-axis and the line x + y = 4. Every cross section perpendicular to the x-axis is a square. Find the volume, and check it the way Mosese does, against \displaystyle \frac{1}{3}Bh.

Solution

The line is y = 4 - x, and the triangle runs from x = 0 to x = 4. The plane at x crosses it from y = 0 up to y = 4 - x, so s(x) = 4 - x and A(x) = (4 - x)^2 = 16 - 8x + x^2. Then

\begin{aligned} V &= \int_0^4 \left(16 - 8x + x^2\right)dx \\ &= \left(16x - 4x^2 + \frac{x^3}{3}\right)\Big|_0^4 \\ &= 64 - 64 + \frac{64}{3} = \frac{64}{3}. \end{aligned}

Mosese’s check: the slice at x = 0 is a 4 by 4 square in the plane x = 0, and the slices shrink steadily to the single point (4, 0, 0). The solid is a pyramid with square base of area B = 16 and apex at distance h = 4 from the base plane. It leans, since the apex is over a corner of the base, but \displaystyle \frac{1}{3}Bh = \frac{1}{3}\cdot 16\cdot 4 = \frac{64}{3} all the same.

Example 2

Day 1: squares between two curves

The base of a solid is the region between y = 3x and y = x^2. Cross sections perpendicular to the x-axis are squares. Find the volume.

Solution

The curves meet where x^2 = 3x, so at x = 0 and x = 3. At x = 1 the line gives 3 and the parabola 1, so the line is on top across (0, 3), and

\begin{aligned} s(x) &= 3x - x^2, \\ A(x) &= (3x - x^2)^2 = 9x^2 - 6x^3 + x^4. \end{aligned}

Expanding first lets the power rule handle each term:

\begin{aligned} V &= \int_0^3 \left(9x^2 - 6x^3 + x^4\right)dx \\ &= \left(3x^3 - \frac{3}{2}x^4 + \frac{x^5}{5}\right)\Big|_0^3 \\ &= 81 - \frac{243}{2} + \frac{243}{5} = \frac{81}{10}. \end{aligned}

A rough check: the widest slice is at \displaystyle x = \frac{3}{2}, where \displaystyle s = \frac{9}{4} and \displaystyle A = \frac{81}{16}, so the volume is at most \displaystyle 3\cdot\frac{81}{16} \approx 15.2. The answer 8.1 is under that, as it must be.

Example 3

Day 1: a volume easier than its base

The base of a solid is the region under \displaystyle y = \sqrt{\sin x} and above the x-axis for 0 \le x \le \pi. Cross sections perpendicular to the x-axis are squares. Find the volume.

Solution

The side is \displaystyle s(x) = \sqrt{\sin x}, which is real because \sin x \ge 0 on [0, \pi]. Squaring removes the root: A(x) = \sin x. So

V = \int_0^\pi \sin x\,dx = -\cos x\,\Big|_0^\pi = 1 - (-1) = 2.

The area of the base itself, \displaystyle \int_0^\pi \sqrt{\sin x}\,dx, has no antiderivative built from the usual functions (stated without proof); the volume is the easier integral because the slice area squares the width.

Example 4

Day 2: semicircles between two parabolas

The base of a solid is the region between y = x^2 and y = 8 - x^2. Cross sections perpendicular to the x-axis are semicircles whose diameters lie across the base. Find the volume.

Solution

The curves meet where x^2 = 8 - x^2, at x = \pm 2, and 8 - x^2 is on top between them. So

\begin{aligned} s(x) &= 8 - 2x^2 = 2(4 - x^2), \\ A(x) &= \frac{\pi}{8}\cdot 4(4 - x^2)^2 = \frac{\pi}{2}(4 - x^2)^2. \end{aligned}

A is even, so the integral over [-2, 2] is twice the integral over [0, 2]:

\begin{aligned} V &= \pi\int_0^2 \left(16 - 8x^2 + x^4\right)dx \\ &= \pi\left(32 - \frac{64}{3} + \frac{32}{5}\right) = \frac{256\pi}{15} \approx 53.6. \end{aligned}

Example 5

Day 2: one base, six solids

The base is the region under \displaystyle y = \sqrt{x} and above the x-axis for 0 \le x \le 4, with slices perpendicular to the x-axis. Find the volume when the slices are each shape in the table.

Solution

Here \displaystyle s(x) = \sqrt{x}. For every shape with area ks^2 the one integral needed is

\int_0^4 s(x)^2\,dx = \int_0^4 x\,dx = 8.

Multiplying by k: squares give 8; equilateral triangles \displaystyle \frac{\sqrt{3}}{4}\cdot 8 = 2\sqrt{3}; semicircles \displaystyle \frac{\pi}{8}\cdot 8 = \pi; isosceles right triangles with a leg on the base 4, and with the hypotenuse on the base 2. Rectangles of height 3 have area \displaystyle 3\sqrt{x}, a different integral:

V = \int_0^4 3\sqrt{x}\,dx = 2x^{3/2}\,\Big|_0^4 = 16.

Example 6

Day 2: Eseta and Viliame slice the same base two ways

The base is the region between y = x^2 and the line y = 9. The cross sections are equilateral triangles. Eseta takes them perpendicular to the x-axis, Viliame perpendicular to the y-axis. Find both volumes.

Solution

Eseta: the region runs from x = -3 to x = 3, and the vertical segment at x goes from the parabola up to the line, s(x) = 9 - x^2. Using that A is even,

\begin{aligned} V &= \frac{\sqrt{3}}{4}\cdot 2\int_0^3 \left(81 - 18x^2 + x^4\right)dx \\ &= \frac{\sqrt{3}}{2}\left(243 - 162 + \frac{243}{5}\right) \\ &= \frac{324\sqrt{3}}{5} \approx 112.2. \end{aligned}

Viliame: the region runs from y = 0 to y = 9. Solving y = x^2 for x gives the two branches \displaystyle x = -\sqrt{y} and \displaystyle x = \sqrt{y}, so \displaystyle s(y) = 2\sqrt{y} and \displaystyle A(y) = \frac{\sqrt{3}}{4}\cdot 4y = \sqrt{3}\,y. Then

\begin{aligned} V &= \int_0^9 \sqrt{3}\,y\,dy = \frac{\sqrt{3}}{2}y^2\,\Big|_0^9 \\ &= \frac{81\sqrt{3}}{2} \approx 70.1. \end{aligned}

Both are right. Eseta’s largest triangle stands on the segment of length 9 through the middle; Viliame’s largest stands on the top edge, of length 6. Same base, two different solids.

Figure 3The base of Example 6. A slice perpendicular to the x-axis stands on a vertical segment; a slice perpendicular to the y-axis stands on a horizontal one.

Example 7

Day 2: Viliame estimates the volume of an obelisk

Viliame measures the area of the horizontal cross section of a stone obelisk at five heights.

height x (m)	area A(x) (m^2)
0	1.44
1.5	1.12
3	0.85
4.5	0.61
6	0.40

Estimate the volume with a left sum, a right sum and their average. If the stone narrows steadily upward, what is certain?

Solution

With \Delta x = 1.5, the left sum uses the first four areas and the right sum the last four:

\begin{aligned} L &= 1.5(1.44 + 1.12 + 0.85 + 0.61) = 6.03, \\ R &= 1.5(1.12 + 0.85 + 0.61 + 0.40) = 4.47. \end{aligned}

Their average is 5.25 cubic meters. If A decreases throughout, then on each piece A(x) lies between its value at the top and at the bottom, so each slab’s volume lies between those areas times 1.5, and 4.47 \le V \le 6.03. The midpoint sum on two pieces of thickness 3, using the heights 1.5 and 4.5, gives 3(1.12 + 0.61) = 5.19, another estimate inside that range.

Boundaries

A(x) must be the area of the slice perpendicular to the axis of integration. Slanted cuts of the same solid have other areas and do not go in this integral.
The formula needs A integrable on [a, b]; continuous is enough. It defines volume for any such solid and agrees with every volume formula from Geometry.
The side s(x) is a length, so it must be top minus bottom (or right minus left). If the boundary curves cross inside [a, b], which one is on top changes, and the interval is split there (Area between curves). If a slice crosses the base in several pieces, A(x) is the total area of the shapes on all of them.
The area ks^2 holds only for shapes scaled with s. A rectangle of fixed height has area hs; a rectangle whose height is a multiple of s has area that multiple times s^2.
A table gives only an estimate. Left and right sums bracket the volume only when A is monotonic between the measurements.

Where this goes next

When the solid is made by spinning a region about a line, each slice is a circle or a ring, and the same integral becomes the The disk method and the The washer method. Slicing along cylinders instead of planes gives the The shell method. The underlying move, a quantity built from thin pieces whose sizes are known, adds up to an integral, returns in Work as an integral and arc length.

Exercises

Day 1: volume from slice areas

A solid lies along the x-axis from x = 0 to x = 2 (centimeters), and its cross section at x has area A(x) = 3x^2 + 1 square centimeters. Find its volume.
A solid lies between x = 0 and x = 3, and its cross section at x is a square of side 1 + x. Find its volume.
A concrete bollard is 40 cm tall. Its horizontal cross sections are squares whose side changes linearly from 20 cm at the top to 30 cm at the bottom. Find its volume.

Day 1: squares on a base

The base is the region under y = x^2 and above the x-axis for 0 \le x \le 2. Cross sections perpendicular to the x-axis are squares. Find the volume.
The base is the triangle bounded by y = 2x, the x-axis and x = 3, with square cross sections perpendicular to the x-axis. Find the volume, and check it with \displaystyle \frac{1}{3}Bh.
The base is the region between \displaystyle y = \sqrt{x} and y = x. Cross sections perpendicular to the x-axis are squares. Find the volume.
The base is the region between y = 1 - x^2 and y = x^2 - 1, with square cross sections perpendicular to the x-axis. Find the volume.
The base is the region under \displaystyle y = \frac{1}{x} and above the x-axis for 1 \le x \le 4, with square cross sections perpendicular to the x-axis. Find the volume.
The base is the region under y = e^{x/2} and above the x-axis for 0 \le x \le \ln 4, with square cross sections perpendicular to the x-axis. Find the volume.
The base is the disk x^2 + y^2 \le 1, with square cross sections perpendicular to the x-axis. Find the volume.

Day 2: other shapes

The base is the triangle bounded by the two axes and y = 2 - x. Cross sections perpendicular to the x-axis are equilateral triangles. Find the volume.
The base is the region between y = x and y = x^2. Cross sections perpendicular to the x-axis are semicircles with diameters across the base. Find the volume.
The base is the region between y = 4 - x^2 and the x-axis. Cross sections perpendicular to the x-axis are isosceles right triangles. Find the volume (a) with a leg on the base, (b) with the hypotenuse on the base.
The base is the region between y = 4x and y = x^2. Cross sections perpendicular to the x-axis are rectangles of height 3. Find the volume.
The base is the region under \displaystyle y = \frac{1}{x} and above the x-axis for 1 \le x \le 2. Cross sections perpendicular to the x-axis are rectangles whose height is three times their base. Find the volume.

Day 2: slices perpendicular to the y-axis

The base is the region under \displaystyle y = \sqrt{x} and above the x-axis for 0 \le x \le 4. Cross sections perpendicular to the y-axis are squares. Find the volume.
The base is the triangle with vertices (0, 0), (4, 0) and (0, 2). Cross sections perpendicular to the y-axis are semicircles. Find the volume.
The base is the region bounded by y = \ln x, the x-axis and x = e. Cross sections perpendicular to the y-axis are rectangles of height 2. Find the volume.

Day 2: tables

Eseta measures the cross sections of a standing stone 4 m tall at heights x = 0, 1, 2, 3, 4 m and finds areas 0.90, 1.05, 0.98, 0.80 and 0.52 square meters. Estimate the volume with a left sum, a right sum and their average.
A solid is 6 cm long. Its cross sections at x = 1, 3 and 5 cm have areas 4.2, 5.0 and 3.6 square centimeters. Estimate its volume with a midpoint sum.

Reasoning

Two solids have the same base. One has square cross sections; the other has square cross sections whose sides are twice as long at every x. How do the volumes compare? Answer the same question for rectangles of fixed height whose bases are doubled.
The base is the triangle with vertices (0, 0), (2, 0) and (0, 2), with square cross sections. Find the volume with slices perpendicular to the x-axis and with slices perpendicular to the y-axis, and explain why the two agree here although they did not in Example 6.
The base is the disk x^2 + y^2 \le r^2, and cross sections perpendicular to the x-axis are semicircles with diameters across the disk. Find the volume, explain why the solid is half a ball, and recover the volume of a ball.
Explain why, for a solid whose cross-sectional area decreases from one end to the other, a left sum overestimates its volume and a right sum underestimates it.

Answers

10 cubic centimeters: \int_0^2 (3x^2 + 1)\,dx = (x^3 + x)\Big|_0^2 = 10.
21: \int_0^3 (1 + 2x + x^2)\,dx = 3 + 9 + 9 = 21.
\displaystyle \frac{76{,}000}{3} \approx 25{,}333 cubic centimeters, about 25.3 liters. Measuring x down from the top, the side is \displaystyle 20 + \frac{x}{4}, so \displaystyle V = \int_0^{40} \left(400 + 10x + \frac{x^2}{16}\right)dx, which is \displaystyle 16{,}000 + 8{,}000 + \frac{4{,}000}{3}.
\displaystyle \frac{32}{5}: s = x^2, \displaystyle \int_0^2 x^4\,dx = \frac{32}{5}.
36: s = 2x, \int_0^3 4x^2\,dx = 36. The solid is a pyramid with apex at the origin and base a 6 by 6 square in the plane x = 3, so \displaystyle \frac{1}{3}\cdot 36\cdot 3 = 36.
\displaystyle \frac{1}{30}: on [0, 1], \displaystyle \sqrt{x} \ge x, so \displaystyle s = \sqrt{x} - x and \int_0^1 \left(x - 2x^{3/2} + x^2\right)dx is \displaystyle \frac{1}{2} - \frac{4}{5} + \frac{1}{3} = \frac{1}{30}.
\displaystyle \frac{64}{15}: s = 2 - 2x^2 on [-1, 1], A = 4 - 8x^2 + 4x^4, and \displaystyle V = 2\left(4 - \frac{8}{3} + \frac{4}{5}\right) = \frac{64}{15}.
\displaystyle \frac{3}{4}: \displaystyle \int_1^4 \frac{1}{x^2}\,dx = -\frac{1}{x}\Big|_1^4 = 1 - \frac{1}{4}.
3: A = \left(e^{x/2}\right)^2 = e^x, and \int_0^{\ln 4} e^x\,dx = 4 - 1 = 3.
\displaystyle \frac{16}{3}: \displaystyle s = 2\sqrt{1 - x^2}, A = 4(1 - x^2), and \int_{-1}^1 4(1 - x^2)\,dx is \displaystyle 4\left(2 - \frac{2}{3}\right) = \frac{16}{3}.
\displaystyle \frac{2\sqrt{3}}{3}: \displaystyle \frac{\sqrt{3}}{4}\int_0^2 (4 - 4x + x^2)\,dx, which is \displaystyle \frac{\sqrt{3}}{4}\cdot\frac{8}{3}.
\displaystyle \frac{\pi}{240}: s = x - x^2 on [0, 1], \displaystyle \int_0^1 (x^2 - 2x^3 + x^4)\,dx = \frac{1}{30}, times \displaystyle \frac{\pi}{8}.
(a) \displaystyle \frac{256}{15}, (b) \displaystyle \frac{128}{15}. With s = 4 - x^2 on [-2, 2], \displaystyle \int_{-2}^2 (4 - x^2)^2\,dx = \frac{512}{15}; multiply by \displaystyle \frac{1}{2} and by \displaystyle \frac{1}{4}.
32: s = 4x - x^2 on [0, 4], \displaystyle V = 3\int_0^4 (4x - x^2)\,dx = 3\left(32 - \frac{64}{3}\right) = 32.
\displaystyle \frac{3}{2}: \displaystyle A = 3s^2 = \frac{3}{x^2}, and \displaystyle \int_1^2 \frac{3}{x^2}\,dx = 3\left(1 - \frac{1}{2}\right).
\displaystyle \frac{256}{15}: at height y, 0 \le y \le 2, the region runs from x = y^2 to x = 4, so s = 4 - y^2 and \int_0^2 (16 - 8y^2 + y^4)\,dy is \displaystyle 32 - \frac{64}{3} + \frac{32}{5} = \frac{256}{15}.
\displaystyle \frac{4\pi}{3}: the hypotenuse is x = 4 - 2y, so s = 4 - 2y for 0 \le y \le 2, and \displaystyle \frac{\pi}{8}\int_0^2 (16 - 16y + 4y^2)\,dy, which is \displaystyle \frac{\pi}{8}\cdot\frac{32}{3}.
2: at height y, 0 \le y \le 1, the region runs from x = e^y to x = e, so V = 2\int_0^1 (e - e^y)\,dy = 2\big(e - (e - 1)\big) = 2.
Left 3.73, right 3.35, average 3.54 cubic meters: \Delta x = 1, L = 0.90 + 1.05 + 0.98 + 0.80 and R = 1.05 + 0.98 + 0.80 + 0.52.
25.6 cubic centimeters: the points 1, 3, 5 are the midpoints of pieces of width 2, so 2(4.2 + 5.0 + 3.6) = 25.6.
Doubling the side multiplies each square’s area by 4, so A(x) and hence V = \int_a^b A(x)\,dx are multiplied by 4. A rectangle of fixed height h has area hs, so doubling s doubles A(x) and the volume.
Both \displaystyle \frac{8}{3}: perpendicular to the x-axis s = 2 - x and \displaystyle \int_0^2 (2 - x)^2\,dx = \frac{8}{3}; perpendicular to the y-axis s = 2 - y and the same integral in y. The triangle is symmetric about the line y = x, so reflecting in it carries one solid onto the other. The base in Example 6 has no such symmetry.
\displaystyle \frac{2}{3}\pi r^3: \displaystyle s = 2\sqrt{r^2 - x^2} is a diameter, so each semicircle has radius \displaystyle \sqrt{r^2 - x^2} and area \displaystyle \frac{\pi}{2}(r^2 - x^2), and \displaystyle \frac{\pi}{2}\int_{-r}^{r} (r^2 - x^2)\,dx is \displaystyle \frac{\pi}{2}\cdot\frac{4r^3}{3}. The upper half of the ball x^2 + y^2 + z^2 \le r^2 meets the plane at x in exactly this semicircle, so the two solids are the same; the whole ball is twice this, \displaystyle \frac{4}{3}\pi r^3.
On each piece [x_{i-1}, x_i] a decreasing A satisfies A(x_i) \le A(x) \le A(x_{i-1}). Integrals keep inequalities, so the slab’s volume \int_{x_{i-1}}^{x_i} A(x)\,dx lies between A(x_i)\,\Delta x and A(x_{i-1})\,\Delta x. Adding over the pieces gives R \le V \le L.