Volumes by slicing
Tonight's idea
In grade 10 you found the area under \(y = x^2\) with thin strips, and the sphere's volume by Cavalieri's principle: equal slices, equal volumes. Tonight the slices get integrated.
Cut a solid by planes perpendicular to an axis. At position \(x\) the slice has cross-section area \(A(x)\); a slab of thickness \(dx\) has volume \(A(x)\,dx\). Add the slabs:
\(V = \displaystyle\int_a^b A(x)\,dx\). The whole job is writing down \(A(x)\).
For a sphere of radius \(r\) centred at the origin, the slice at \(x\) is a disc. Its radius is the height of the circle \(x^2 + y^2 = r^2\) there, \(\sqrt{r^2 - x^2}\), so \(A(x) = \pi(r^2 - x^2)\) and
\[V = \int_{-r}^{r}\pi(r^2 - x^2)\,dx = \pi\Big[r^2x - \frac{x^3}{3}\Big]_{-r}^{r} = \pi\Big(2r^3 - \frac{2r^3}{3}\Big) = \frac43\pi r^3.\]
A solid made by spinning a curve \(y = f(x)\) about the \(x\)-axis (a solid of revolution) always has disc slices of radius \(f(x)\), so \(V = \pi\int_a^b f(x)^2\,dx\); the sphere is the case \(f(x) = \sqrt{r^2 - x^2}\). Slices need not be discs: any shape whose area you can write down will do.
Watch one: the cone from similar triangles
A cone has base radius \(R\) and height \(h\). Put the apex at the origin and measure \(y\) down the axis. The slice at depth \(y\) is a disc whose radius \(\rho\) comes from similar triangles: \(\rho/y = R/h\), so \(\rho = Ry/h\) and
\[V = \int_0^h \pi\Big(\frac{Ry}{h}\Big)^2 dy = \frac{\pi R^2}{h^2}\cdot\frac{h^3}{3} = \frac13\pi R^2h.\]
The famous third is \(\int y^2\,dy = y^3/3\), the same integral as the area under a parabola, now stacked into a solid.
Watch thin strips close in on an exact area, one dimension down: mathdetours.com/demos/riemann.html. The no-integral route to the same formulas is grade 10's Cavalieri sheet.
Your turn
- (a) Rotate \(y = \sqrt{x}\), \(0 \le x \le 4\), about the \(x\)-axis and find the volume. (b) Rotate \(y = 2\), \(0 \le x \le 5\), and check the answer against the cylinder formula.
- A square pyramid has base side \(s\) and height \(h\). Measuring \(y\) from the apex, explain why the slice at depth \(y\) is a square of side \(sy/h\), then integrate to get \(V = \tfrac13 s^2h\).
- A vase is made by rotating \(y = 1 + x^2/4\), \(0 \le x \le 2\) (centimetres), about the \(x\)-axis. Expand the square and find the volume exactly and to one decimal place. Then design your own: choose a profile \(f(x)\) you can square and integrate, and compute its volume.
- A hemispherical bowl of radius 10 cm holds water to a depth of 4 cm. Set up the slicing integral (think about where along the axis the water starts and stops) and find the volume of water.
Stretch: the napkin ring
Drill a cylindrical hole straight through the centre of a sphere of radius \(r\) so that the ring left behind has height \(h\). Slice the ring at height \(x\): each slice is an annulus. Show that its area does not involve \(r\), and find the ring's volume. The size of the sphere did not matter.
Answers
- (a) \(\pi\int_0^4 x\,dx = \pi\big[x^2/2\big]_0^4 = 8\pi \approx 25.1\). (b) \(\pi\int_0^5 4\,dx = 20\pi\), and \(\pi r^2h = \pi\cdot 4\cdot 5 = 20\pi\).
- Every slice is a scaled copy of the base, scaled by \(y/h\) (similar triangles down any face), so its side is \(sy/h\) and \(A(y) = s^2y^2/h^2\). \(V = \dfrac{s^2}{h^2}\displaystyle\int_0^h y^2\,dy = \dfrac{s^2}{h^2}\cdot\dfrac{h^3}{3} = \tfrac13 s^2h\). Same integral as the cone, different slice shape.
- \((1 + x^2/4)^2 = 1 + x^2/2 + x^4/16\). \(V = \pi\big[x + x^3/6 + x^5/80\big]_0^2 = \pi\big(2 + \tfrac43 + \tfrac25\big) = \tfrac{56}{15}\pi \approx 11.7\) cm³.
- Put the centre of the full sphere at the origin and measure \(x\) downward; the bowl's bottom is at \(x = 10\) and the water surface at \(x = 6\). Slices are discs of radius \(\sqrt{100 - x^2}\): \(V = \pi\displaystyle\int_6^{10}(100 - x^2)\,dx = \pi\big[100x - x^3/3\big]_6^{10} = \pi\big(\tfrac{2000}{3} - 528\big) = \tfrac{416}{3}\pi \approx 436\) cm³. (In general a cap of depth \(d\) has volume \(\tfrac13\pi d^2(3r - d)\).)
Stretch. The hole's radius satisfies \(r_0^2 = r^2 - h^2/4\). At height \(x\) (with \(|x| \le h/2\)) the outer radius squared is \(r^2 - x^2\), so the annulus has area \(\pi(r^2 - x^2) - \pi(r^2 - h^2/4) = \pi(h^2/4 - x^2)\): no \(r\). \(V = \pi\int_{-h/2}^{h/2}(h^2/4 - x^2)\,dx = \pi\big(\tfrac{h^3}{4} - \tfrac{h^3}{12}\big) = \tfrac16\pi h^3\), the volume of a sphere of diameter \(h\). A 2 cm high ring cut from the Earth has the same volume as a 2 cm high ring cut from an orange.