Let there be a cone 
in radius of its base and 
in height.
Find the volume 
of the cone, using what are so-called i. the disc (/washer) method and ii. the (cylindrical) shell method, if the names my memory serves me right.
Answer. 
.
Roughwork.
Visualize the cone.
i. By disc method,
![\begin{aligned} \frac{R}{h} & = \frac{r}{h-z} = \tan\theta \\ r & = R\bigg( 1-\frac{z}{h}\bigg) \\ V & = \pi R^2\int_{0}^{h}\bigg( 1-\frac{z}{h}\bigg)^2\,\mathrm{d}z \\ \cdots & \cdots\cdots\cdots\cdots\cdots\cdots\cdots \\ \textrm{let }& z'=1-\frac{z}{h} \\ \frac{\mathrm{d}z'}{\mathrm{d}z} & = -\frac{1}{h} \\ \mathrm{d}z & = -h\,\mathrm{d}z' \\ z=0 & \Leftrightarrow z'=1 \\ z=h & \Leftrightarrow z'= 0 \\ \cdots & \cdots\cdots\cdots\cdots\cdots\cdots\cdots \\ V& = \pi R^2\int_{1}^{0} z'^2(-h\,\mathrm{d}z') \\ & = -\pi R^2h \bigg[ \frac{z'^3}{3}\bigg]\bigg|_{1}^{0} \\ & = -\pi R^2h \bigg(\frac{(0)^3}{3} - \frac{(1)^3}{3}\bigg) \\ & = \frac{\pi R^2h}{3}\\ \end{aligned}](https://physicspupil.com/wp-content/ql-cache/quicklatex.com-c82832fa803b3f7f367a1777e5229fd8_l3.svg "Rendered by QuickLaTeX.com")
ii. By shell method,
we are obtaining the volume of a solid of revolution about the 
-axis by integrating the slices (/cross sections) ranging between ![\theta\in [0,2\pi ]](https://physicspupil.com/wp-content/ql-cache/quicklatex.com-15eb5ad510855b5fc2c3bc9b1b9abfe4_l3.svg "Rendered by QuickLaTeX.com")
.
It is left the reader as an exercise.
This problem is not to be attempted.