Quick questions on Volumes of revolution explained: H2 Mathematics Calculus

Rotating the region under $y = \mathrm{f}(x)$ between $x = a$ and $x = b$ a full turn about the $x$-axis sweeps out thin discs of radius $y$ and thickness $dx$. Summing $\pi y^2\,dx$ gives

Rotating about the $y$-axis instead, the discs have radius $x$ and thickness $dy$, so

When the region lies between two curves $y = \mathrm{f}(x)$ (outer) and $y = \mathrm{g}(x)$ (inner), rotating about the $x$-axis gives a washer (a disc with a hole):

The trickiest setup step is often the limits, which come from where the region starts and ends, not from numbers handed to you. For rotation about the $x$-axis, the limits are the $x$-values bounding the region; for a region between two curves, find them by solving the curves' intersection. To rotate the region between $y = x$ and $y = x^2$, set $x = x^2$ to get $x = 0$ and $x = 1$, which become the integration limits. When rotating about the $y$-axis, the limits are $y$-values instead, so read or compute the region's lowest and highest $y$.

The radius is the distance from the axis to the curve; for the $x$-axis it is $y$, for the $y$-axis it is $x$.

Match the limits to the variable of integration ($x$ or $y$).

The region under $y = x$ from $0$ to $2$ is rotated about the $x$-axis. Find the volume. [3 marks]

State the formula for the volume when rotating about the $y$-axis. [1 mark]

When rotating a region between two curves about the $x$-axis, what is integrated? [1 mark]