Quick questions on Areas between curves and volumes of revolution in QCE Specialist Mathematics Unit 4

If $f(x) \geq g(x)$ on $[a, b]$, the area enclosed between them is

Rotating the region under $y = f(x)$ from $x = a$ to $x = b$ about the $x$-axis sweeps out a solid of circular cross-sections of radius $f(x)$. Each thin disc has volume $\pi [f(x)]^2\, dx$, so

Rotating the region between $x = g(y)$ and the $y$-axis from $y = c$ to $y = d$ about the $y$-axis gives

When the rotated region lies between an outer curve $R(x)$ and an inner curve $r(x)$, the cross-section is an annulus (washer):

The reliable workflow is always the same: sketch the region, mark the limits where the boundary curves meet, decide the axis of rotation, and then identify the radius. For an $x$-axis rotation the radius is the vertical distance from the axis to the curve, namely the $y$-value; for a $y$-axis rotation it is the horizontal distance, the $x$-value, so you must first rearrange the curve into the form $x = g(y)$. Getting the limits from the intersection points and the radius from the correct distance accounts for most of the marks; the integration itself is usually routine.

When the two curves cross inside the interval, the upper and lower roles swap, and a single integral of $f - g$ would let positive and negative pieces cancel. Find every intersection in the interval, split the region at each one, and integrate the positive difference (upper minus lower) on each piece separately, then add the parts. Equivalently, integrate the absolute value of the difference. For example, the area between $y = \sin x$ and the $x$-axis over $[0, 2\pi]$ must be split at $x = \pi$, because the curve dips below the axis on $(\pi, 2\pi)$.