5.12 Areas and volumes of revolutionIB Maths: Applications and Interpretation HL: Revision notes

For a continuous curve $y=f(x)$ with $f(x)\ge0$ on $a\le x\le b$, the area between the curve and the $x$-axis is $A=\int_a^b y\,dx.$ Use your GDC to evaluate a definite integral, or integrate by hand when the function is simple. Give answers to 3 s.f. unless exact values are asked for. Example: the area under $y=x^2$ from $x=0$ to $x=3$ is $\int_0^3 x^2\,dx=\left[\frac{x^3}{3}\right]_0^3=9$.

Where the curve lies below the $x$-axis, $\int y\,dx$ is negative. An integral gives a signed area: positive above the axis, negative below. To find the total area enclosed between a curve and the $x$-axis, find where the curve crosses the axis (the roots), split the interval there, and add the magnitudes: $A=\int_a^b|f(x)|\,dx.$ Example: $y=x^2-4x$ is below the axis for $0<x<4$. $\int_0^4(x^2-4x)\,dx=-\frac{32}{3}$, so the area is $\frac{32}{3}$. For $0\le x\le6$, $\int_0^6(x^2-4x)\,dx=0$ because the areas cancel, but the total area is $\frac{32}{3}+\frac{32}{3}=\frac{64}{3}$.

When a region is bounded by the $y$-axis, a curve and horizontal lines $y=c$ and $y=d$, integrate with respect to $y$: $A=\int_c^d x\,dy,$ where $x$ is written as a function of $y$. Example: for $y=x^2$ ($x\ge0$), $x=\sqrt{y}$, so the area between the curve and the $y$-axis for $0\le y\le4$ is $\int_0^4\sqrt{y}\,dy=\left[\frac23y^{3/2}\right]_0^4=\frac{16}{3}$. If part of the curve is to the left of the $y$-axis ($x<0$), that part gives a negative integral, so split at $x=0$ in the same way.

When a region is rotated through $360^\circ$ about the $x$-axis, it sweeps out a solid of revolution. Each thin slice is a disc of radius $y$ and thickness $dx$, so $V=\int_a^b\pi y^2\,dx.$ Square $y$ before integrating: for $y=\sqrt{x}$, $y^2=x$. Example: $y=\sqrt{x}$ from $x=0$ to $4$: $V=\pi\int_0^4 x\,dx=\pi\left[\frac{x^2}{2}\right]_0^4=8\pi$. Volumes are never negative; if $y<0$ the square is still positive.

For rotation about the $y$-axis, the discs have radius $x$ and thickness $dy$: $V=\int_c^d\pi x^2\,dy.$ Rewrite the curve with $x^2$ as a function of $y$ and use $y$-limits. Example: $y=x^2$ for $0\le y\le9$ gives $x^2=y$, so $V=\pi\int_0^9 y\,dy=\pi\left[\frac{y^2}{2}\right]_0^9=\frac{81\pi}{2}$. Check which axis the question names. The same curve gives different volumes about the $x$-axis and the $y$-axis.

In Paper 2, set up the integral and use your GDC to evaluate it. Write the integral (including $\pi$ and limits) in your working so that method marks are available. Worked example: the region between $y=x^3-6x^2+8x$ and the $x$-axis for $0\le x\le2$ is rotated about the $x$-axis. $V=\pi\int_0^2\left(x^3-6x^2+8x\right)^2dx=30.6$ to 3 s.f. In context, state units: if lengths are in cm, areas are in cm$^2$ and volumes in cm$^3$. Round measurements sensibly (3 s.f.).