Volumes of revolution practice questions — IB Maths AA

$V = \pi \int_0^2 (4x^2 + 4x + 1) \, dx = \pi\!\left[\dfrac{4x^3}{3} + 2x^2 + x\right]_0^2 = \pi\!\left(\dfrac{32}{3} + 8 + 2\right) = \pi \cdot \dfrac{32 + 30}{3} = \dfrac{62\pi}{3}$.

$V = \pi \int_0^1 (x^2 - x^4) \, dx = \pi\!\left[\dfrac{x^3}{3} - \dfrac{x^5}{5}\right]_0^1 = \pi\!\left(\dfrac{1}{3} - \dfrac{1}{5}\right) = \dfrac{2\pi}{15}$.

$V = \pi \int_0^1 x^4 \, dx = \pi \cdot \dfrac{1}{5} = \dfrac{\pi}{5}$.

$V = \pi \int_0^{\pi/2} \cos^2 x \, dx = \pi \int_0^{\pi/2} \dfrac{1 + \cos 2x}{2} \, dx = \dfrac{\pi}{2}\!\left[x + \dfrac{\sin 2x}{2}\right]_0^{\pi/2} = \dfrac{\pi}{2} \cdot \dfrac{\pi}{2} = \dfrac{\pi^2}{4}$.

The graph is a semicircle; rotating about the $x$-axis sweeps out a sphere of radius $R$. (Volume $= \dfrac{4}{3}\pi R^3$.)

$V = \pi \int_0^4 (\sqrt{x})^2 \, dx = \pi \int_0^4 x \, dx = \pi \cdot \dfrac{16}{2} = 8\pi$.

$V = \pi \int_0^1 (e^x)^2 \, dx = \pi \int_0^1 e^{2x} \, dx = \pi \cdot \dfrac{e^{2x}}{2}\Big|_0^1 = \dfrac{\pi}{2}(e^2 - 1)$.

$V = \pi \displaystyle\int_a^b [f(x)]^2 \, dx$ — the disk-method formula.

Cylinder of radius $2$ and height $5$: $V = \pi \cdot 4 \cdot 5 = 20\pi$.

$V = \pi \int_0^3 x^2 \, dx = \pi \cdot \dfrac{27}{3} = 9\pi$. (Equivalently, $V_{\text{cone}} = \dfrac{1}{3}\pi r^2 h = \dfrac{1}{3}\pi \cdot 9 \cdot 3 = 9\pi$.)