Volumes by Slicing: Disks, Washers & Cross-Sections

The axis is horizontal: vertical slices, $dx$. The parabola meets the $x$-axis at $x = \pm 2$.

$r = 4 - x^2$.

$V = \pi\displaystyle\int_{-2}^{2}(4 - x^2)^2\,dx = \pi\int_{-2}^{2}(16 - 8x^2 + x^4)\,dx$.

$\pi\Big[16x - \tfrac{8x^3}{3} + \tfrac{x^5}{5}\Big]_{-2}^{2} = 2\pi\left(32 - \tfrac{64}{3} + \tfrac{32}{5}\right) = \tfrac{512\pi}{15}$.

Vertical slices, $dx$, from $0$ to $\pi$.

$r = \sqrt{\sin x}$.

$V = \pi\displaystyle\int_0^{\pi}\sin x\,dx$: squaring removed the root.

$\pi\big[-\cos x\big]_0^{\pi} = \pi(1 + 1) = 2\pi$.

The axis is vertical: horizontal slices, $dy$, from $y = 0$ to $y = 8$. Write the curve as $x = y^{1/3}$.

$r = y^{1/3}$, from the $y$-axis to the curve.

$V = \pi\displaystyle\int_0^8 y^{2/3}\,dy$.

$\pi\Big[\tfrac{3}{5}y^{5/3}\Big]_0^8 = \pi \cdot \tfrac{3}{5} \cdot 32 = \tfrac{96\pi}{5}$.

$x = x^2$ at $x = 0$ and $x = 1$; on $[0, 1]$, $x \ge x^2$.

Vertical slices, $dx$, from $0$ to $1$.

$R = x$ (farther from the axis), $r = x^2$.

$V = \pi\displaystyle\int_0^1 (x^2 - x^4)\,dx$.

$\pi\left(\tfrac{1}{3} - \tfrac{1}{5}\right) = \tfrac{2\pi}{15}$.

$x^2 = 2x$ at $x = 0$ and $x = 2$; on $[0, 2]$, $2x \ge x^2$.

The axis $y = -2$ is horizontal: vertical slices, $dx$.

The axis is below, so radius $= y - (-2)$: $R = 2x + 2$, $r = x^2 + 2$.

$V = \pi\displaystyle\int_0^2\big[(2x + 2)^2 - (x^2 + 2)^2\big]\,dx = \pi\int_0^2 (8x - x^4)\,dx$.

$\pi\left(16 - \tfrac{32}{5}\right) = \tfrac{48\pi}{5}$.

Same region: $x$ from $0$ to $2$, with $2x$ above $x^2$.

Vertical slices, $dx$.

The axis is above, so radius $= 5 - y$. The lower curve is farther: $R = 5 - x^2$, $r = 5 - 2x$.

$V = \pi\displaystyle\int_0^2\big[(5 - x^2)^2 - (5 - 2x)^2\big]\,dx = \pi\int_0^2 (x^4 - 14x^2 + 20x)\,dx$.

$\pi\left(\tfrac{32}{5} - \tfrac{112}{3} + 40\right) = \tfrac{136\pi}{15}$.

In terms of $y$ the curves are $x = \sqrt{y}$ and $x = \tfrac{y}{2}$; they meet at $y = 0$ and $y = 4$.

The $y$-axis is vertical: horizontal slices, $dy$, from $0$ to $4$.

$R = \sqrt{y}$ (the parabola is farther from the $y$-axis), $r = \tfrac{y}{2}$.

$V = \pi\displaystyle\int_0^4\left(y - \frac{y^2}{4}\right)\,dy$.

$\pi\left(8 - \tfrac{16}{3}\right) = \tfrac{8\pi}{3}$.

Slices perpendicular to the $x$-axis, $x$ from $-2$ to $2$.

$s = \sqrt{4 - x^2} - \left(-\sqrt{4 - x^2}\right) = 2\sqrt{4 - x^2}$.

$A = \dfrac{\sqrt{3}}{4}s^2 = \sqrt{3}\,(4 - x^2)$.

$V = \sqrt{3}\displaystyle\int_{-2}^{2}(4 - x^2)\,dx = \sqrt{3} \cdot \tfrac{32}{3} = \tfrac{32\sqrt{3}}{3}$.

Horizontal slices at height $y$, from $0$ to $3$.

$s = e^{-y}$ (given).

$A = s^2 = e^{-2y}$.

$V = \displaystyle\int_0^3 e^{-2y}\,dy = \tfrac{1}{2}\left(1 - e^{-6}\right)$ m$^3$.

Slices perpendicular to the $x$-axis, $x$ from $0$ to $4$.

$s = \sqrt{x} - 0 = \sqrt{x}$.

$A = \dfrac{\pi}{8}s^2 = \dfrac{\pi x}{8}$.

$V = \displaystyle\int_0^4 \frac{\pi x}{8}\,dx = \frac{\pi}{8} \cdot 8 = \pi$.

Horizontal slices at height $y$, from $0$ (base) to $9$ (tip).

Similar triangles: the side shrinks linearly from $6$ to $0$, so $s = 6\left(1 - \frac{y}{9}\right) = 6 - \frac{2y}{3}$.

$A = \left(6 - \frac{2y}{3}\right)^2$.

$V = \displaystyle\int_0^9\left(6 - \frac{2y}{3}\right)^2\,dy = 108$, matching $\tfrac{1}{3}(\text{base area})(\text{height}) = \tfrac{1}{3}(36)(9)$.