Exercises — Volume of revolution — shell method

Axis = $y$-axis, so strips are vertical (parallel to the $y$-axis) with thickness $dx$.

• Radius: distance from the $y$-axis (the line $x=0$) to a strip at position $x$ is $r=x$.
• Height: the strip runs from $y=0$ up to $y=x^3$, so $h=x^3$.
• Limits: $x$ runs from $0$ to $2$. $V={\int }_0^{2}2\pi x\cdot x^{3}dx.$

• Radius from the $y$-axis: $r=x$.
• Height of the strip: $h=\sqrt{x}$.
• Limits: $x$ from $1$ to $4$. $V={\int }_1^{4}2\pi x\sqrt{x}dx.$ Notice we did not invert to $x=y^2$. That is exactly why shells are convenient here.

$r=x$, $h=x^2$, $x\in [0,3]$. $V={\int }_0^{3}2\pi x\cdot x^{2}dx=2\pi {\int }_0^3{x}^{3}dx=2\pi {\left[\frac{x^4}{4}\right]}_0^3=2\pi \cdot \frac{81}{4}=\frac{81\pi }{2}.$ What we did: multiplied circumference $2\pi x$ by height $x^2$, integrated the shells from the axis outward.

$r=x$, $h=\sqrt{x}=x^{1/2}$, $x\in [0,4]$. $V={\int }_0^{4}2\pi x\cdot x^{1/2}dx=2\pi {\int }_0^4{x}^{3/2}dx=2\pi {\left[\frac{2}{5}{x}^{5/2}\right]}_0^4.$ Now $4^{5/2}=(4^{1/2}{)}^5=2^5=32$, so $V=2\pi \cdot \frac{2}{5}\cdot 32=\frac{128\pi }{5}.$

$r=x$, $h=x^2+1$, $x\in [0,2]$. $V={\int }_0^{2}2\pi x(x^2+1)dx=2\pi {\int }_0^2(x^3+x)dx=2\pi {\left[\frac{x^4}{4}+\frac{x^2}{2}\right]}_0^2.$ Plug in $x=2$: $\frac{16}{4}+\frac{4}{2}=4+2=6$, so $V=12\pi$.

The axis $x=-1$ lies to the left of the whole region ($0\le x\le 2$). The distance from a strip at $x$ to the line $x=-1$ is $r=x-(-1)=x+1(\text{positive throughout, since }x\ge 0).$ Height stays $h=x^2$. $V={\int }_0^{2}2\pi (x+1)x^{2}dx=2\pi {\int }_0^2(x^3+x^2)dx=2\pi {\left[\frac{x^4}{4}+\frac{x^3}{3}\right]}_0^2.$ At $x=2$: $\frac{16}{4}+\frac{8}{3}=4+\frac{8}{3}=\frac{20}{3}$, so $V=2\pi \cdot \frac{20}{3}=\frac{40\pi }{3}.$

The axis $x=2$ is to the right of the region. Distance from a strip at $x$ to $x=2$ is $r=2-x(\text{positive, since }x\le 2).$ Height $h=x^2$. $V={\int }_0^{2}2\pi (2-x)x^{2}dx=2\pi {\int }_0^2(2x^2-x^3)dx=2\pi {\left[\frac{2x^3}{3}-\frac{x^4}{4}\right]}_0^2.$ At $x=2$: $\frac{2\cdot 8}{3}-\frac{16}{4}=\frac{16}{3}-4=\frac{4}{3}$, so $V=2\pi \cdot \frac{4}{3}=\frac{8\pi }{3}.$

Axis is horizontal, so strips are horizontal with thickness $dy$, radius $r=y$ (distance to the $x$-axis), and height = horizontal width of the region at that $y$.

• For a given $y$, the region is bounded on the left by the curve $x=\sqrt{y}$ and on the right by the line $x=2$. So width $=x_R-x_L=2-\sqrt{y}$.
• $y$ runs from $0$ (bottom) to $4$ (since $y=x^2$ reaches $4$ at $x=2$). $V={\int }_0^{4}2\pi y(2-\sqrt{y})dy=2\pi {\int }_0^4(2y-y^{3/2})dy=2\pi {\left[y^2-\frac{2}{5}{y}^{5/2}\right]}_0^4.$ At $y=4$: $16-\frac{2}{5}\cdot 32=16-\frac{64}{5}=\frac{80-64}{5}=\frac{16}{5}$, so $V=2\pi \cdot \frac{16}{5}=\frac{32\pi }{5}.$

On $(0,1)$ the line $y=x$ sits above the parabola $y=x^2$, so height $h=x-x^2$. Axis $x=2$ is to the right of the region, so radius $r=2-x$. $V={\int }_0^{1}2\pi (2-x)(x-x^2)dx.$ Expand the integrand: $(2-x)(x-x^2)=2x-2x^2-x^2+x^3=2x-3x^2+x^3.$ $V=2\pi {\int }_0^1(2x-3x^2+x^3)dx=2\pi {\left[x^2-x^3+\frac{x^4}{4}\right]}_0^1.$ At $x=1$: $1-1+\frac{1}{4}=\frac{1}{4}$, so $V=2\pi \cdot \frac{1}{4}=\frac{\pi }{2}.$

The parabola $y=2x-x^2$ is above the $x$-axis for $0\le x\le 2$, so $h=2x-x^2$. Axis is the $y$-axis, so $r=x$. $V={\int }_0^{2}2\pi x(2x-x^2)dx=2\pi {\int }_0^2(2x^2-x^3)dx=2\pi {\left[\frac{2x^3}{3}-\frac{x^4}{4}\right]}_0^2.$ At $x=2$: $\frac{2\cdot 8}{3}-\frac{16}{4}=\frac{16}{3}-4=\frac{4}{3}$, so $V=2\pi \cdot \frac{4}{3}=\frac{8\pi }{3}.$

On $0\le x\le 1$, the upper curve is $y=2-x^2$ and lower is $y=x^2$, so height $h=(2-x^2)-x^2=2-2x^2.$ Axis $x=1$ is to the right of this piece, so radius $r=1-x$ (positive on $[0,1)$, zero at $x=1$). $V={\int }_0^{1}2\pi (1-x)(2-2x^2)dx=2\pi {\int }_0^{1}2(1-x)(1-x^2)dx.$ Expand $(1-x)(1-x^2)=1-x^2-x+x^3$, times $2$: $2-2x-2x^2+2x^3$. $V=2\pi {\int }_0^1(2-2x-2x^2+2x^3)dx=2\pi {\left[2x-x^2-\frac{2x^3}{3}+\frac{x^4}{2}\right]}_0^1.$ At $x=1$: $2-1-\frac{2}{3}+\frac{1}{2}=\frac{12-6-4+3}{6}=\frac{5}{6}$, so $V=2\pi \cdot \frac{5}{6}=\frac{5\pi }{3}.$

Radius $r=x$, height $h=x^n$, limits $0$ to $a$. $V={\int }_0^{a}2\pi x\cdot x^{n}dx=2\pi {\int }_0^a{x}^{n+1}dx=2\pi {\left[\frac{x^{n+2}}{n+2}\right]}_0^a=\frac{2\pi a^{n+2}}{n+2}.$ Check the algebra against L2: with $n=3$, $a=2$ this gives $\frac{2\pi \cdot 2^5}{5}=\frac{64\pi }{5}$. Hmm — the boxed claim says $a^{n+3}/(n+3)$, which would give $\frac{2\pi \cdot 2^6}{6}$. The boxed statement is a deliberate trap: the correct exponent is $n+2$, not $n+3$, because $x\cdot x^n=x^{n+1}$ and integrating adds one more, giving $x^{n+2}$. The right general formula is $\boxed{V=\frac{2\pi a^{n+2}}{n+2}}.$ (Sanity: $n=1$, $a=1$ gives $\frac{2\pi }{3}$, the volume of the cone-like solid under $y=x$ spun about the $y$-axis — correct.)

Shells (from Exercise 2.1 pattern, but $a=2$): $r=x$, $h=x^2$, $V_{\text{shell}}={\int }_0^{2}2\pi x\cdot x^{2}dx=2\pi {\left[\frac{x^4}{4}\right]}_0^2=8\pi .$ Disks/washers: slice perpendicular to the $y$-axis, so slices are horizontal, thickness $dy$. At height $y$, the solid is a disk of radius $R=2$ with a bite taken out by the curve $x=\sqrt{y}$ (inner radius $r_{\text{in}}=\sqrt{y}$), because the region only fills from $x=\sqrt{y}$ out to $x=2$. $y$ runs $0$ to $4$: $V_{\text{washer}}={\int }_0^4\pi (2^2-(\sqrt{y}{)}^2)dy=\pi {\int }_0^4(4-y)dy=\pi {\left[4y-\frac{y^2}{2}\right]}_0^4=\pi (16-8)=8\pi .$ Both give $\boxed{8\pi }$. Shells kept us in $x$; washers forced the inversion $x=\sqrt{y}$. See Volume of revolution — disk and washer method.

$r$ when rotating about $x=c$ ::: $|x-c|$, distance from strip to the axis Thickness when the axis is horizontal ::: $dy$ (strips run parallel to the axis) Height for a region between two curves ::: top curve minus bottom curve Volume of $y=x^2$, $0\le x\le 3$, about $y$-axis ::: $\frac{81\pi }{2}$ Volume of $y=\sqrt{x}$, $0\le x\le 4$, about $y$-axis ::: $\frac{128\pi }{5}$ General $y=x^n$, $0\le x\le a$, about $y$-axis ::: $\frac{2\pi a^{n+2}}{n+2}$