Volumes of revolution: disks, washers and shells

Worked solutions

Problem 1

1. A slice at $x$ is a disk of radius $\sqrt x$, area $\pi(\sqrt x)^2 = \pi x$.The region touches the axis, so there is no hole.
2. $V = \displaystyle\int_0^4 \pi x\,dx = \pi\Big[\tfrac{x^2}2\Big]_0^4$Add the disks from $x = 0$ to $x = 4$.
3. $8\pi \approx 25.13$$\pi\cdot\tfrac{16}2$. The solid is a paraboloid, and its volume is half that of the cylinder of radius 2 and length 4 around it, $16\pi$.

Problem 2

1. The line goes from $(0, 0)$ to $(h, r)$, so the solid is a cone with tip at the origin, height $h$ and base radius $r$.The radius grows linearly from 0 to $r$.
2. $V = \displaystyle\int_0^h \pi\Big(\frac rh\,x\Big)^2dx = \frac{\pi r^2}{h^2}\int_0^h x^2\,dx$Disks of radius $\tfrac rh x$; the constants come out of the integral.
3. $\dfrac{\pi r^2}{h^2}\cdot\dfrac{h^3}3$$\int_0^h x^2\,dx = \tfrac{h^3}3$.
4. $\tfrac13\pi r^2h$A third of the cylinder $\pi r^2h$, the familiar formula.

Problem 3

1. The upper half circle is $y = \sqrt{R^2 - x^2}$ for $-R \le x \le R$.Solve for $y \ge 0$.
2. $V = \displaystyle\int_{-R}^R \pi\big(R^2 - x^2\big)\,dx$Disks of radius $\sqrt{R^2 - x^2}$; squaring removes the root.
3. $\pi\Big[R^2x - \tfrac{x^3}3\Big]_{-R}^R = \pi\Big(\tfrac23R^3 - \big(-\tfrac23R^3\big)\Big)$At $R$: $R^3 - \tfrac{R^3}3 = \tfrac23R^3$; at $-R$ it is the negative.
4. $\tfrac43\pi R^3$$\tfrac43 = \tfrac23 + \tfrac23$. Rotating the half circle, not the whole circle, avoids counting every slice twice.

Problem 4

1. On $(0, 1)$, $x > x^2$, so the outer radius is $R = x$ and the inner radius is $r = x^2$.The region does not touch the axis between its ends, so each slice is a washer.
2. $V = \displaystyle\int_0^1 \pi\big(x^2 - x^4\big)\,dx$$\pi(R^2 - r^2)$: square each radius first.
3. $\pi\Big(\tfrac13 - \tfrac15\Big)$$\int_0^1 x^2\,dx = \tfrac13$ and $\int_0^1 x^4\,dx = \tfrac15$.
4. $\tfrac{2\pi}{15} \approx 0.419$The cone swept by the line, $\tfrac\pi3$, minus the solid swept by the parabola, $\tfrac\pi5$.

Problem 5

1. Slices perpendicular to the $y$-axis are horizontal, so integrate in $y$, from $y = 0$ to $y = 4$.The axis is vertical.
2. At height $y$ the region runs from the $y$-axis to $x = \sqrt y$, so the disk has radius $\sqrt y$.Solve $y = x^2$ for $x \ge 0$.
3. $V = \displaystyle\int_0^4 \pi y\,dy = \pi\Big[\tfrac{y^2}2\Big]_0^4$$\pi(\sqrt y)^2 = \pi y$.
4. $8\pi \approx 25.13$Shells give the same: $\int_0^2 2\pi x\,(4 - x^2)\,dx = 2\pi(8 - 4) = 8\pi$.

Problem 6

1. A vertical strip at $x$ runs from $y = x^2$ up to $y = x$, so its height is $h(x) = x - x^2$.Top minus bottom, as for an area.
2. The strip is at distance $\rho(x) = x$ from the $y$-axis.Shells about a vertical axis are integrated in $x$.
3. $V = \displaystyle\int_0^1 2\pi x\,(x - x^2)\,dx = 2\pi\int_0^1 (x^2 - x^3)\,dx$$2\pi\rho h\,dx$ for each shell.
4. $2\pi\Big(\tfrac13 - \tfrac14\Big) = \tfrac\pi6 \approx 0.524$Washers in $y$ agree: the outer radius is $\sqrt y$ and the inner $y$, and $\int_0^1 \pi(y - y^2)\,dy = \tfrac\pi6$.

Problem 7

1. Shells: the strip at $x$ has height $x - x^2$ and is $2 - x$ from the axis.The region lies to the left of $x = 2$, so the distance is $2 - x$, not $x$.
2. $V = \displaystyle\int_0^1 2\pi(2 - x)(x - x^2)\,dx = 2\pi\int_0^1 (2x - 3x^2 + x^3)\,dx = 2\pi\Big(1 - 1 + \tfrac14\Big) = \tfrac\pi2$$(2 - x)(x - x^2) = 2x - 2x^2 - x^2 + x^3$.
3. Washers: at height $y$ the region runs from $x = y$ to $x = \sqrt y$. The far edge $x = y$ is $2 - y$ from the axis and the near edge $x = \sqrt y$ is $2 - \sqrt y$.Horizontal slices perpendicular to the vertical axis; outer radius from the far edge, inner from the near one.
4. $\displaystyle\int_0^1 \pi\Big((2 - y)^2 - (2 - \sqrt y)^2\Big)\,dy = \pi\int_0^1 \big(4\sqrt y - 5y + y^2\big)\,dy = \pi\Big(\tfrac83 - \tfrac52 + \tfrac13\Big)$$(4 - 4y + y^2) - (4 - 4\sqrt y + y)$.
5. $\tfrac\pi2 \approx 1.571$, both ways$\tfrac83 + \tfrac13 = 3$ and $3 - \tfrac52 = \tfrac12$. The solid is three times the one about the $y$-axis (Problem 6), because the region is farther from this axis.

Problem 8

1. The axis is horizontal, so use washers in $x$. Every point of the region is above $y = -1$, so a point at height $y$ is $y + 1$ from the axis.A radius is a distance from the axis, not a $y$-coordinate.
2. $R = x + 1$ and $r = x^2 + 1$.The line is the far edge and the parabola the near one.
3. $V = \displaystyle\int_0^1 \pi\Big((x + 1)^2 - (x^2 + 1)^2\Big)\,dx = \pi\int_0^1 \big(2x - x^2 - x^4\big)\,dx$$(x^2 + 2x + 1) - (x^4 + 2x^2 + 1)$.
4. $\pi\Big(1 - \tfrac13 - \tfrac15\Big)$Integrate term by term.
5. $\tfrac{7\pi}{15} \approx 1.466$$\tfrac{15 - 5 - 3}{15}$. More than three times the volume about the $x$-axis (Problem 4), $\tfrac{2\pi}{15}$.

Problem 9

1. In $y$, the region runs from $x = y^2$ on the left to $x = y + 2$ on the right, for $0 \le y \le 2$.The same region as the area-between-curves page's Problem 5: the curves meet at $(4, 2)$.
2. Shells about the $x$-axis are horizontal strips: length $(y + 2) - y^2$, distance $y$ from the axis.Strips parallel to the axis.
3. $V = \displaystyle\int_0^2 2\pi y\,(y + 2 - y^2)\,dy = 2\pi\Big[\tfrac{y^3}3 + y^2 - \tfrac{y^4}4\Big]_0^2 = 2\pi\Big(\tfrac83 + 4 - 4\Big)$One integral of a polynomial.
4. Washers in $x$: on $[0, 2]$ the slice is a disk of radius $\sqrt x$; on $[2, 4]$ it is a washer with $R = \sqrt x$ and $r = x - 2$.The bottom of the region changes at $x = 2$, where the line crosses the axis, so washers need two integrals.
5. $\displaystyle\int_0^2 \pi x\,dx + \int_2^4 \pi\big(x - (x - 2)^2\big)\,dx = 2\pi + \pi\Big(6 - \tfrac83\Big) = 2\pi + \tfrac{10\pi}3$$\int_2^4 x\,dx = 6$ and $\int_2^4 (x - 2)^2\,dx = \tfrac83$.
6. $\tfrac{16\pi}3 \approx 16.76$, both waysPick the method that keeps the region one piece in the variable you integrate.

Problem 10

1. The base runs from $x = -1$ to $x = 1$, and at $x$ it is $1 - x^2$ tall.$1 - x^2 = 0$ at $x = \pm1$.
2. The cross-section at $x$ is a square of side $1 - x^2$, area $A(x) = (1 - x^2)^2$.Nothing is rotated: the slice's area comes from the shape it is told to be.
3. $V = \displaystyle\int_{-1}^1 \big(1 - 2x^2 + x^4\big)\,dx = 2\Big(1 - \tfrac23 + \tfrac15\Big)$The integrand is even, so double the integral from 0 to 1.
4. $\tfrac{16}{15} \approx 1.067$$2\cdot\tfrac{8}{15}$. With semicircles of diameter $1 - x^2$ instead, $A = \tfrac\pi8(1 - x^2)^2$ and the volume is $\tfrac\pi8\cdot\tfrac{16}{15} = \tfrac{2\pi}{15}$.