Corrected exercises: volumes by cylindrical shells (MATH 141)

The strip at $x$ is PARALLEL to the axis of rotation, the $y$-axis, so turning it sweeps a thin cylindrical shell. Its radius is the distance from the strip to the axis: the axis is the line $x = 0$ and the strip stands at $x \ge 0$, so the radius is $x$. Its height is the length of the strip, from the $x$-axis up to the curve, $f(x) = x(x - 3)^2$. Its thickness is $\Delta x$. Cut the shell along a vertical line and unroll it: it becomes a thin slab whose length is the circumference $2\pi x$, whose height is $f(x)$ and whose thickness is $\Delta x$, so its volume is about $2\pi x \, f(x)\,\Delta x$. Adding the shells and letting $\Delta x \to 0$ gives $V = \int_0^3 2\pi x f(x)\,dx$. The three factors were READ on the figure; memorizing $2\pi\int x f(x)\,dx$ as a string of letters is exactly what fails as soon as the axis moves.

$V = 2\pi\int_0^3 x \cdot x(x - 3)^2\,dx$. Expand first: $x^2(x^2 - 6x + 9) = x^4 - 6x^3 + 9x^2$. Then $V = 2\pi\left[\frac{x^5}{5} - \frac{3x^4}{2} + 3x^3\right]_0^3$, and differentiating the bracket gives back $x^4 - 6x^3 + 9x^2$. At $x = 3$: $\frac{243}{5} - \frac{243}{2} + 81 = \frac{486 - 1215 + 810}{10} = \frac{81}{10}$, and the bracket is $0$ at $x = 0$. So $V = 2\pi \cdot \frac{81}{10} = \frac{81\pi}{5}$. Expanding the polynomial before integrating is faster than any substitution here, and it keeps the arithmetic in fractions a marker can follow.

A washer comes from a HORIZONTAL slice, perpendicular to the axis, and its radii are the $x$-values where the line $y = c$ crosses the boundary. Now $f'(x) = (x - 3)^2 + 2x(x - 3) = 3(x - 3)(x - 1)$: $f$ rises from $0$ to $f(1) = 4$ on $[0, 1]$, then falls back to $0$ on $[1, 3]$. So every line $y = c$ with $0 < c < 4$ meets the curve TWICE inside the region, at an inner radius and an outer radius, which are two roots of the cubic $x^3 - 6x^2 + 9x - c = 0$. Writing those roots as formulas in $c$ means solving a general cubic, which nobody does on a MATH 141 paper. Shells never ask for $x$ as a function of $y$: that is exactly why the method exists.

First: $2\pi\int_0^3 (x^3 - 6x^2 + 9x)\,dx = 2\pi\left[\frac{x^4}{4} - 2x^3 + \frac{9x^2}{2}\right]_0^3 = 2\pi\left(\frac{81}{4} - 54 + \frac{81}{2}\right) = 2\pi \cdot \frac{27}{4} = \frac{27\pi}{2}$. It cannot be a volume: with $x$ and $y$ in centimetres, $f(x)\,dx$ is in square centimetres, and so is $2\pi\int f(x)\,dx$. The dropped factor is the RADIUS, a length. A shell integrand is always a product of three lengths, radius times height times thickness, and counting them takes five seconds. Second: $\pi\int_0^3 x f(x)\,dx = \frac{81\pi}{10}$, exactly half the true value. The units are right, but the unrolled length of a shell is the circumference $2\pi x$, not $\pi x$: the $\pi r^2$ of the area of a disk has leaked into the wrong formula. Each slip costs most of the marks, because the set-up is where they are.

The region fits in the rectangle $[0, 3] \times [0, 4]$, so the solid fits in the cylinder of radius $3$ and height $4$, of volume $\pi \cdot 3^2 \cdot 4 = 36\pi$. And $\frac{81}{5} = 16.2 < 36$: consistent, and a ratio near one half is plausible for a ring-shaped ridge that peaks at radius $1$ and slopes down to the rim. A value above $36\pi$, a negative value, or a value without $\pi$ would prove the set-up wrong before any recomputation.

Shells about the $x$-axis use strips PARALLEL to the $x$-axis, so horizontal strips, and the variable is the one perpendicular to the axis, $y$. Every boundary must therefore be written as $x$ in terms of $y$. From $y = \sqrt{x - 1}$ with $y \ge 0$: $y^2 = x - 1$, so $x = 1 + y^2$, the LEFT end of the strip. The RIGHT end is the line $x = 5$. The strips run from $y = 0$ up to the height where the curve meets $x = 5$, that is $y = \sqrt{4} = 2$. So $0 \le y \le 2$.

The radius is the distance from the strip at height $y$ to the $x$-axis: $y$. The height of the shell is the length of the strip, right end minus left end: $5 - (1 + y^2) = 4 - y^2$. So $V = 2\pi\int_0^2 y(4 - y^2)\,dy = 2\pi\left[2y^2 - \frac{y^4}{4}\right]_0^2 = 2\pi(8 - 4) = 8\pi$. The word height is misleading here, since the strip lies flat: it is the length of the cylinder, measured along the axis.

With disks, the slice is perpendicular to the axis: a vertical slice at $x$ sweeps a disk of radius $\sqrt{x - 1}$. $V = \pi\int_1^5 (x - 1)\,dx = \pi\left[\frac{(x - 1)^2}{2}\right]_1^5 = \pi \cdot 8 = 8\pi$. The same solid gives the same number, as it must. Here disks were even quicker: shells are a choice, not an obligation, and exercise 6 turns that choice into a rule.

With $u = x - 1$, $du = dx$, $x = u + 1$ and bounds $0$ to $4$: $\int_1^5 x\sqrt{x - 1}\,dx = \int_0^4 (u + 1)u^{1/2}\,du = \left[\frac{2}{5}u^{5/2} + \frac{2}{3}u^{3/2}\right]_0^4 = \frac{64}{5} + \frac{16}{3} = \frac{272}{15}$. The student's number is $\frac{544\pi}{15}$. The integral is perfectly correct for ANOTHER solid: a vertical strip at $x$, radius $x$, height $\sqrt{x - 1}$, is a shell about the $y$-axis. The orientation of the strip decides the axis: a shell strip is parallel to the axis it turns around. Vertical strips mean a vertical axis, whatever the question says, and the whole question is lost.

$2\pi\int_0^2 (y + y^3)\,dy = 2\pi\left[\frac{y^2}{2} + \frac{y^4}{4}\right]_0^2 = 2\pi(2 + 4) = 12\pi$. She took the height of the strip to be the $x$-coordinate of the curve, $1 + y^2$, which is the distance from the $y$-axis to the curve: that is the strip of the region to the LEFT of the curve, between $x = 0$ and $x = 1 + y^2$. The two regions together fill the rectangle $[0, 5] \times [0, 2]$, and indeed $8\pi + 12\pi = 20\pi = \pi \cdot 2^2 \cdot 5$, the cylinder of radius $2$ and length $5$. The height of a horizontal strip is always right end minus left end, read on the figure.

Vertical strips, parallel to the vertical axis, at $0 \le x \le 2$. Radius $x$, height $4 - x^2$. $V = 2\pi\int_0^2 x(4 - x^2)\,dx = 2\pi\left[2x^2 - \frac{x^4}{4}\right]_0^2 = 2\pi(8 - 4) = 8\pi$.

The whole region lies to the LEFT of the line $x = 3$, since $x \le 2$. The distance from the strip at $x$ to the axis is therefore $3 - x$, a positive number between $1$ and $3$. $V = 2\pi\int_0^2 (3 - x)(4 - x^2)\,dx = 2\pi\int_0^2 (12 - 4x - 3x^2 + x^3)\,dx = 2\pi\left[12x - 2x^2 - x^3 + \frac{x^4}{4}\right]_0^2 = 2\pi(24 - 8 - 8 + 4) = 24\pi$. With the radius $x - 3$ the integrand changes sign and the answer is $-24\pi$. A negative volume is not a small solid: it is the proof that the radius is not a distance. The rule: radius equals the larger coordinate minus the smaller one, axis or strip, whichever is larger on the figure.

The region lies to the RIGHT of $x = -1$. The distance is $x - (-1) = x + 1$. $V = 2\pi\int_0^2 (x + 1)(4 - x^2)\,dx = 2\pi\left[2x^2 - \frac{x^4}{4} + 4x - \frac{x^3}{3}\right]_0^2 = 2\pi\left(8 - 4 + 8 - \frac{8}{3}\right) = 2\pi \cdot \frac{28}{3} = \frac{56\pi}{3}$. The classic slip is to read the line $x = -1$ as subtract $1$: the radius $x - 1$ gives $2\pi\left(4 - 8 + \frac{8}{3}\right) = -\frac{8\pi}{3}$, negative again, because $x - 1$ is not even the right size, let alone the right sign.

$A = \int_0^2 (4 - x^2)\,dx = 8 - \frac{8}{3} = \frac{16}{3}$. By linearity, $2\pi\int_0^2 (x + 1)h(x)\,dx = 2\pi\int_0^2 x\,h(x)\,dx + 2\pi\int_0^2 h(x)\,dx = 8\pi + 2\pi A = 8\pi + \frac{32\pi}{3} = \frac{56\pi}{3}$, which is c). In the same way $2\pi\int_0^2 (3 - x)h(x)\,dx = 6\pi A - 8\pi = 32\pi - 8\pi = 24\pi$, which is b). Moving the axis a distance $d$ further from the region adds $2\pi d A$ to the volume: every shell gets a radius longer by $d$. This costs one line and checks two answers at once.

Compare the two radii strip by strip: $3 - x > x + 1$ exactly when $x < 1$. On the part $0 \le x < 1$ the strips are the TALL ones, with heights between $3$ and $4$, and there the axis $x = 3$ gives the larger radius; for $x > 1$ the strips are short and the advantage goes to $x = -1$, but on much less height. The tall strips win: $24\pi = \frac{72\pi}{3} > \frac{56\pi}{3}$. The same comparison explains why both exceed $8\pi = \frac{24\pi}{3}$, every radius being larger than $x$.

The parabola meets the line where $y^2 = 4$, at $y = -2$ and $y = 2$, so $-2 \le y \le 2$. The strip at height $y$ goes from $x = y^2$ on the left to $x = 4$ on the right: its length is $4 - y^2$. The axis $y = 3$ lies ABOVE every point of $R$, since $y \le 2 < 3$, so the distance from the strip to the axis is $3 - y$. No separate formula is needed below the $x$-axis: for $y = -2$ the radius is $3 - (-2) = 5$, larger than for $y = 2$, which is exactly what the figure shows. The $x$-axis plays no role at all in this question; only the axis of rotation does.

$V = 2\pi\int_{-2}^{2} (3 - y)(4 - y^2)\,dy = 2\pi\left[3\int_{-2}^{2} (4 - y^2)\,dy - \int_{-2}^{2} y(4 - y^2)\,dy\right]$. The second integrand is ODD, $g(-y) = -g(y)$, on an interval symmetric about $0$, so its integral is $0$. The first is $3\left[4y - \frac{y^3}{3}\right]_{-2}^{2} = 3\left(\frac{16}{3} + \frac{16}{3}\right) = 32$. So $V = 2\pi \cdot 32 = 64\pi$. Using the symmetry must be SAID on the paper: the odd function and the symmetric interval are the two hypotheses, and naming them earns the step.

The zero is a statement about a SIGNED integral, not about a solid. In $\int_{-2}^{2} y(4 - y^2)\,dy$ the factor $y$ is negative below the $x$-axis, so there it is not a distance, and the two halves cancel. Inside b) this is harmless: it is the odd part of $(3 - y)(4 - y^2)$, and dropping it is the symmetry shortcut. Read as a VOLUME about the $x$-axis it is meaningless: that axis cuts through $R$, the lower half and the upper half sweep the same solid, and the volume is $2\pi\int_0^2 y(4 - y^2)\,dy = 8\pi$, from the upper half alone. A volume can be $0$ only for a region of zero area: a zero or a negative number always points at a radius that is not a distance.

Washers come from slices perpendicular to the axis, so vertical slices at $0 \le x \le 4$, from $y = -\sqrt{x}$ to $y = \sqrt{x}$. Measured from the axis $y = 3$, the far end is at distance $3 + \sqrt{x}$ and the near end at $3 - \sqrt{x}$: outer and inner radii. $V = \pi\int_0^4 \left[(3 + \sqrt{x})^2 - (3 - \sqrt{x})^2\right]dx = \pi\int_0^4 12\sqrt{x}\,dx = 12\pi \cdot \frac{2}{3} \cdot 8 = 64\pi$. The two methods agree.

The line $y = -2$ lies BELOW $R$ (it touches it only at the corner $(4, -2)$), so the radius is $y - (-2) = y + 2$. $V = 2\pi\int_{-2}^{2} (y + 2)(4 - y^2)\,dy = 2\pi\left[0 + 2 \cdot \frac{32}{3}\right] = \frac{128\pi}{3}$, the odd part vanishing again. Check by linearity, as in exercise 3: the two radii differ by $(3 - y) - (y + 2) = 1 - 2y$, and $2\pi\int_{-2}^{2} (1 - 2y)(4 - y^2)\,dy = 2\pi\left(\frac{32}{3} - 0\right) = \frac{64\pi}{3}$, which is indeed $64\pi - \frac{128\pi}{3}$.

$x = x^2 - 2x$ gives $x^2 - 3x = 0$, so $x = 0$ or $x = 3$. At $x = 1$ the line gives $1$ and the parabola $-1$: the line is on top on $[0, 3]$. Height of the strip: $h(x) = x - (x^2 - 2x) = 3x - x^2$, positive on $(0, 3)$ as a length must be.

Radius $x$, height $3x - x^2$: $V = 2\pi\int_0^3 (3x^2 - x^3)\,dx = 2\pi\left[x^3 - \frac{x^4}{4}\right]_0^3 = 2\pi\left(27 - \frac{81}{4}\right) = 2\pi \cdot \frac{27}{4} = \frac{27\pi}{2}$.

The first student is transferring a reflex from signed area to volume. The $x$-axis is not the axis of rotation here, so it is just a line drawn across the region: the strip at $x$ is ONE segment, from $y = x^2 - 2x$ up to $y = x$, and it sweeps one shell whose height is its full length. Nothing is subtracted, and splitting at the $x$-axis only multiplies the chances of a slip. The second student lost the parentheses around the bottom curve: $x - x^2 - 2x = -x - x^2$, a height that is NEGATIVE on the whole interval. His integral is $2\pi\int_0^3 x(-x - x^2)\,dx = 2\pi\left(-9 - \frac{81}{4}\right) = -\frac{117\pi}{2}$. A negative volume is the alarm; writing the bottom curve in parentheses, $x - (x^2 - 2x)$, is the habit that prevents it.

The region lies to the right of $x = -1$, so the radius is $x + 1$. By linearity, $2\pi\int_0^3 (x + 1)h(x)\,dx = \frac{27\pi}{2} + 2\pi A$, with $A = \int_0^3 (3x - x^2)\,dx = \frac{27}{2} - 9 = \frac{9}{2}$. So $V = \frac{27\pi}{2} + 9\pi = \frac{45\pi}{2}$.

The height is symmetric about $x = \frac{3}{2}$: $h(3 - x) = 3(3 - x) - (3 - x)^2 = (3 - x)x = h(x)$. For the axis $x = 4$ the radius is $4 - x$, and the substitution $x = 3 - t$ turns $\int_0^3 (4 - x)h(x)\,dx$ into $\int_0^3 (1 + t)h(t)\,dt$, the integral of d). So the two volumes are equal: $\frac{45\pi}{2}$. Geometrically, both lines stand $\frac{5}{2}$ from the line of symmetry of $R$, one on each side. Direct check: $2\pi\int_0^3 (4 - x)h(x)\,dx = 8\pi A - \frac{27\pi}{2} = 36\pi - \frac{27\pi}{2} = \frac{45\pi}{2}$.

Washers about the $y$-axis come from horizontal slices at height $y$, $0 \le y \le 2$, with outer radius $1$ and inner radius the $x$ such that $x^3 + x = y$. The function $x^3 + x$ is increasing ($3x^2 + 1 > 0$), so that $x$ exists, but writing it requires solving a cubic in $x$ for every $y$: no formula a MATH 141 student can use. Shells need no inverse. Vertical strips, radius $x$, height $x^3 + x$: $V = 2\pi\int_0^1 (x^4 + x^2)\,dx = 2\pi\left(\frac{1}{5} + \frac{1}{3}\right) = 2\pi \cdot \frac{8}{15} = \frac{16\pi}{15}$.

About the $x$-axis the vertical slice is now PERPENDICULAR to the axis: it sweeps a disk of radius $x^3 + x$. $V = \pi\int_0^1 (x^3 + x)^2\,dx = \pi\int_0^1 (x^6 + 2x^4 + x^2)\,dx = \pi\left(\frac{1}{7} + \frac{2}{5} + \frac{1}{3}\right) = \pi \cdot \frac{15 + 42 + 35}{105} = \frac{92\pi}{105}$. Shells here would need horizontal strips, hence the inverse again. Same region, two axes, two different methods, and one reason in both cases: $R_1$ is easy in $x$ and impossible in $y$. Whatever the axis, the method to use is the one whose strip is VERTICAL.

Shells: vertical strips, radius $x$, height $\ln x$, so $V = 2\pi\int_1^e x\ln x\,dx$. By parts with $u = \ln x$, $dv = x\,dx$: $\int_1^e x\ln x\,dx = \left[\frac{x^2}{2}\ln x\right]_1^e - \int_1^e \frac{x}{2}\,dx = \frac{e^2}{2} - \frac{e^2 - 1}{4} = \frac{e^2 + 1}{4}$, so $V = \frac{\pi(e^2 + 1)}{2}$. Washers: $\ln x$ inverts to $x = e^y$, so the horizontal slice at $0 \le y \le 1$ runs from $x = e^y$ to $x = e$: outer radius $e$, inner radius $e^y$. $V = \pi\int_0^1 (e^2 - e^{2y})\,dy = \pi\left(e^2 - \frac{e^2 - 1}{2}\right) = \frac{\pi(e^2 + 1)}{2}$. Same volume. The washers cost one line; the shells cost an integration by parts. Rough size, from $e^2 \approx 7.4$: about $3.14 \times 4.2 \approx 13$.