Corrected exercises: volumes by cylindrical shells (MATH 141)

a) 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.

b) $V = 2\pi\int_0^3 x \cdot x(x - 3)^2\,dx$. 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.

c) 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.

d) 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.

e) 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.

a) 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 $0 \le y \le 2$.

b) 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.

c) 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.

d) 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.

e) $2\pi\int_0^2 (y + y^3)\,dy = 2\pi\left[\frac{y^2}{2} + \frac{y^4}{4}\right]_0^2 = 2\pi(2 + 4) = 12\pi$. 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.

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

b) 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.

c) 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.

d) $A = \int_0^2 (4 - x^2)\,dx = 8 - \frac{8}{3} = \frac{16}{3}$. 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.

e) 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 $x$.

a) 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.

b) $V = 2\pi\int_{-2}^{2} (3 - y)(4 - y^2)\,dy = 2\pi\left[3\int_{-2}^{2} (4 - y^2)\,dy - \int_{-2}^{2} y(4 - y^2)\,dy\right]$. 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.

c) 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.

d) 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.

e) 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 $64\pi - \frac{128\pi}{3}$.

a) $x = x^2 - 2x$ 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.

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

c) 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.

d) 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 $V = \frac{27\pi}{2} + 9\pi = \frac{45\pi}{2}$.

e) 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: $2\pi\int_0^3 (4 - x)h(x)\,dx = 8\pi A - \frac{27\pi}{2} = 36\pi - \frac{27\pi}{2} = \frac{45\pi}{2}$.

a) 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 $x^3 + x$: $V = 2\pi\int_0^1 (x^4 + x^2)\,dx = 2\pi\left(\frac{1}{5} + \frac{1}{3}\right) = 2\pi \cdot \frac{8}{15} = \frac{16\pi}{15}$.

b) 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.

c) 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 $3.14 \times 4.2 \approx 13$.

d) Shells about the $x$-axis: horizontal strips at $0 \le y \le 1$, from $x = e^y$ to $x = e$, length $e - e^y$, radius $y$. $V = 2\pi\int_0^1 y(e - e^y)\,dy = 2\pi\left(\frac{e}{2} - \int_0^1 ye^y\,dy\right)$, and by parts ($u = y$, $dv = e^y\,dy$) $\int_0^1 ye^y\,dy = \left[ye^y - e^y\right]_0^1 = 0 - (-1) = 1$. So $V = 2\pi\left(\frac{e}{2} - 1\right) = \pi(e - 2)$. Disks: $V = \pi\int_1^e (\ln x)^2\,dx$, which needs parts TWICE: $\int (\ln x)^2\,dx = x(\ln x)^2 - 2x\ln x + 2x$, so $V = \pi\left[(e - 2e + 2e) - 2\right] = \pi(e - 2)$. Same volume, and this time the shells are cheaper, one integration by parts against two.

e) Before choosing a method, ask in which variable the boundary curves are easy to write, and in which variable the resulting integral is easy to do. The axis only decides which strip each method uses: vertical strips give shells about a vertical axis and disks or washers about a horizontal one, horizontal strips the other way round. So pick the variable first, and let the axis tell you the name of the method.

a) Vertical strips, radius $x$, height $2x^2 - x^4$: $V = 2\pi\int_0^{\sqrt{2}} (2x^3 - x^5)\,dx = 2\pi\left[\frac{x^4}{2} - \frac{x^6}{6}\right]_0^{\sqrt{2}} = 2\pi\left(\frac{4}{2} - \frac{8}{6}\right) = 2\pi \cdot \frac{2}{3} = \frac{4\pi}{3}$. Powers of $\sqrt{2}$ are the only arithmetic: $(\sqrt{2})^4 = 4$ and $(\sqrt{2})^6 = 8$.

b) $2x^2 - x^4 = y$ is a quadratic in $X = x^2$: $X^2 - 2X + y = 0$, so $X = 1 \pm \sqrt{1 - y}$. For $0 < y < 1$ this gives TWO points of the boundary: $x_{in}^2 = 1 - \sqrt{1 - y}$ on the rising branch and $x_{out}^2 = 1 + \sqrt{1 - y}$ on the falling branch. The slice at height $y$ is a washer with area $\pi(x_{out}^2 - x_{in}^2) = 2\pi\sqrt{1 - y}$, and the squares of the radii are exactly what the quadratic gives, so no square root of a square root ever appears. $V = 2\pi\int_0^1 \sqrt{1 - y}\,dy = 2\pi\left[-\frac{2}{3}(1 - y)^{3/2}\right]_0^1 = \frac{4\pi}{3}$. The same number, but it took a quadratic in $x^2$, two branches and a substitution, where the shells took one line.