Corrected exercises: volumes by slicing, disks and washers (MATH 141)

The slice at $x$ is a disk of radius $\sin x$, the height of the region above the axis, so $dV = \pi \sin^2 x\,dx$. Then $V = \pi \int_0^\pi \sin^2 x\,dx$. An even power of sine calls for the half-angle formula: $\sin^2 x = \frac{1 - \cos 2x}{2}$, so $V = \frac{\pi}{2} \int_0^\pi (1 - \cos 2x)\,dx = \frac{\pi}{2}\left[x - \frac{\sin 2x}{2}\right]_0^\pi = \frac{\pi}{2}(\pi - 0) = \frac{\pi^2}{2}$. The integral is set up in $x$ because the axis is horizontal: the slices are perpendicular to the axis, so they are stacked ALONG it, and the variable that moves along the $x$-axis is $x$. Rough size: $\frac{3.14^2}{2} \approx 4.9$ cubic units.

Radius $\sec x$, so $V = \pi \int_0^{\pi/3} \sec^2 x\,dx = \pi \left[\tan x\right]_0^{\pi/3} = \pi\left(\sqrt{3} - 0\right) = \pi\sqrt{3}$. Here squaring the radius is exactly what makes the integral immediate: $\sec x$ alone would need the $\int \sec x\,dx = \ln|\sec x + \tan x|$ formula, $\sec^2 x$ is a derivative of the table. That is the general lesson of the chapter: the integrand of a disk volume is $f^2$, and it is $f^2$, not $f$, whose technique you must choose.

Radius $\frac{1}{\sqrt{4 - x^2}}$, so the square is $\frac{1}{4 - x^2}$, a RATIONAL function: the root has disappeared, and with it any reason for an arcsine. The denominator factors, $4 - x^2 = (2 - x)(2 + x)$, and the decomposition, posed before looking for its constants, is $\frac{1}{4 - x^2} = \frac{A}{2 - x} + \frac{B}{2 + x}$. Clearing denominators, $1 = A(2 + x) + B(2 - x)$; $x = 2$ gives $A = \frac{1}{4}$ and $x = -2$ gives $B = \frac{1}{4}$. So $V = \frac{\pi}{4} \int_0^1 \left(\frac{1}{2 - x} + \frac{1}{2 + x}\right)dx = \frac{\pi}{4}\left[\ln\frac{2 + x}{2 - x}\right]_0^1 = \frac{\pi}{4}(\ln 3 - \ln 1) = \frac{\pi}{4}\ln 3$. The answer $\frac{\pi^2}{6}$ is $\pi \int_0^1 \frac{dx}{\sqrt{4 - x^2}} = \pi \arcsin\frac{1}{2} = \pi \cdot \frac{\pi}{6}$: the student saw the familiar arcsine form in $f$ and integrated $f$ itself, forgetting that the slice is a disk. The check: $\frac{\pi}{4}\ln 3 \approx 0.86$ and $\frac{\pi^2}{6} \approx 1.64$ are not even close.

The first student forgot to square: $\pi \int \sin x\,dx$ is $\pi$ times the AREA of the region. If $x$ and $y$ are lengths in cm, $\int \sin x\,dx$ is in cm$^2$, and multiplying by the pure number $\pi$ leaves cm$^2$: not a volume. The second squared the wrong thing: $\left(\int \sin x\,dx\right)^2$ is an area squared, in cm$^4$. The volume $\int \pi f(x)^2\,dx$ is in cm$^2 \times$ cm $=$ cm$^3$, because the square is taken SLICE BY SLICE, before the sum: each slice is a disk, $\pi r^2$, and only then are the slices added. The sum of the squares is not the square of the sum, and here $2\pi$, $4\pi$ and $\frac{\pi^2}{2}$ are three different numbers.

The solid fits inside the cylinder of radius $\max \sin x = 1$ and length $\pi$, of volume $\pi \cdot 1^2 \cdot \pi = \pi^2$. Our answer $\frac{\pi^2}{2}$ is smaller, as it must be, and the ratio is exactly $\frac{1}{2}$ because $V = \pi \cdot \pi \cdot \left(\frac{1}{\pi}\int_0^\pi \sin^2 x\,dx\right)$ and the average value of $\sin^2 x$ over a half period is $\frac{1}{2}$. A bounding cylinder costs five seconds and catches a missing $\pi$ or a missing square every time: a disk volume larger than its bounding cylinder is always wrong.

$2x = x^2$ gives $x(x - 2) = 0$, so $x = 0$ or $x = 2$, at the points $(0, 0)$ and $(2, 4)$. At $x = 1$, $2x = 2 > 1 = x^2$, so the line is on top on $(0, 2)$. Testing one point between the intersections is the whole justification: a polynomial difference $2x - x^2$ cannot change sign between two consecutive roots.

The axis is horizontal, so the slices are vertical and the variable is $x$. At $x$, the slice goes from the parabola up to the line. Measured from the axis $y = 0$, the far end is at distance $R(x) = 2x$ and the near end at distance $r(x) = x^2$. Then $V = \pi \int_0^2 \left[(2x)^2 - (x^2)^2\right]dx = \pi \int_0^2 (4x^2 - x^4)\,dx = \pi\left[\frac{4x^3}{3} - \frac{x^5}{5}\right]_0^2 = \pi\left(\frac{32}{3} - \frac{32}{5}\right) = \frac{64\pi}{15}$. The integrand $4x^2 - x^4 = x^2(4 - x^2)$ is $\ge 0$ on $[0, 2]$, as $R^2 - r^2$ must be.

$\pi \int_0^2 (2x - x^2)^2\,dx = \pi \int_0^2 (4x^2 - 4x^3 + x^4)\,dx = \pi\left(\frac{32}{3} - 16 + \frac{32}{5}\right) = \frac{16\pi}{15}$, a quarter of the true volume. His slice at $x = 1$ has area $\pi(2 - 1)^2 = \pi$, the area of a disk of radius $1$. The true slice there is the washer between radii $2$ and $1$, of area $\pi(4 - 1) = 3\pi$. The expression $\pi(R - r)^2$ treats the THICKNESS of the region as a radius, which is what you get if you slide the region down onto the axis before turning it: a smaller solid. Square each radius, then subtract: $R^2 - r^2 = (R - r)(R + r)$, and the factor $R + r$ he lost is exactly what the distance to the axis contributes.

With disks and washers the slice is always perpendicular to the axis. The axis is now vertical, so the slices are horizontal, stacked along the $y$-axis, and the variable is $y$, from $0$ to $4$. Solve each curve for $x$: the line $y = 2x$ becomes $x = \frac{y}{2}$ and the parabola, on the right half, becomes $x = \sqrt{y}$. At height $y$, the slice runs from $x = \frac{y}{2}$ to $x = \sqrt{y}$, and since $\sqrt{y} \ge \frac{y}{2}$ on $[0, 4]$ (at $y = 1$: $1 > \frac{1}{2}$), the far end from the $y$-axis is the PARABOLA: $R(y) = \sqrt{y}$, $r(y) = \frac{y}{2}$. Then $V = \pi \int_0^4 \left(y - \frac{y^2}{4}\right)dy = \pi\left[\frac{y^2}{2} - \frac{y^3}{12}\right]_0^4 = \pi\left(8 - \frac{16}{3}\right) = \frac{8\pi}{3}$. The same region, a different axis, and the roles of the curves are REVERSED: about the $x$-axis the line was outer, about the $y$-axis the parabola is. That is the thread of the chapter: a radius is read from the axis, never from the words on top.

With $R = \frac{y}{2}$ and $r = \sqrt{y}$ she computes $\pi \int_0^4 \left(\frac{y^2}{4} - y\right)dy = -\frac{8\pi}{3}$. A volume cannot be negative, and her integrand $\frac{y^2}{4} - y$ is negative on the whole interval $(0, 4)$: every one of her washers has an inner radius larger than the outer one. The sign is the free alarm of the washer method. Taking the absolute value at the end would give the right number for the wrong reason and lose the method marks; the fix is to redo the radii from the axis.

The axis $y = -1$ lies BELOW the region. At $x$, the slice runs from $y = 0$ to $y = 4 - x^2$. Distances to the axis: to the top, $R(x) = (4 - x^2) - (-1) = 5 - x^2$; to the bottom, $r(x) = 0 - (-1) = 1$. The region is separated from the axis by a strip of width $1$, so the slices are washers even though the region sits on the $x$-axis. Then $V = \pi \int_{-2}^{2} \left[(5 - x^2)^2 - 1\right]dx = \pi \int_{-2}^{2} (24 - 10x^2 + x^4)\,dx$. The integrand is even, so $V = 2\pi \int_0^2 (24 - 10x^2 + x^4)\,dx = 2\pi\left(48 - \frac{80}{3} + \frac{32}{5}\right) = 2\pi \cdot \frac{720 - 400 + 96}{15} = \frac{832\pi}{15}$. Treating it as a disk of radius $4 - x^2$ gives the volume about the $x$-axis, $\frac{512\pi}{15}$, a different solid.

The axis $y = 4$ lies ABOVE the region and touches it at the vertex. Now the far boundary from the axis is the BOTTOM of the slice: $R(x) = 4 - 0 = 4$. The near boundary is the arch: $r(x) = 4 - (4 - x^2) = x^2$. So $V = \pi \int_{-2}^{2} (16 - x^4)\,dx = 2\pi\left(32 - \frac{32}{5}\right) = \frac{256\pi}{5}$. The student who writes $r = 4 - x^2$ measured the HEIGHT of the arch above the $x$-axis, the distance to the wrong line. With the axis above the region, the lower curve gives the outer radius: that inversion is the classic loss of this question, and it is invisible unless each radius is written as axis minus curve (or curve minus axis) before simplifying.

The axis is vertical, so the slices are horizontal and the variable is $y \in [0, 4]$. Solve $y = 4 - x^2$ for $x$: $x = \pm\sqrt{4 - y}$, and the slice at height $y$ runs from $x = -\sqrt{4 - y}$ to $x = \sqrt{4 - y}$. Both ends are to the LEFT of $x = 2$, so the distances are $R(y) = 2 - (-\sqrt{4 - y}) = 2 + \sqrt{4 - y}$ and $r(y) = 2 - \sqrt{4 - y}$. Write $s = \sqrt{4 - y}$: $R^2 - r^2 = (2 + s)^2 - (2 - s)^2 = 8s$. So $V = 8\pi \int_0^4 \sqrt{4 - y}\,dy$, and the substitution $u = 4 - y$, $du = -dy$, with $y = 0 \to u = 4$ and $y = 4 \to u = 0$, gives $8\pi \int_0^4 \sqrt{u}\,du = 8\pi \cdot \frac{2}{3} \cdot 4^{3/2} = 8\pi \cdot \frac{16}{3} = \frac{128\pi}{3}$. Expanding $R^2$ and $r^2$ separately works too, but the difference of squares removes every root in one line: when both radii share a term, factor $R^2 - r^2 = (R - r)(R + r)$ first.

The inner radius says it. In a) $r = 1$ for every $x$: the solid is pierced by a tunnel of radius $1$ all along the axis, a thick ring. In b) $r = x^2$ vanishes only at $x = 0$: the arch touches the axis at its vertex, so the hole pinches to a point there, and the solid is a spool whose waist closes. In c) $r = 2 - \sqrt{4 - y}$ vanishes only at $y = 0$, where the region touches $x = 2$ at the point $(2, 0)$: again the hole closes at one point, at the base. A region that touches the axis along a whole segment gives disks; a region that touches it at isolated points gives washers whose hole closes there; a region separated from the axis gives a genuine tunnel.

The curve $y = \ln x$ runs from $(1, 0)$ to $(e, 1)$, so the heights in $\mathcal{R}$ go from $0$ to $1$. At height $y$, the left end of the slice is on the curve, where $\ln x = y$, that is $x = e^y$, and the right end is on the line $x = e$. So $\mathcal{R} = \{(x, y) : 0 \le y \le 1,\ e^y \le x \le e\}$. The limits in $y$ are $0$ and $1$, NOT $1$ and $e$: those are the $x$ limits, and carrying them into a $dy$ integral is the most frequent set-up error of the chapter.

Distances to the $y$-axis at height $y$: the far end is the line, $R(y) = e$, the near end is the curve, $r(y) = e^y$. The region never touches the $y$-axis, so the solid has a tunnel: washers. $V = \pi \int_0^1 (e^2 - e^{2y})\,dy = \pi\left[e^2 y - \frac{e^{2y}}{2}\right]_0^1 = \pi\left(e^2 - \frac{e^2}{2} + \frac{1}{2}\right) = \frac{\pi(e^2 + 1)}{2}$. Size check with $e \approx 2.72$: $\frac{3.14 \times 8.4}{2} \approx 13$, less than the cylinder of radius $e$ and height $1$, $\pi e^2 \approx 23$.

Horizontal axis, vertical slices, disks of radius $\ln x$ for $1 \le x \le e$: $V = \pi \int_1^e (\ln x)^2\,dx$. Integrate by parts with $u = (\ln x)^2$, $dv = dx$, so $du = \frac{2\ln x}{x}\,dx$ and $v = x$: $\int (\ln x)^2\,dx = x(\ln x)^2 - 2\int \ln x\,dx$. Again $u = \ln x$, $dv = dx$: $\int \ln x\,dx = x\ln x - x$. Hence a primitive is $x(\ln x)^2 - 2x\ln x + 2x$, whose derivative is $(\ln x)^2 + 2\ln x - 2\ln x - 2 + 2 = (\ln x)^2$. Evaluate: at $e$, $e - 2e + 2e = e$; at $1$, $0 - 0 + 2 = 2$. So $V = \pi(e - 2) \approx 2.26$. The parts are forced: $dv = dx$ is the only choice that makes $\ln x$ disappear by differentiation.

The axis $x = e$ is the right edge of $\mathcal{R}$, and the region touches it along the whole segment from $(e, 0)$ to $(e, 1)$: every horizontal slice has one end ON the axis, so there is no hole and the slices are disks of radius $e - e^y$. $V = \pi \int_0^1 (e - e^y)^2\,dy = \pi \int_0^1 (e^2 - 2e\,e^y + e^{2y})\,dy = \pi\left(e^2 - 2e(e - 1) + \frac{e^2 - 1}{2}\right) = \frac{\pi}{2}\left(4e - e^2 - 1\right)$. Size check: $4e - e^2 - 1 \approx 10.87 - 7.39 - 1 = 2.48$, so $V \approx 3.9$, positive as it must be.

The axis $y = 1$ lies above $\mathcal{R}$. Vertical slices from $y = 0$ to $y = \ln x$: the FAR end from the axis is the bottom, $R(x) = 1 - 0 = 1$, the near end is the curve, $r(x) = 1 - \ln x$. Then $R^2 - r^2 = 1 - (1 - \ln x)^2 = 2\ln x - (\ln x)^2$, and $V = \pi\left(2\int_1^e \ln x\,dx - \int_1^e (\ln x)^2\,dx\right) = \pi\left(2 \cdot 1 - (e - 2)\right) = \pi(4 - e)$, using $\int_1^e \ln x\,dx = [x\ln x - x]_1^e = 0 - (-1) = 1$ and the integral of c). Once more the lower boundary gives $R$ because the axis is above: the height of the curve is never a radius by itself.

Put the base at height $0$ and the apex at height $4$, and slice perpendicular to the vertical axis. The slices are squares, and their side decreases linearly from $6$ at $y = 0$ to $0$ at $y = 4$: by similar triangles, $s(y) = 6\left(1 - \frac{y}{4}\right)$. So $A(y) = 36\left(1 - \frac{y}{4}\right)^2$ and $V = \int_0^4 36\left(1 - \frac{y}{4}\right)^2dy$. With $u = 1 - \frac{y}{4}$, $du = -\frac{dy}{4}$, the bounds become $1$ and $0$: $V = 36 \cdot 4 \int_0^1 u^2\,du = 144 \cdot \frac{1}{3} = 48$. And $\frac{1}{3} \times 36 \times 4 = 48$: the one third of the formula is the $\int_0^1 u^2\,du$ of every slicing proof.

The slice at $x$ stands on the chord of the disk at abscissa $x$, which runs from $y = -\sqrt{4 - x^2}$ to $y = \sqrt{4 - x^2}$: its length is $2\sqrt{4 - x^2}$, the WHOLE chord. For an isosceles right triangle with hypotenuse $h$, each leg is $\frac{h}{\sqrt{2}}$ and the area is $\frac{1}{2}\left(\frac{h}{\sqrt{2}}\right)^2 = \frac{h^2}{4}$. So $A(x) = \frac{4(4 - x^2)}{4} = 4 - x^2$, and $V = \int_{-2}^{2} (4 - x^2)\,dx = 2\left(8 - \frac{8}{3}\right) = \frac{32}{3}$. No $\pi$ in the answer, and none should be expected: the base is round, but the slices are triangles.

Now the leg is the chord, $\ell = 2\sqrt{4 - x^2}$, and the area is $\frac{\ell^2}{2} = 2(4 - x^2)$. So $V = 2 \cdot \frac{32}{3} = \frac{64}{3}$. The ratio is $2$ at EVERY $x$, since $\frac{\ell^2/2}{h^2/4} = 2$ for the same chord, and integrating a ratio that is constant slice by slice keeps it. Reading which side lies on the base is therefore worth the whole question: the two statements differ by one word.

A semicircle on the chord has radius $\sqrt{4 - x^2}$, half the chord, and area $\frac{1}{2}\pi\left(\sqrt{4 - x^2}\right)^2 = \frac{\pi}{2}(4 - x^2)$. So $V = \frac{\pi}{2} \cdot \frac{32}{3} = \frac{16\pi}{3}$. The same polynomial $4 - x^2$ appears in b), c) and d): only the SHAPE constant changes ($\frac{1}{4}$, $\frac{1}{2}$, $\frac{\pi}{8}$ times the chord squared). Computing $\int_{-2}^{2}(4 - x^2)\,dx = \frac{32}{3}$ once and multiplying is the efficient way through such a question.