10Volumes by disks and washers

Independent check: the pyramid answer agrees with the classical $\tfrac13(\text{base})(\text{height})=32$, so the formula really did work on a non-rotational solid.

Rotation is a way of learning what the slice is. It is not what makes the integral legal.

Independent check with the identity $R^{2}-r^{2}=(R-r)(R+r)$: both bands have $R-r=1$, so each area is $\pi(R+r)$, giving $3\pi$ and $5\pi$. The gain is $\pi$ times the gain in $R+r$, which is $2$.

A thin ring of fixed width has area $\pi(R+r)$ times that width, so its area grows in a straight line with the distance to the axis — linearly, not quadratically.

Independent check by counting intersections: a horizontal line at height $y=0.2$ cuts the arch twice and cuts each of the other three graphs once, which is exactly what one branch against two means.

A boundary that is not monotone on the interval will always give more than one branch, and horizontal slicing needs the region between them.

Independent check on a case where it fails: writing $R=x^{2}$ and $r=\sqrt{x}$ on $[0,1]$ gives $x^{4}-x$, which is negative throughout — and indeed those two labels are the wrong way round there.

This is the cheapest error detector in the section. Glance at the sign of the integrand before integrating.

Independent check by comparison with the cylinder of radius $1$ and length $1$, volume $\pi$: the solid takes a seventh of it, matching the pattern $\tfrac{1}{2n+1}$ for $y=x^{n}$, here $n=3$.

For $y=x^{n}$ on $[0,a]$ about the $x$-axis, the solid always fills the fraction $\tfrac{1}{2n+1}$ of its cylinder. Handy as a check, useless as a substitute for the set-up.

Independent check by elementary geometry: this solid is a cone of radius $2$ and height $4$, so $\tfrac13\pi r^{2}h=\tfrac13\pi\cdot4\cdot4=\tfrac{16\pi}{3}$.

When the region is bounded by straight lines the answer is usually a cone, a cylinder or a frustum, and the elementary formula is then a complete independent check.

Independent check by splitting the solid. The outer piece is the cone swept by $y=x$ on $[0,1]$, radius $1$ and height $1$, so $\tfrac{\pi}{3}$ by the elementary formula; the drilled core is $\pi\int_0^1x^{4}dx=\tfrac{\pi}{5}$. The difference is $\tfrac{2\pi}{15}$.

Whenever the outer boundary is a straight line through the origin, the outer solid is a cone and half the check is free.

Independent check against the square-section solid on the same base, whose volume is $\tfrac{16}{3}$. Slice for slice, an equilateral triangle of side $s$ has $\tfrac{\sqrt3}{4}$ of the area of the square of side $s$, and indeed $\tfrac{4\sqrt3/3}{16/3}=\tfrac{\sqrt3}{4}$.

Whenever the slice shape changes but the base does not, the volumes stay in the same ratio as the two slice areas. That turns one computed solid into a check on the next.

Independent check by splitting off the shift: about the $x$-axis the same region gives $\pi\int_0^1x^{4}dx=\tfrac{\pi}{5}=\tfrac{3\pi}{15}$, and the extra from moving the axis down two units is $\pi\int_0^14x^{2}dx=\tfrac{4\pi}{3}=\tfrac{20\pi}{15}$. Together, $\tfrac{23\pi}{15}$.

The shift contributed its own clean integral. Splitting the answer that way is both a check and a way of seeing where the extra volume came from.

Independent check with no identity in it. On $[0,\pi/2]$ the graphs of $\cos^{2}$ and $\sin^{2}$ are mirror images, so their integrals are equal, and they add to $\int_0^{\pi/2}1\,dx=\pi/2$; each is therefore $\pi/4$, and $V=4\pi\cdot\tfrac{\pi}{4}=\pi^{2}$. Size check: the solid sits inside the cylinder of radius $2$ and length $\pi/2$, of volume $2\pi^{2}$, and the answer is exactly half of it, which is what the squared radius $4\cos^{2}x$ should give, since it averages $2$ over $[0,\pi/2]$, half of its largest value $4$.

Two habits are worth keeping from this one: the coefficient goes into the square, and a squared sine or cosine is a signal to reach for an identity rather than to doubt the set-up.

Independent check by bracketing: the solid sits inside the cylinder of radius $9$ and height $3$, volume $243\pi$, and contains the cone of radius $9$ and height $3$, volume $81\pi$, because $9-y^{2}\ge 9-3y$ on $[0,3]$. Our $\tfrac{648}{5}=129.6$ lies between $81$ and $243$ in units of $\pi$.

A region touching the axis along a straight boundary gives disks even when the axis is a shifted line. Whether there is a hole is a question about the picture, not about which letter the axis is written with.

Independent check by complement: the cylinder of radius $4$ and height $2$ has volume $32\pi$, and the rest of it is the solid swept by the region above the curve, $\pi\int_0^2\bigl[y^{2}\bigr]^{2}dy=\pi\int_0^2 y^{4}dy=\tfrac{32\pi}{5}$. And $32\pi-\tfrac{32\pi}{5}=\tfrac{128\pi}{5}$.

A radius that is itself a power is the easiest one to carry into the integrand unsquared, because it already looks like a squared thing.

Independent check at two known values of $k$: at $k=2$ the formula gives $\pi\bigl(\tfrac{64}{3}-8\bigr)=\tfrac{40\pi}{3}$, which is the answer computed directly earlier for that axis; and substituting $k=\tfrac94$ back gives $\pi(24-8)=16\pi$ as required.

Because the squared terms in $R^{2}-r^{2}$ cancel, volume is linear in the position of the axis. Doubling the volume therefore does not mean doubling the distance.

Independent check by building the solid as a difference of two solids rather than as a ring of slices. The outer solid is $\pi\int_0^{\pi/2}(1+\cos x)^{2}dx=\pi\bigl(\tfrac{3\pi}{4}+2\bigr)$ and the drilled-out core is a cylinder of radius $1$ and length $\pi/2$, of volume $\pi^{2}/2$; subtracting gives $2\pi+\pi^{2}/4$ again. Numerically that is $8.75$, and $8.75-2.47=6.28=2\pi$, which confirms part (b) independently of the argument used to get it.

The clean split depends on the region lying above both lines, above $y=0$ and above $y=-1$. When that holds, replacing the axis $y=0$ by $y=-k$ leaves the squared part $f^{2}-g^{2}$ untouched and adds a cross term $2k(f-g)$, twice the shift times the height of the region; that is why a shifted-axis answer so often falls into a clean $\pi$ piece and a clean $\pi^{2}$ piece, and why a dropped cross term is easy to spot. Both halves of the condition earn their keep. Put the axis on the far side of the region and the squared part changes sign rather than surviving, which is what part (d) of ex-10-exam is recording. Let the region cross $y=0$ and the reading fails outright: in ex-10-14 the same region gives $\pi^{2}$ about the $x$-axis and $8\pi$ about $y=-1$, not $\pi^{2}$ plus a cross term.

Independent check by going forward again: $\int_0^2\bigl(3x^{2}+2\bigr)dx=8+4=12$, and $V(2)=8+4=12$. The two agree.

Any statement about a running total of volume is a statement about an integral, and the Fundamental Theorem turns it into a statement about the slice.

Independent check by the second route shown: evaluating the integral first gives $V(t)=\tfrac{\pi t^{4}}{2}$, whose derivative is $2\pi t^{3}$, and at $t=1$ that is $2\pi$.

Whenever the limit of a volume integral moves faster than the variable you are differentiating in, the chain rule supplies the missing factor.

Independent check at two ends: $V(1)=0$, which is right because the region is empty there, and $V(2)=\tfrac{\pi}{3}\cdot\tfrac78=\tfrac{7\pi}{24}$, already $87.5$ percent of the limit, which matches how fast $x^{-4}$ collapses.

The volume is bounded even though the solid is not. Length and volume are different questions, and only one of them was asked.

Independent check with no calculus in it: $A(r)=10r-r^{2}$ is a downward parabola with roots at $r=0$ and $r=10$, so its vertex sits halfway between them at $r=5$ and the value there is $25$. A second opinion from a rival shape: the half disk, $\theta=\pi$, forces $r=20/(2+\pi)\approx 3.89$ and gives $A=\tfrac{\pi}{2}r^{2}\approx 23.8$, which is smaller, as it must be.

At the optimum the arc has length $r\theta=10$ and the two radii together also measure $10$ — the same even split between the straight part and the curved part that a fixed-perimeter rectangle shows between its two dimensions. Every one of these problems runs the same way: spend the constraint on one variable, then maximise the area formula of the named shape. The only thing that changes is which row of the geometry table you need.

Independent check at the join, which is where a wrong split shows itself. The first integrand collapses to $\bigl(3+u\bigr)^{2}-\bigl(3-u\bigr)^{2}=12u$ with $u=\sqrt{x-1}$, so it equals $12$ at $x=2$, and the second gives $(6-2)^{2}-(3-1)^{2}=16-4=12$ there as well. At both ends the integrands vanish, at $x=1$ because $R$ has shrunk to the point $(1,2)$ and at $x=5$ because $(6-5)^{2}-(3-2)^{2}=0$, and both are right because $R$ closes up there. A second and genuinely independent route confirms the value: a strip taken parallel to the axis instead of across it turns the whole of part (d) into $2\pi\int_1^4(5-y)\bigl[(y+1)-\bigl(y^{2}-4y+5\bigr)\bigr]dy$, one integral rather than two, and to three decimals both come to $70.686$. That single integral belongs to the next section, which is worth knowing in advance, since a real paper puts both parts inside the same question.

Independent check by bracketing. The widest cut is at $y=-\tfrac12$, where $w=\tfrac94$ and $A=\tfrac{81}{64}$, so across a range of length $3$ the volume cannot exceed $3\cdot\tfrac{81}{64}=\tfrac{243}{64}\approx 3.80$, and $\tfrac{81}{40}=2.025$ sits below it. A second check from the other side: the base has area $\int_{-2}^{1}\bigl(2-y-y^{2}\bigr)dy=\tfrac92$, so the mean width is $\tfrac32$, and a solid of that constant width would have volume $\tfrac14\bigl(\tfrac32\bigr)^{2}\cdot 3=\tfrac{27}{16}\approx 1.69$. The true value has to be larger, because the mean of $w^{2}$ exceeds the square of the mean of $w$, and $2.025>1.69$.

Independent check for continuity at the two joins, which is exactly where a wrong split announces itself. At $y=5$ the first integrand gives $(2-1)^{2}=1$ and the second gives $1$; at $y=8$ the second gives $1$ and the third gives $1-(1-1)^{2}=1$; at $y=9$ the third gives $1-1=0$, which is right, because $R$ has shrunk to the single point $x=0$ and its circle has no area. The middle piece is checkable on its own: it contributes $3\pi$, precisely the cylinder of radius $1$ and height $3$ that the $y$-axis edge sweeps between $y=5$ and $y=8$. Finally the three pieces add to $16\pi$, whereas the tempting single set-up $\pi\int_0^9\bigl(\sqrt{9-y}-1\bigr)^{2}dy$ gives $13.5\pi$ and is wrong twice over: above $y=5$ that expression is no longer the far end, and above $y=8$ it is negative and the square hides it.

Independent check by average height, using no integral at all. The roof is a plane, so the mean of $\sqrt3(y+3)$ over the disk equals its value at the centre $(0,0)$, namely $3\sqrt3$, and the volume is that mean times the base area: $3\sqrt3\cdot 9\pi=27\sqrt3\,\pi$. A crude bound agrees as well: the tallest point of the roof is $6\sqrt3$, standing over $(0,3)$, so the solid fits inside a cylinder of volume $9\pi\cdot 6\sqrt3=54\sqrt3\,\pi$, and the answer is exactly half of that, which is what a flat roof through a tangent line has to give.