10Volumes by disks and washers

Look at what the formula actually mentions. Does the word rotation appear anywhere in its hypotheses?

The hypotheses are: the solid lies between two planes, and the slice at $x$ has area $A(x)$ with $A$ continuous. Rotation is not among them.

A pyramid has square slices and no rotation anywhere, and the formula gave its volume as $32$ for base $4$ and height $6$.

So the claim is false: rotation is one way to know $A(x)$, not a condition for the formula.

Do not reach for a rule about scaling. Write each volume down from the ring area and compare the two numbers.

A slice is a ring of area $\pi\bigl(R^{2}-r^{2}\bigr)$, and here both radii are constants, so each volume is that area times the length $1$.

Band 1 has $R=2$, $r=1$, so $A_1=\pi(4-1)=3\pi$. Band 2 has $R=3$, $r=2$, so $A_2=\pi(9-4)=5\pi$.

So $V_1=3\pi$ and $V_2=5\pi$: the volume rose by $2\pi$, and did not double.

For each option, ask what the radius would have to be a function of, and whether the boundary can be written that way.

A vertical axis forces horizontal slices and thickness $dy$, so every radius has to be a function of $y$; the boundary must be solvable for $x$.

$y=x^{2}$ gives $x=\sqrt{y}$ in one step, and $y=2x$ gives $x=y/2$; but $y=x-x^{2}$ gives two branches at every height, $x=\tfrac12\bigl(1\pm\sqrt{1-4y}\bigr)$.

So the arch under $y=x-x^{2}$ is the one that resists: a single horizontal slice needs both branches at once.

Ask what squaring does to the order of two non-negative numbers.

For $0\le u\le v$ we have $u^{2}\le v^{2}$, because squaring is increasing on the non-negative numbers.

Here $0\le r(x)\le R(x)$ at every point of $[a,b]$, so $\bigl[r(x)\bigr]^{2}\le\bigl[R(x)\bigr]^{2}$.

So the claim is true, and a negative integrand is proof that the two radii were assigned the wrong way round.

The region rests on the axis along the whole interval, so ask what the radius of the slice at $x$ is before writing anything.

Disk method: $V=\pi\int_a^b\bigl[f(x)\bigr]^{2}dx$, with limits from the range of the thickness variable.

Here $f(x)=x^{3}$ on $0\le x\le 1$, so $\bigl[f(x)\bigr]^{2}=x^{6}$.

So $V=\pi\int_0^1x^{6}dx=\dfrac{\pi}{7}$.

The axis is vertical, so decide first which way the slices run and what their thickness is called.

A vertical axis forces horizontal slices, thickness $dy$, radii written in $y$, and limits taken from the range of $y$.

Here $y=2x$ becomes $x=y/2$, and the region reaches from $y=0$ up to $y=4$; the horizontal strip runs from the axis out to the line, so $r(y)=y/2$.

So $V=\pi\int_0^4\dfrac{y^{2}}{4}dy=\dfrac{16\pi}{3}$.

Nothing can be assigned until you know where the region starts and stops. Set the two boundaries equal first.

Washer method: $V=\pi\int_a^b\bigl(R^{2}-r^{2}\bigr)dx$, with $R$ the distance to the far edge and $r$ to the near edge.

Here $x=x^{2}$ gives $x=0$ and $x=1$; at $x=\tfrac12$ the line is at $0.5$ and the parabola at $0.25$, so the line is on top and the axis is $y=0$ below both.

So $V=\pi\int_0^1\bigl(x^{2}-x^{4}\bigr)dx=\dfrac{2\pi}{15}$.

The side of the triangle is a chord of the base circle. Ask how long that chord is at position $x$, from the bottom of the disk to the top.

Slicing gives $V=\int_a^b A(x)\,dx$, and an equilateral triangle of side $s$ has area $\tfrac{\sqrt3}{4}s^{2}$.

At position $x$ the disk runs from $y=-\sqrt{1-x^{2}}$ to $y=+\sqrt{1-x^{2}}$, so the chord has length $s(x)=2\sqrt{1-x^{2}}$, and $x$ runs from $-1$ to $1$.

So $A(x)=\sqrt3\bigl(1-x^{2}\bigr)$ and $V=\dfrac{4\sqrt3}{3}$.

Draw the line $y=-2$ and ask which edge of the region is farther from it. Both radii are distances measured upwards.

$R=\lvert y_{\text{far}}-k\rvert$ and $r=\lvert y_{\text{near}}-k\rvert$ with $k=-2$ here.

The region runs from $y=0$ up to $y=x^{2}$ for $0\le x\le 1$, and the axis is below both, so the far edge is the curve.

So $R=x^{2}+2$, $r=2$, and $V=\pi\int_0^1\bigl(x^{4}+4x^{2}\bigr)dx=\dfrac{23\pi}{15}$.

Draw the line $x=9$. Does the region touch it, and if so, along what part of its boundary?

A vertical axis forces horizontal slices, thickness $dy$, and radii measured as $\lvert x-9\rvert$; a region touching the axis gives $r=0$, that is, disks.

At height $y$ the region runs from $x=y^{2}$ to $x=9$, with $0\le y\le 3$; the far edge is the curve, at distance $9-y^{2}$, and the near edge lies on the axis itself.

So the integral is $\pi\int_0^3\bigl(9-y^{2}\bigr)^{2}dy=\dfrac{648\pi}{5}$.

Read each step against the one before it and ask whether it follows. Do not recompute the whole problem yet.

The washer integrand is $\bigl[R\bigr]^{2}-\bigl[r\bigr]^{2}$: each radius is squared, whatever it happens to be made of.

Step 2 hands over $R=4$ and $r=y^{2}$. Squaring those gives $16$ and $y^{4}$, not $16$ and $y^{2}$.

So step 3 is the first wrong one; it should read $V=\pi\int_0^2\bigl(16-y^{4}\bigr)dy=\dfrac{128\pi}{5}$.

Do the ordinary set-up but refuse to put a number in for the position of the axis. Which edge of the region is farther from a line lying above it?

$R=\lvert y_{\text{far}}-k\rvert$ and $r=\lvert y_{\text{near}}-k\rvert$, then $V=\pi\int_a^b\bigl(R^{2}-r^{2}\bigr)dx$.

Here the far edge is $y=0$, at distance $k$, and the near edge is $y=\sqrt{x}$, at distance $k-\sqrt{x}$, for $0\le x\le 4$.

So $V(k)=\pi\Bigl(\dfrac{32k}{3}-8\Bigr)$, and $V(k)=16\pi$ gives $k=\dfrac94$.

You are given a volume as a function of where you stop, and asked for an area. Which of the two is the derivative of the other?

Slicing says $V(t)=\int_0^t A(x)\,dx$, so by the Fundamental Theorem $V'(t)=A(t)$ wherever $A$ is continuous.

Here $V(t)=t^{3}+2t$, so $V'(t)=3t^{2}+2$, and the position asked for is $t=2$.

So $A(2)=3\cdot4+2=14$.

The upper limit is not $t$ but a function of $t$. Ask what that does to the derivative before computing anything.

If $V(t)=\int_0^{g(t)}h(u)\,du$ then $V'(t)=h\bigl(g(t)\bigr)\cdot g'(t)$: the Fundamental Theorem followed by the chain rule.

Here $h(u)=\pi u$ and $g(t)=t^{2}$, so $g'(t)=2t$, and the value wanted is at $t=1$.

So $V'(t)=\pi t^{2}\cdot 2t=2\pi t^{3}$, and $V'(1)=2\pi$.

Part (a) is an ordinary disk problem with a letter for the right-hand end. Part (b) asks what happens to that expression as the end runs away.

Disk method gives $V(b)=\pi\int_1^b\bigl[f(x)\bigr]^{2}dx$, and a limit at infinity is read off the resulting expression.

Here $f(x)=x^{-2}$, so $\bigl[f(x)\bigr]^{2}=x^{-4}$, and the region starts at $x=1$.

So $V(b)=\dfrac{\pi}{3}\Bigl(1-\dfrac{1}{b^{3}}\Bigr)$, whose limit is $\dfrac{\pi}{3}$.