11Volumes by shells and transcendental-function foundations

Do not look for a proof. A single region that can be handled both ways is enough to settle a claim that says always.

The deciding rule is that a strip parallel to the axis becomes a shell and a strip across the axis becomes a disk or a washer.

Take the region under $y=\sqrt{x}$ on $[0,4]$ and the vertical axis. Horizontal strips run across that axis, so they give washers with outer radius $4$ and inner radius $y^{2}$, over $0\le y\le 2$.

So the claim is false: with horizontal strips the same rotation is a washer problem.

No integration is needed to compare them. Ask how far each square travels in one full turn, and how that compares between the two.

The shell element is $2\pi(\text{radius})(\text{height})(\text{thickness})$, so at equal height and thickness the volume swept is proportional to the distance from the axis.

$P$ occupies $0\le x\le 1$ with average radius $\tfrac12$, while $Q$ occupies $10\le x\le 11$ with average radius $\tfrac{21}{2}$; heights and widths are both $1$.

So $V_P=\pi$, $V_Q=21\pi$, and the ratio is $21$.

Do not test with a positive number; the claim is obviously fine there. Ask where each side is defined, and try a number where the two answers differ.

The power law $\ln(x^{r})=r\ln x$ is proved from the definition, which requires the base to be positive; the definition of $\ln$ has domain $(0,\infty)$.

Put $x=-1$ into the claim: the left side is $\ln((-1)^{2})=\ln 1=0$, while the right side is $2\ln(-1)$, which does not exist.

So the claim is false as stated; with bars, $\ln(x^{2})=2\ln\lvert x\rvert$, it is true for every $x\neq 0$.

The question restricts what you are allowed to use. List the facts about $e^{x}$ that exist at this point in the course; there is essentially one.

$\exp$ is the inverse of $\ln$, so $\ln(e^{x})=x$; and $\ln$ has derivative $1/x$, which came from the Fundamental Theorem applied to the defining integral.

Set $y=e^{x}$. The inverse relation gives $\ln y=x$, and differentiating both sides with respect to $x$ gives $y'/y=1$.

Multiplying by $y$ gives $y'=e^{x}$, which is the required identity.

Draw the strip before writing anything. Which way does it run relative to the axis, and what is its thickness called?

Shells about the vertical axis: $V=2\pi\int_a^b x\,f(x)\,dx$, where $f(x)$ is the height of the strip at $x$.

Here $f(x)=x^{3}$ and $x$ runs from $0$ to $2$, so the integrand is $x\cdot x^{3}=x^{4}$.

The integral is $2\pi\bigl[\tfrac{x^{5}}{5}\bigr]_0^2=\tfrac{64\pi}{5}$.

Do not write an integral yet. Point at the strip at $x=0$ and at the line $x=3$, and say how far apart they are; then do the same at $x=2$.

Shell method about a vertical line $x=c$: $V=2\pi\int_a^b(\text{distance to the axis})\times(\text{height})\,dx$, with the distance written so that it is non-negative on $[a,b]$.

The region occupies $0\le x\le 2$, entirely to the left of $x=3$, so the radius is $3-x$; the height is still $x^{3}$.

So $V=2\pi\int_0^2(3-x)x^{3}dx=\tfrac{56\pi}{5}$.

Two things are missing before any integral can be written: where the region starts and stops, and which of the two curves is on top there.

Shells about the vertical axis with a region between two curves: $V=2\pi\int_a^b x\bigl[\text{top}(x)-\text{bottom}(x)\bigr]dx$.

The curves $y=\sqrt x$ and $y=x^{2}$ meet at $x=0$ and $x=1$, and on $(0,1)$ the root is above the square, so the height is $\sqrt x-x^{2}$.

So $V=2\pi\int_0^1 x(\sqrt x-x^{2})dx=\tfrac{3\pi}{10}$.

Before choosing a method, look at whether one part of the integrand is the derivative of another part.

Substitution: with $u=g(x)$ you get $\int f(g(x))g'(x)dx=\int f(u)du$, and in a definite integral the limits move too.

Here $g(x)=\ln x$ has $g'(x)=1/x$, and the integrand is exactly $\ln x\cdot\frac1x$. The limits $x=1,e$ become $u=0,1$.

So the integral is $\int_0^1 u\,du=\tfrac12$.

Nothing needs substituting. Write the antiderivative first and look at what the upper limit does to it.

The exponential is its own antiderivative: $\int e^{x}dx=e^{x}+C$; and $e^{\ln a}=a$ for every $a>0$.

Evaluating between the given limits $0$ and $\ln 3$ gives $e^{\ln 3}-e^{0}$, and $e^{\ln 3}=3$, $e^{0}=1$.

So the value is $3-1=2$.

Ask what each of the two standard rules assumes about the thing it differentiates, and check whether this function satisfies either assumption.

Logarithmic differentiation: take $\ln$ of both sides, differentiate implicitly so the left becomes $y'/y$, then multiply back by $y$ written in $x$.

Here $\ln y=\sqrt x\ln x$; differentiating the right-hand side by the product rule gives $\frac{\ln x}{2\sqrt x}+\frac{1}{\sqrt x}$.

So $y'=x^{\sqrt x}\cdot\frac{\ln x+2}{2\sqrt x}$, and at $x=4$ that is $4(\ln 4+2)\approx 13.5$.

Do not evaluate any of the four. Decide two things first: what the radius of the strip at $x$ is, and which variable the limits belong to.

Shells about a vertical line $x=c$: $V=2\pi\int(\text{distance to axis})(\text{height})\,dx$, with the limits being the range of $x$ because $dx$ is the thickness.

The region occupies $0\le x\le 4$ and the axis is at $x=6$, so the radius is $6-x$ and the height is $\sqrt x$; the limits are $0$ and $4$.

So the integral is $2\pi\int_0^4(6-x)\sqrt x\,dx$, and it equals $\tfrac{192\pi}{5}$.

First decide the method by comparing the direction of your strip with the direction of the axis. Only then start measuring lengths.

Washers about a horizontal line $y=d$: $V=\pi\int\bigl[R(x)^{2}-r(x)^{2}\bigr]dx$, where $R$ and $r$ are distances from that line to the far and near boundaries.

Here the axis is $y=-1$, the far boundary is $y=\sqrt x$ and the near one is $y=0$, so $R=\sqrt x+1$ and $r=1$, with $x$ from $0$ to $4$.

So the integral is $\pi\int_0^4\bigl[(\sqrt x+1)^{2}-1\bigr]dx=\tfrac{56\pi}{3}$.

Take the parts in order and, for each one, write the radius before anything else. In part (a) watch what the radius does to the height.

Shells about a vertical line $x=c$: $V=2\pi\int(\text{distance to axis})(\text{height})\,dx$; and $\int\frac{dx}{x}=\ln\lvert x\rvert+C$.

For the region under $y=1/x$ on $[1,b]$: about the vertical axis the radius is $x$ and the height $1/x$, so the integrand is $1$. About $x=1$ the radius is $x-1$, so the integrand is $(x-1)/x=1-\frac1x$.

Hence $V=2\pi(b-1)$, then $b=6$, then $V=2\pi\bigl[x-\ln x\bigr]_1^6=2\pi(5-\ln 6)$.

Write the product of radius and height before doing anything else, and simplify it.

Shells about the vertical axis: $V=2\pi\int_a^b x\,f(x)\,dx$, and here $f(x)=1/x$.

The integrand is $x\cdot\frac1x=1$, on the interval from $x=1$ to $x=e$.

So $V=2\pi\bigl[x\bigr]_1^{e}=2\pi(e-1)$.

Two decisions before any formula: which way does your strip run compared with the axis, and does the resulting solid have a hole in it?

A strip across the axis gives disks or washers, $V=\pi\int\bigl[R^{2}-r^{2}\bigr]$, and $r=0$ exactly when the region touches the axis all along.

Here the axis $y=4$ is the top boundary of the region itself, so $r=0$ and $R(x)=4-x^{2}$, with $x$ running from $0$ to $2$.

So $V=\pi\int_0^2(4-x^{2})^{2}dx=\tfrac{256\pi}{15}$.

Compare the numerator with the derivative of the denominator before choosing anything.

If $u=g(x)$ then $\int\frac{g'(x)}{g(x)}dx=\ln\lvert g(x)\rvert+C$, and a constant factor can always be moved outside.

Here $g(x)=x^{2}+1$ with $g'(x)=2x$, and the integrand has only $x$, so the integral is $\tfrac12\int_1^2\frac{du}{u}$ after the limits move.

That gives $\tfrac12\ln 2$.

Identify the outer structure before differentiating: is this a product, a quotient, or a composition? Name the pieces.

Product rule $(uv)'=u'v+uv'$, together with $\frac{d}{dx}e^{w}=e^{w}w'$ and $\frac{d}{dx}\ln w=w'/w$.

With $u=e^{x^{2}}$ and $v=\ln(3x)$: $u'=2x e^{x^{2}}$ and $v'=3/(3x)=1/x$.

So $g'(x)=2xe^{x^{2}}\ln(3x)+e^{x^{2}}/x$, and at $x=1$ this is $e(2\ln 3+1)\approx 8.69$.

The exponential factor is never zero and never negative. What does that let you ignore when you look for sign changes?

A local maximum sits where $f'$ changes from positive to negative, and $f'$ can only change sign at a critical number.

Here $f'(x)=e^{-x}x(2-x)$, and since $e^{-x}>0$ the sign is that of $x(2-x)$, which is positive on $(0,2)$ and negative outside it.

So the sign goes $+\to-$ at $x=2$: a local maximum there.