Arc length and surface area of revolution

Worked solutions

Problem 1

1. $y' = 2$, so $\sqrt{1 + y'^2} = \sqrt5$A line has constant slope.
2. $L = \displaystyle\int_0^3\sqrt5\,dx = 3\sqrt5$The integrand is a constant.
3. The ends are $(0, 1)$ and $(3, 7)$, at distance $\sqrt{3^2 + 6^2} = \sqrt{45} = 3\sqrt5$The two agree.
4. $3\sqrt5 \approx 6.708$For a line, the arc length formula is the distance formula.

Problem 2

1. $y' = \tfrac32x^{1/2}$, so $1 + y'^2 = 1 + \tfrac94x$Square the derivative.
2. $L = \displaystyle\int_0^4\Big(1 + \frac94x\Big)^{1/2}dx$Substitute $u = 1 + \tfrac94x$, $du = \tfrac94\,dx$; $u$ runs from $1$ to $10$.
3. $= \dfrac49\cdot\dfrac23\big[u^{3/2}\big]_1^{10} = \dfrac8{27}\big(10^{3/2} - 1\big)$The power rule in $u$.
4. $\dfrac8{27}\big(10\sqrt{10} - 1\big) \approx 9.073$The chord from $(0, 0)$ to $(4, 8)$ has length $\sqrt{80} \approx 8.944$, a little shorter.

Problem 3

1. $y' = \sinh x = \dfrac{e^x - e^{-x}}2$Differentiate each exponential.
2. $1 + \sinh^2x = \cosh^2x$Expand: $1 + \tfrac{e^{2x} - 2 + e^{-2x}}4 = \tfrac{e^{2x} + 2 + e^{-2x}}4$, a perfect square.
3. $L = \displaystyle\int_0^{\ln2}\cosh x\,dx = \sinh(\ln2) - \sinh0$$\cosh x > 0$, so the square root is $\cosh x$ itself.
4. $\sinh(\ln2) = \dfrac{2 - \frac12}{2} = \dfrac34$$e^{\ln2} = 2$ and $e^{-\ln2} = \tfrac12$.

Problem 4

1. $y' = \dfrac{x^2}2 - \dfrac1{2x^2}$Differentiate term by term.
2. $1 + y'^2 = 1 + \dfrac{x^4}4 - \dfrac12 + \dfrac1{4x^4} = \dfrac{x^4}4 + \dfrac12 + \dfrac1{4x^4} = \Big(\dfrac{x^2}2 + \dfrac1{2x^2}\Big)^2$The middle term of $y'^2$ is $-2\cdot\tfrac{x^2}2\cdot\tfrac1{2x^2} = -\tfrac12$, and adding $1$ flips its sign.
3. $L = \displaystyle\int_1^2\Big(\frac{x^2}2 + \frac1{2x^2}\Big)dx = \Big[\frac{x^3}6 - \frac1{2x}\Big]_1^2$The square root is gone.
4. $= \Big(\dfrac86 - \dfrac14\Big) - \Big(\dfrac16 - \dfrac12\Big) = \dfrac{13}{12} + \dfrac13$Evaluate at both limits.
5. $\dfrac{17}{12} \approx 1.417$The curve was chosen so that $1 + y'^2$ is a square; most curves are not so kind.

Problem 5

1. $\dfrac{dx}{dy} = (y - 1)^{1/2}$The curve is given as $x$ in terms of $y$, so differentiate in $y$.
2. $1 + \Big(\dfrac{dx}{dy}\Big)^2 = 1 + (y - 1) = y$The square root will be $\sqrt y$.
3. $L = \displaystyle\int_1^4\sqrt y\,dy = \frac23\big[y^{3/2}\big]_1^4$The limits are values of $y$.
4. $\dfrac23(8 - 1) = \dfrac{14}3 \approx 4.667$Solving for $y$ and integrating in $x$ would have given the same length with a much harder integral.

Problem 6

1. $y' = \dfrac{-x}{\sqrt{1 - x^2}}$, so $1 + y'^2 = \dfrac{1 - x^2 + x^2}{1 - x^2} = \dfrac1{1 - x^2}$A common denominator.
2. $L = \displaystyle\int_0^{1/2}\frac{dx}{\sqrt{1 - x^2}} = \arcsin\tfrac12 - \arcsin0$A standard antiderivative.
3. $= \dfrac\pi6$$\sin\tfrac\pi6 = \tfrac12$.
4. $\dfrac\pi6 \approx 0.5236$The curve is the unit circle, and the arc from $(0, 1)$ to $\big(\tfrac12, \tfrac{\sqrt3}2\big)$ subtends an angle of $\tfrac\pi6$ at the center. On a circle of radius $1$, arc length is the angle.

Problem 7

1. $L = \displaystyle\int_0^1\sqrt{1 + 4x^2}\,dx$$y' = 2x$. This one is not a perfect square.
2. $x = \tfrac12\tan\theta$, $dx = \tfrac12\sec^2\theta\,d\theta$, $\sqrt{1 + 4x^2} = \sec\theta$; $\theta$ runs from $0$ to $\arctan2$Trig substitution for $\sqrt{1 + (2x)^2}$.
3. $L = \dfrac12\displaystyle\int_0^{\arctan2}\sec^3\theta\,d\theta = \frac14\Big[\sec\theta\tan\theta + \ln\lvert\sec\theta + \tan\theta\rvert\Big]_0^{\arctan2}$The standard integral of $\sec^3$, done by parts.
4. At $\theta = \arctan2$, $\tan\theta = 2$ and $\sec\theta = \sqrt5$The triangle with opposite $2$ and adjacent $1$.
5. $L = \dfrac{\sqrt5}2 + \dfrac14\ln\big(2 + \sqrt5\big) \approx 1.4789$Between the chord $\sqrt2 \approx 1.4142$ and the path along the axes, $2$.

Problem 8

1. $ds = \sqrt{1 + y'^2}\,dx = \dfrac{r}{\sqrt{r^2 - x^2}}\,dx$As in Problem 6, with radius $r$: $1 + y'^2 = \dfrac{r^2}{r^2 - x^2}$.
2. $2\pi y\,ds = 2\pi\sqrt{r^2 - x^2}\cdot\dfrac{r}{\sqrt{r^2 - x^2}}\,dx = 2\pi r\,dx$The radius $y$ cancels the square root.
3. $S = \displaystyle\int_{-r}^r 2\pi r\,dx = 2\pi r\cdot2r$A constant integrand.
4. $S = 4\pi r^2$Every slice of the sphere of the same thickness $dx$ has the same area, $2\pi r\,dx$, wherever it is cut: Archimedes' hat-box theorem.

Problem 9

1. $y' = \dfrac1{2\sqrt x}$, so $\sqrt{1 + y'^2} = \sqrt{1 + \dfrac1{4x}}$The slant factor.
2. $2\pi y\sqrt{1 + y'^2} = 2\pi\sqrt x\sqrt{1 + \dfrac1{4x}} = 2\pi\sqrt{x + \tfrac14}$Multiply under one square root.
3. $S = 2\pi\displaystyle\int_0^2\Big(x + \frac14\Big)^{1/2}dx = \frac{4\pi}3\Big[\Big(x + \frac14\Big)^{3/2}\Big]_0^2$The power rule.
4. $= \dfrac{4\pi}3\Big(\dfrac{27}8 - \dfrac18\Big) = \dfrac{4\pi}3\cdot\dfrac{13}4$$\big(\tfrac94\big)^{3/2} = \tfrac{27}8$ and $\big(\tfrac14\big)^{3/2} = \tfrac18$.
5. $S = \dfrac{13\pi}3 \approx 13.614$The surface is a paraboloid, open at the wide end.

Problem 10

1. The distance from the point $(x, x^2)$ to the $y$-axis is $x$, so $r = x$About the $y$-axis, the radius is horizontal.
2. $S = \displaystyle\int_0^1 2\pi x\sqrt{1 + 4x^2}\,dx$$ds = \sqrt{1 + 4x^2}\,dx$ as in Problem 7.
3. $u = 1 + 4x^2$, $du = 8x\,dx$: $S = \dfrac{2\pi}8\displaystyle\int_1^5 u^{1/2}\,du = \frac\pi4\cdot\frac23\big[u^{3/2}\big]_1^5$This time the $x$ in the radius makes the substitution work.
4. $S = \dfrac\pi6\big(5\sqrt5 - 1\big) \approx 5.330$A bowl: like the surface in Problem 9 it is a paraboloid, but here its axis is the $y$-axis, the axis of turning.