Exercises — Surface area of revolution

(a) Rotating about the x-axis, the radius = distance of the point to the x-axis = its height $=y=\sqrt{r^2-x^2}$. (b) By the chain rule on $\sqrt{r^2-x^2}=(r^2-x^2{)}^{1/2}$: $y^{\prime }=\frac{1}{2}(r^2-x^2{)}^{-1/2}\cdot (-2x)=\frac{-x}{\sqrt{r^2-x^2}}.$ WHY chain rule? The inside $r^2-x^2$ is itself a function of $x$, so we differentiate the outer power and multiply by the inner derivative $-2x$.

$y^{\prime }=\frac{r}{h}$ — a straight line has one fixed slope, so yes, constant. $\sqrt{1+(y^{\prime }{)}^2}=\sqrt{1+\frac{r^2}{h^2}}=\sqrt{\frac{h^2+r^2}{h^2}}=\frac{\sqrt{h^2+r^2}}{h}.$ WHAT this means geometrically: a straight generating line makes a perfect cone, so the slant is the same everywhere — matching frustum/cone geometry.

Rotating about the y-axis means points circle the vertical line $x=0$. The radius is the horizontal distance $=x=3$. Distance travelled in one spin $=2\pi r=2\pi (3)=6\pi$. Trap-proofing: the height $7$ is irrelevant to the radius here — only distance to the axis matters.

Radius $=y=\frac{r}{h}x$; slant factor $=\frac{\sqrt{h^2+r^2}}{h}$ (from 1.2). $S={\int }_0^{h}2\pi \cdot \frac{r}{h}x\cdot \frac{\sqrt{h^2+r^2}}{h}dx=\frac{2\pi r\sqrt{h^2+r^2}}{h^2}{\int }_0^{h}xdx.$ ${\int }_0^{h}xdx=\frac{h^2}{2}$, so $S=\frac{2\pi r\sqrt{h^2+r^2}}{h^2}\cdot \frac{h^2}{2}=\pi r\sqrt{h^2+r^2}=\pi rL.✓$

$y^{\prime }=3x^2$, so $1+(y^{\prime }{)}^2=1+9x^4$. $S={\int }_0^{1}2\pi x^3\sqrt{1+9x^4}dx.$ Choose substitution because the outside factor $x^3$ is (a multiple of) the derivative of the inside $1+9x^4$. Let $u=1+9x^4$, $du=36x^{3}dx\Rightarrow x^{3}dx=\frac{du}{36}$. Limits: $x=0\to u=1$; $x=1\to u=10$. $S=2\pi {\int }_1^{10}\sqrt{u}\frac{du}{36}=\frac{\pi }{18}\cdot \frac{2}{3}[u^{3/2}{]}_1^{10}=\frac{\pi }{27}\left(10^{3/2}-1\right).$ Numerically $\approx 3.563$.

$y=x^{1/2}$, $y^{\prime }=\frac{1}{2}{x}^{-1/2}$, so $(y^{\prime }{)}^2=\frac{1}{4x}$ and $1+(y^{\prime }{)}^2=\frac{4x+1}{4x}$. Then $y\sqrt{1+(y^{\prime }{)}^2}=x^{1/2}\cdot \frac{\sqrt{4x+1}}{2\sqrt{x}}=\frac{\sqrt{4x+1}}{2}$. WHY beautiful: the $\sqrt{x}$ cancels, leaving a clean root. $S={\int }_0^{4}2\pi \cdot \frac{\sqrt{4x+1}}{2}dx=\pi {\int }_0^4\sqrt{4x+1}dx.$ Let $u=4x+1$, $du=4dx$; $x=0\to u=1$, $x=4\to u=17$: $S=\pi {\int }_1^{17}\sqrt{u}\frac{du}{4}=\frac{\pi }{4}\cdot \frac{2}{3}[u^{3/2}{]}_1^{17}=\frac{\pi }{6}\left(17^{3/2}-1\right)\approx \mathrm{36.18.}$

Axis = y-axis ⇒ radius = horizontal distance = $x$. We could integrate in $y$ (using $x=\sqrt{y}$) but integrating in $x$ is cleaner because $y^{\prime }=2x$ is simple and the radius is literally $x$. $S={\int }_0^{1}2\pi x\sqrt{1+(2x{)}^2}dx={\int }_0^{1}2\pi x\sqrt{1+4x^2}dx.$ Sub $u=1+4x^2$, $du=8xdx\Rightarrow xdx=\frac{du}{8}$; $x=0\to u=1$, $x=1\to u=5$: $S=2\pi {\int }_1^5\sqrt{u}\frac{du}{8}=\frac{\pi }{4}\cdot \frac{2}{3}[u^{3/2}{]}_1^5=\frac{\pi }{6}\left(5^{3/2}-1\right)\approx \mathrm{5.33.}$ This matches Example 3 in the parent note. ✅

The curve is given as $x$ in terms of $y$, and we rotate about the y-axis (radius $=x$). Integrating in $y$ is natural: the limits are $y$-limits and $\frac{dx}{dy}$ is easy. $\frac{dx}{dy}=3y^2,ds=\sqrt{1+(3y^2{)}^2}dy=\sqrt{1+9y^4}dy.$ Radius $=x=y^3$: $S={\int }_0^{1}2\pi y^3\sqrt{1+9y^4}dy.$ Same integral shape as 2.2! Sub $u=1+9y^4$, $du=36y^{3}dy$; $u:1\to 10$: $S=\frac{\pi }{27}\left(10^{3/2}-1\right)\approx \mathrm{3.563.}$ Insight: rotating $y=x^3$ about the x-axis (2.2) and rotating $x=y^3$ about the y-axis are mirror-images, so equal $S$ is expected.

Why parametric $ds$ here? This curve loops back on itself (a full circle), so it is not a single function $y=f(x)$ — for each $x$ there are two $y$'s. The general element $ds=\sqrt{(dx{)}^2+(dy{)}^2}$ from the top of the page saves us: dividing and multiplying by $dt$ gives the parametric form $ds=\sqrt{{\left(\frac{dx}{dt}\right)}^2+{\left(\frac{dy}{dt}\right)}^2}dt.$ Compute the two derivatives: $\frac{dx}{dt}=-\sin t,\frac{dy}{dt}=\cos t,\sqrt{(x^{\prime }{)}^2+(y^{\prime }{)}^2}=\sqrt{{\sin }^{2}t+{\cos }^{2}t}=1.$ WHY this is $1$: the point moves around a unit circle at unit speed, so $ds=1dt$. Radius from x-axis $=y=2+\sin t$: $S={\int }_0^{2\pi }2\pi (2+\sin t)\cdot 1dt=2\pi {\left[2t-\cos t\right]}_0^{2\pi }=2\pi (4\pi )=8{\pi }^2.$ (${\int }_0^{2\pi }\sin tdt=0$.) So $S=8{\pi }^2\approx 78.96$. Cross-check with Pappus: $S=2\pi \bar{r}L$ where $L=2\pi$ (arc length of the whole unit circle, $2\pi \cdot 1$) and $\bar{r}$ is the height of the curve's centroid. The centroid of a full circle is its geometric centre, which we placed at height $2$ (that is exactly what "$y=2+\sin t$, a circle centred at height $2$" means — averaging $2+\sin t$ over one full turn gives $2$, since $\sin t$ averages to $0$). So $\bar{r}=2$ and $S=2\pi (2)(2\pi )=8{\pi }^2$. ✅

Radius = distance to the line $y=-1$, which is $|y-(-1)|=y+1=x+1$ (here $y=x\ge 0$, so positive). $y^{\prime }=1\Rightarrow ds=\sqrt{1+1}dx=\sqrt{2}dx$. $S={\int }_0^{2}2\pi (x+1)\sqrt{2}dx=2\sqrt{2}\pi {\int }_0^2(x+1)dx.$ ${\int }_0^2(x+1)dx=[\frac{x^2}{2}+x{]}_0^2=2+2=4$. $S=2\sqrt{2}\pi \cdot 4=8\sqrt{2}\pi \approx \mathrm{35.54.}$

From the parent note's sphere derivation, the whole integrand collapses to a constant: $y\sqrt{1+(y^{\prime }{)}^2}=\sqrt{r^2-x^2}\cdot \frac{r}{\sqrt{r^2-x^2}}=r.$ WHY it collapses: the height shrinks near the poles exactly as fast as the slant grows, so their product is flat. $S={\int }_a^{b}2\pi rdx=2\pi r(b-a).$ The stunning conclusion: two bands of equal width cut anywhere on the sphere have equal surface area — near the equator (short, fat) or near a pole (tall, skinny). Set $a=-r,b=r$ to recover $4\pi r^2$.

From 2.3's antiderivative, $S(\epsilon )=\frac{\pi }{6}\left((4\epsilon +1{)}^{3/2}-1\right)$ over $[0,\epsilon ]$. As $\epsilon \to 0^{+}$: $(4\epsilon +1{)}^{3/2}\to 1$, so $S\to \frac{\pi }{6}(1-1)=0$. ✅ Interpretation: a vanishingly short curve near the origin sweeps a vanishingly small surface — the formula behaves correctly at the degenerate end. Note the curve is vertical at $x=0$ (slope $\to \infty$), yet because the radius $y=\sqrt{x}\to 0$ too, the product stays finite and the area still goes to $0$.

The bug: the radius must be the distance to the axis, which is $|y|=|x^3|$, never negative. Area cannot be negative; symmetry that cancels signed integrals is illegal here. Split into the correct positive contributions. By symmetry the two halves have equal area: $S=2{\int }_0^{1}2\pi x^3\sqrt{1+9x^4}dx.$ From 2.2, ${\int }_0^{1}2\pi x^3\sqrt{1+9x^4}dx=\frac{\pi }{27}(10^{3/2}-1)$, so $S=2\cdot \frac{\pi }{27}\left(10^{3/2}-1\right)=\frac{2\pi }{27}\left(10^{3/2}-1\right)\approx \mathrm{7.125.}$ Every point contributes real, positive skin — the surface is a symmetric "double-horn" about the origin.

"Two-pie radius, slide along the rope." Radius = distance to the axis (mind shifts & signs), rope = arc length $ds$. Match $ds$ to the variable your limits live in. Distances are never negative.