Exercises — Volume of revolution — disk method, washer method

WHAT is happening: the region sits on the x-axis (its bottom edge is $y=0$, the axis itself). When it spins, each slice is a full solid coin — no hole. So this is a disk.

WHY disk: a hole only appears when the region is held away from the axis. Here the region touches the axis, so the coin's centre is filled.

Set-up: radius $=$ distance from axis to curve $=f(x)=x^2$. Thickness $=dx$. So $V=\pi {\int }_0^2(x^2{)}^{2}dx=\pi {\int }_0^2{x}^{4}dx.$ (We stop at set-up — recognition only.)

WHAT: two curves bound the region — an upper line $y=4$ and a lower parabola $y=x^2$. On $[0,2]$ the whole region floats above the axis with a gap between it and $y=0$... but the gap that matters is between the region and the axis. Since the lower edge $y=x^2$ is above the axis (except at $x=0$), spinning creates a hole. This is a washer.

WHY: the far curve from the axis is $y=4$ (bigger $y$ = farther), so $R=4$. The near curve is $y=x^2$, so $r=x^2$.

Set-up: $V=\pi {\int }_0^2(4^2-(x^2{)}^2)dx=\pi {\int }_0^2(16-x^4)dx.$

$V=\pi {\int }_0^2{x}^{4}dx=\pi {\left[\frac{x^5}{5}\right]}_0^2=\pi \cdot \frac{32}{5}=\frac{32\pi }{5}.$ WHAT we did: integrated $x^4$ using the power rule $\int x^{n}dx=\frac{x^{n+1}}{n+1}$, then plugged in the limits. WHY: the definite integral is the limit of the coin-sum as thickness $\to 0$ (Definite Integral as a Limit of Sums). $\boxed{V=\frac{32\pi }{5}}$

WHY disk: region rests on the axis → solid coins. $V=\pi {\int }_1^3{\left(\frac{1}{x}\right)}^{2}dx=\pi {\int }_1^3{x}^{-2}dx.$ WHAT the square did: ${\left(\frac{1}{x}\right)}^2=x^{-2}$ — squaring the radius turned the fraction into a clean power we can integrate. $=\pi {\left[\frac{x^{-1}}{-1}\right]}_1^3=\pi {\left[-\frac{1}{x}\right]}_1^3=\pi \left(-\frac{1}{3}+1\right)=\frac{2\pi }{3}.$ $\boxed{V=\frac{2\pi }{3}}$

$V=\pi {\int }_0^2(16-x^4)dx=\pi {\left[16x-\frac{x^5}{5}\right]}_0^2.$ Compute: $16(2)=32$ and $\frac{2^5}{5}=\frac{32}{5}$, so $V=\pi \left(32-\frac{32}{5}\right)=\pi \cdot \frac{160-32}{5}=\frac{128\pi }{5}.$ $\boxed{V=\frac{128\pi }{5}}$

WHAT to decide first: on $(0,1)$, which curve is farther from the axis? Test $x=\frac{1}{4}$: $\sqrt{\frac{1}{4}}=\frac{1}{2}$ but $\frac{1}{4}=0.25$. So $\sqrt{x}>x$ here — the root is on top. Therefore $R=\sqrt{x}$ (outer) and $r=x$ (inner).

WHY this matters: get them backwards and you'd integrate a negative integrand and get a negative "volume." The far curve must be $R$. $V=\pi {\int }_0^1((\sqrt{x}{)}^2-x^2)dx=\pi {\int }_0^1(x-x^2)dx.$ $=\pi {\left[\frac{x^2}{2}-\frac{x^3}{3}\right]}_0^1=\pi \left(\frac{1}{2}-\frac{1}{3}\right)=\pi \cdot \frac{1}{6}=\frac{\pi }{6}.$ $\boxed{V=\frac{\pi }{6}}$

WHAT changed: the axis is no longer $y=0$; it is the line $y=-1$ (see figure, the dashed line below the region). The radius is the distance from that line to the curve.

WHY re-measure: radius always means "axis-to-curve distance." From $y=-1$ up to $y=x$ the distance is $x-(-1)=x+1$. Since $x+1>0$ on $[0,3]$, no absolute value needed.

Is there a hole? The region's bottom is $y=0$, which is a distance $1$ above the axis $y=-1$ — so the region does not touch the axis. That means a washer, with inner radius $r=0-(-1)=1$. $V=\pi {\int }_0^3((x+1{)}^2-1^2)dx=\pi {\int }_0^3(x^2+2x+1-1)dx=\pi {\int }_0^3(x^2+2x)dx.$ $=\pi {\left[\frac{x^3}{3}+x^2\right]}_0^3=\pi \left(\frac{27}{3}+9\right)=\pi (9+9)=18\pi .$ $\boxed{V=18\pi }$

STEP 1 — where do they meet? Set $2x=x^2\Rightarrow x^2-2x=0\Rightarrow x(x-2)=0$, so $x=0$ and $x=2$. These are the limits.

STEP 2 — which is outer? Test $x=1$: $y=2x$ gives $2$, $y=x^2$ gives $1$. The line is on top: $R=2x$, $r=x^2$. $V=\pi {\int }_0^2((2x{)}^2-(x^2{)}^2)dx=\pi {\int }_0^2(4x^2-x^4)dx.$ $=\pi {\left[\frac{4x^3}{3}-\frac{x^5}{5}\right]}_0^2=\pi \left(\frac{32}{3}-\frac{32}{5}\right).$ $=\pi \cdot 32\left(\frac{1}{3}-\frac{1}{5}\right)=\pi \cdot 32\cdot \frac{2}{15}=\frac{64\pi }{15}.$ $\boxed{V=\frac{64\pi }{15}}$

WHY switch to $y$: we spin about the y-axis, so slices are horizontal coins of thickness $dy$; their radius is a horizontal distance, measured in $x$. So we must write $x$ as a function of $y$.

STEP 1 — invert: $y=x^2\Rightarrow x=\sqrt{y}$ (first quadrant, $x\ge 0$). The radius of each coin is $x=\sqrt{y}$. As $y$ runs from $0$ (bottom) to $4$ (top line), the region sweeps a solid — it touches the y-axis, so disk. $V=\pi {\int }_0^4(\sqrt{y}{)}^{2}dy=\pi {\int }_0^{4}ydy=\pi {\left[\frac{y^2}{2}\right]}_0^4=\pi \cdot 8=8\pi .$ $\boxed{V=8\pi }$

Sanity via Shell Method (optional cross-check): as vertical shells, radius $x$, height $4-x^2$, $V=2\pi {\int }_0^{2}x(4-x^2)dx=2\pi [2x^2-\frac{x^4}{4}{]}_0^2=2\pi (8-4)=8\pi$. ✓ Two methods, same number — confidence.

WHAT the solid is: spinning a semicircle about its diameter sweeps out a full ball of radius $R$. Each slice is a solid disk (the region touches the axis at $x=\pm R$).

Radius of a slice: $f(x)=\sqrt{R^2-x^2}$, so $[f(x){]}^2=R^2-x^2$ — the square erases the square root cleanly (this is why disk-method problems love circles). $V=\pi {\int }_{-R}^R(R^2-x^2)dx.$

Use symmetry: the integrand is even (same at $x$ and $-x$), so integrate $0\to R$ and double: $V=2\pi {\int }_0^R(R^2-x^2)dx=2\pi {\left[R^{2}x-\frac{x^3}{3}\right]}_0^R.$ $=2\pi \left(R^3-\frac{R^3}{3}\right)=2\pi \cdot \frac{2R^3}{3}=\frac{4}{3}\pi R^3.$ $\boxed{V=\frac{4}{3}\pi R^3}$ WHY this is the finish line: the answer is a formula in $R$, valid for every sphere — you've re-derived a geometry fact using nothing but the coin-slicing idea.

STEP 1 — distances from the axis $y=2$. Farther curve from $y=2$ is the lower one, since lower $y$ = greater distance from $y=2$. Compare on $[-1,1]$: $y=x^2$ ranges $0$ to $1$, and $y=1$ is the top edge. The parabola $x^2\le 1$ is below the line, so the parabola is farther from $y=2$: outer radius $R=2-x^2$; inner radius $r=2-1=1$. $V=\pi {\int }_{-1}^1((2-x^2{)}^2-1^2)dx.$

STEP 2 — expand. $(2-x^2{)}^2=4-4x^2+x^4$, so the integrand is $4-4x^2+x^4-1=3-4x^2+x^4$. $V=\pi {\int }_{-1}^1(3-4x^2+x^4)dx.$

STEP 3 — even integrand, double $0\to 1$: $V=2\pi {\int }_0^1(3-4x^2+x^4)dx=2\pi {\left[3x-\frac{4x^3}{3}+\frac{x^5}{5}\right]}_0^1.$ $=2\pi \left(3-\frac{4}{3}+\frac{1}{5}\right).$ Common denominator $15$: $3=\frac{45}{15}$, $\frac{4}{3}=\frac{20}{15}$, $\frac{1}{5}=\frac{3}{15}$, giving $\frac{45-20+3}{15}=\frac{28}{15}$. $V=2\pi \cdot \frac{28}{15}=\frac{56\pi }{15}.$ $\boxed{V=\frac{56\pi }{15}}$ (A direct u-substitution $u=2-x^2$ is possible but messier than expanding — expansion wins here.)