Week 6 Notes: Volumes by Rotation, Cross-Sections, and Arc Length (Problem Bank)

Volume by Rotation: Disk/Washer Method

• Core ideas

  • Disk (solid of revolution about the x-axis or a horizontal line when the region is described as y = f(x) ≥ 0):
  • If revolving a region between y = f(x) and y = 0 around the x-axis from x = a to b, the volume is
    $V = \pi \int_{a}^{b} [f(x)]^{2}\,dx.$
  • Washer (region with a hole; outer radius R(x), inner radius r(x), rotated about the x-axis or a horizontal axis):
  • If the region is between y = f{outer}(x) and y = f{inner}(x) and revolved about the x-axis over [a,b],
    $V = \pi \int<em>{a}^{b} \big( [f</em>{outer}(x)]^{2} - [f_{inner}(x)]^{2} \big)\,dx.$
  • Cylindrical shells (usually for rotation about a y-axis or a vertical line; or when integrating with respect to y):
  • For rotation about the y-axis, with x as the radius and height given by the region,
    $V = 2\pi \int_{a}^{b} x\, f(x)\,dx.$
  • More generally, around a vertical line x = a, with radius r = |x - a| and height h(x),
    $V = 2\pi \int<em>{a}^{b} (\text{radius}) \times (\text{height})\,dx = 2\pi \int</em>{a}^{b} |x-a|\,h(x)\,dx.$
  • Cross-sections perpendicular to the axis of rotation
  • If cross-sections are squares with side s determined by the region, area A = s^2.
  • If cross-sections are circles (diameter D given by a function), area A = \frac{\pi}{4} D^{2}.
  • If cross-sections are equilateral triangles with side s, area A = \frac{\sqrt{3}}{4} s^{2}.
  • Geometry-guided shortcuts
  • When y = sqrt(1 - x^2) on [-1,1], rotating about x-axis yields a sphere; volume checks like V = \frac{4}{3}\pi r^{3} can appear (e.g., semicircle base + perpendicular cross-sections).

• Arc length recap

  • Arc length of y = f(x) on [a,b]:
    $L = \int_{a}^{b} \sqrt{1 + [f'(x)]^{2}}\,dx.$
  • Example: length of y = \ln(\sec x) for 0 ≤ x ≤ π/4
  • f(x) = \ln(\sec x); f'(x) = \tan x; so
    $L = \int<em>{0}^{\pi/4} \sqrt{1 + \tan^{2} x}\,dx = \int</em>{0}^{\pi/4} \sec x\,dx = \big[\ln|\sec x + \tan x|\big]_{0}^{\pi/4} = \ln(\sqrt{2}+1).$

• Worked exam-style highlights (key results and setups drawn from the problem bank)

1) Region: x^2 ≤ y ≤ 1 with cross-sections square perpendicular to the plane (edge on the region). Volume setup is by washers. (General template; problem 1 is a typical cross-section problem from the bank.)

2) Region: 0 ≤ x ≤ π/4, 0 ≤ y ≤ sec^2 x rotated around the x-axis.

• Setup: use disk/washer method (since rotating around a horizontal axis). Outer radius y = sec^2 x, so
  $V = \pi \int<em>{0}^{\pi/4} [\sec^2 x]^{2}\,dx = \pi \int</em>{0}^{\pi/4} \sec^{4} x\,dx.$

3) Region: 0 ≤ x ≤ π/4, 0 ≤ y ≤ sec^2 x rotated around the y-axis.

• Setup: use shells about the y-axis: radius = x, height = y = sec^2 x,
  $V = 2\pi \int_{0}^{\pi/4} x\,\sec^{2} x\,dx.$

4) Solid from rotating 0 ≤ y ≤ x/(x^2+1), 0 ≤ x ≤ 1 about the line x = −1.

• Setup: shells about x = −1: radius = x+1, height = y = \frac{x}{x^2+1},
  $V = 2\pi \int_{0}^{1} (x+1) \cdot \frac{x}{x^2+1}\,dx.$

5) Volume relation to deduce ∫_1^4 (f(x) - x)^2 dx given f(x) ≥ 0 on [1,4]

• Given: rotate region between f and x-axis around x-axis yields Vx = 32π and around y-axis yields Vy = 48π.
• Use disk/washer and shell formulas:
  • ∫_1^4 f(x)^2 dx = 32
  • ∫_1^4 x f(x) dx = 24
• Then
  $\int<em>{1}^{4} [f(x) - x]^2 dx = \int</em>{1}^{4} f^2(x) - 2x f(x) + x^2 \,dx = 32 - 2\cdot 24 + \int_{1}^{4} x^2 dx.$
• Compute ∫1^4 x^2 dx = [x^3/3]1^4 = (64 - 1)/3 = 63/3 = 21.
• Therefore the integral equals $32 - 48 + 21 = 5.$

6) Region in QI bounded by x-axis, y-axis, y = \sqrt{1 - x^4}; rotate around the x-axis.

• V = Disk/Washer with radius y:$V = \pi \int<em>{0}^{1} [\sqrt{1 - x^4}]^2 dx = \pi \int</em>{0}^{1} (1 - x^4) dx = \pi \left( x - \frac{x^5}{5} \right)_{0}^{1} = \frac{4\pi}{5}.$

7) Same region rotated around the y-axis (shells about y-axis):

• V = \ 2\pi \int_{0}^{1} x \cdot \sqrt{1 - x^4}\,dx.
• Substitution: let u = x^2, du = 2x dx; then V = \pi \int_{0}^{1} \sqrt{1 - u^2}\,du = \pi \cdot \frac{\pi}{4} = \frac{\pi^{2}}{4}.

8) Length problem: length of x = 2/3 y^{3/2} for 0 ≤ y ≤ 3.

• (Not reproduced here; focus in notes on volume/length methods.)

9) Length of y = \sqrt{1 - x^2} for 0 ≤ x ≤ 1.

• This is a quarter-circle of radius 1; arc length equals a sector angle of π/2 for radius 1, so length = (π/2)·1 = π/2. (Alternatively compute via integral; geometry gives L = π/2.)

10) Cable along y = (2/3) x^{3/2} from origin with total length 42; find endpoint (x,y).

• Solve by arc-length relation along the curve with x as a distance parameter: dx/ds etc. In this problem setup, you would solve ∫_0^X sqrt{1 + (dy/dx)^2} dx = 42 and then compute y = (2/3) X^{3/2}.

Week 6 additional practice: sample solutions and setups

11) Perimeter for region R between f(x) = x^2 and g(x) = 4 − 2x − x^2; find perimeter

• Set up integrals for the intersection points and total length around; DO NOT evaluate per instruction.

12) Shells around x-axis for rotation of region bound by x = 1 + y^2, x = 0, y = 1, y = 2

• Use horizontal shells: height in x-direction is xright − xleft = (1 + y^2) − 0, radius = y, from y=1 to y=2
• Volume: $V = 2\pi \int_{1}^{2} y \cdot (1 + y^2) \, dy.$

13) Let f be continuous on [2,8], g(x) = f(x) + x^2. Rotation about the y-axis of region between f and g from x=2 to 8.

• Shells about y-axis: height = g(x) − f(x) = x^2; radius = x;
  $V = 2\pi \int<em>{2}^{8} x \cdot x^2 \, dx = 2\pi \int</em>{2}^{8} x^3 \, dx.$

14) Gabriel’s Flugelhorn: region 0 ≤ x < ∞, 0 ≤ y ≤ x e^{-x} rotated about the x-axis.

• Volume: $V = \pi \int<em>{0}^{\infty} [x e^{-x}]^{2} dx = \pi \int</em>{0}^{\infty} x^{2} e^{-2x} dx = \pi \cdot \frac{2!}{2^{3}} = \frac{\pi}{4}.$

15) Rotation of region 0 ≤ x ≤ 1, 0 ≤ y ≤ 2(x+1)(x^{2}+1) about x = 1.
- Shells about x = 1: radius = 1 − x, height = 2(x+1)(x^{2}+1);
$V = 2\pi \int_{0}^{1} (1 - x) \cdot 2(x+1)(x^{2}+1) \, dx = 16\pi/5.$

16) Arc length of y = \sqrt{4 - x^2} for 0 ≤ x ≤ 1.

• This is an arc of a circle with radius 2; compute via arc-length formula or geometry; length = \frac{\pi}{3}.

17) Region R with 0 ≤ x ≤ 1 and \sqrt{1 - x^{6}} ≤ y ≤ 3x^{2} + \sqrt{1 - x^{6}}.

• (Part (a) volumes about y-axis; Part (b) about x-axis; setup included in notes.)

18) Region in QI bounded by axes, y = x^{2} e^{x^{2}}, x = 1. Volume about the y-axis.

• Shells: height = y = x^{2} e^{x^{2}}, radius = x, bounds 0 to 1
• $V = 2\pi \int_{0}^{1} x \cdot x^{2} e^{x^{2}} dx.$

19) Region bounded by y = x, y = 2 - x, y = x^{2} (in the shown diagram).

• (a) Shells about x-axis; (b) Washers about x-axis; (c) compute one integral to get the volume.

20) Region with \sqrt{1 - x^{2}} ≤ y ≤ 2\sqrt{1 - x^{2}} for -1 ≤ x ≤ 1; rotate about x-axis and about x = 1.

• (a) About the x-axis: V = \pi \int_{-1}^{1} [ (2\sqrt{1 - x^{2}})^{2} - (\sqrt{1 - x^{2}})^{2} ] dx = 4\pi.
• (b) About x = 1: shell method with radius (1 - x); height = \sqrt{1 - x^{2}};
  $V = 2\pi \int_{-1}^{1} (1 - x) \sqrt{1 - x^{2}} \, dx = \pi^{2}.$

21) Region 0 ≤ x ≤ 1, 0 ≤ y ≤ 5 - 3x (x^{2} + 1)(x + 4) around x = −1.

• Shells: radius = x + 1, height = (5 - 3x)(x^{2} + 1)(x + 4);
  $V = 2\pi \int_{0}^{1} (x+1) (5 - 3x)(x^{2} + 1)(x + 4) \, dx.$

22) Region 0 ≤ y ≤ 2 sec^{2}(x) for 0 ≤ x ≤ π/4; rotate about x = π/2.

• Shells: radius = π/2 − x, height = 2 sec^{2} x;
  $V = 2\pi \int<em>{0}^{π/4} (π/2 - x) \cdot [2\,sec^{2} x] \, dx = 4\pi \int</em>{0}^{π/4} (π/2 - x) \sec^{2} x \, dx.$

23) Region 0 ≤ y ≤ 2\sqrt{x^{4} - 16} for 2 ≤ x ≤ √5; rotate about the y-axis.

• Shells: radius = x, height = 2\sqrt{x^{4} - 16};
  $V = 2\pi \int<em>{2}^{\sqrt{5}} x \cdot 2\sqrt{x^{4} - 16} \, dx = 4\pi \int</em>{2}^{\sqrt{5}} x \sqrt{x^{4} - 16} \, dx.$
• Substitute u = x^{2} to obtain a solvable integral: V = 2\pi \int{4}^{5} \sqrt{u^{2} - 16}\, du = 2\pi\left[ \frac{1}{2} \big( u\sqrt{u^{2}-16} - 16 \ln|u + \sqrt{u^{2}-16}| \big) \right]{4}^{5} = 15\pi - 16\pi \ln 2.

24) Region: bounded by x = 0, y = 0, y = 2 + \sqrt{x}, y = 2x - 4 (F18-E2). (a) Set up volume around the x-axis; (b) set up around the line x = −2.

• Setup-only problems; DO NOT evaluate here.

25) Region bounded by y = x, y = \ln x, y = 0, y = 1; rotate about the y-axis.

• For y ∈ [0,1], the left and right x-bounds are x = y and x = e^{y}; rotating about y-axis gives
  $V = \pi \int_{0}^{1} [e^{y}]^{2} - [y]^{2} \, dy.$

26) Find b so that volume from rotating R (0 ≤ x ≤ b, 0 ≤ y ≤ e^{−x^{2}}) around the y-axis equals \frac{1}{2}π.

• Shells around y-axis: $V = 2\pi \int_{0}^{b} x e^{−x^{2}} dx = π [1 - e^{−b^{2}}] = \tfrac{1}{2}π.$
• Solve e^{−b^{2}} = 1/2 => b = \sqrt{\ln 2}.

27) Region in first quadrant: y = e^{x}, x = \ln 2, rotated about x = \ln 2.

• Shells about x = ln 2: height = e^{x}, radius = ln 2 − x, x from 0 to ln 2,
  $V = 2\pi \int_{0}^{\ln 2} (\ln 2 - x) e^{x} dx.$

28) For fixed r > 0, region between y = 0 and y = \sqrt{r^{2} - x^{2}} rotated about the x-axis.

• This is a disk cross-section problem giving the volume of a sphere of radius r: V = \frac{4}{3} \pi r^{3}.

29) Base is a semicircle of radius 4; cross-sections perpendicular to the base parallel to the diameter are equilateral triangles.

• For a given y in [0,4], diameter across the semicircle is s(y) = 2\sqrt{16 - y^{2}}; side of the equilateral triangle equals s(y).
• Area A(y) = \frac{\sqrt{3}}{4} s(y)^{2} = \frac{\sqrt{3}}{4} \cdot 4\,(16 - y^{2}) = \sqrt{3}\,(16 - y^{2}).
• Volume: $V = \int<em>{0}^{4} A(y) dy = \sqrt{3} \int</em>{0}^{4} (16 - y^{2}) dy = \sqrt{3} \left[ 16y - \frac{y^{3}}{3} \right]_{0}^{4} = \frac{128\sqrt{3}}{3}.$

30) Base over region 0 ≤ x ≤ π/3 and 0 ≤ y ≤ sec x; cross-sections perpendicular to the x-axis are squares with one side on the xy-plane.

• Side length s(x) = sec x; Area A(x) = s(x)^{2} = sec^{2} x;
  $V = \int<em>{0}^{\pi/3} \sec^{2} x \, dx = \tan x|</em>{0}^{\pi/3} = \sqrt{3}.$

31) Region 0 ≤ x ≤ 8, 0 ≤ y ≤ \sqrt{x^{2} + 10x}; rotate about x = −5.

• Shells: radius = x + 5, height = \sqrt{x^{2} + 10x};
  $V = 2\pi \int_{0}^{8} (x + 5) \sqrt{x^{2} + 10x} \, dx.$
• Substitution u = x + 5 yields a solvable integral, giving V = 1152\pi.

32) Region 0 ≤ x ≤ \frac{1}{2}\pi and \sin x ≤ y ≤ 1 + \cos x; rotate about the x-axis.

• Washer method with y-values: V = \pi \int_{0}^{\frac{\pi}{2}} [ (1 + \cos x)^{2} - (\sin x)^{2} ] dx
• Evaluate: (1 + 2\cos x + \cos 2x) integrated from 0 to π/2 gives π/2 + 2; thus V = π(\tfrac{\pi}{2} + 2) = \frac{\pi^{2}}{2} + 2\pi.

33) Given f(x) = 0 with volumes: around x-axis for 0 ≤ x ≤ 4 is 22π and around y-axis is 14π. Determine the volume for region 0 ≤ x ≤ 4 and 0 ≤ y ≤ 3x + 2f(x) spun around the x-axis.

• Use linearity of the integrals: V = π ∫_0^4 (3x + 2f(x))^{2} dx
• Expand: = π [ 9∫ x^{2} dx + 12 ∫ x f(x) dx + 4 ∫ f(x)^{2} dx ]
• Use given: ∫ f^{2} dx = 22, ∫ x f dx = 7, ∫ x^{2} dx = [x^{3}/3]_0^4 = 64/3.
• Therefore V = π [ 9·(64/3) + 12·7 + 4·22 ] = π [ 192 + 84 + 88 ] = 364π.

34) Length: y = \ln(x + \sqrt{x^{2} - 1}) on [1, \sqrt{5}].

• Recognize this as arcosh; arc-length formula yields L = ∫_{1}^{\sqrt{5}} \sqrt{1 + (1/(x^{2} - 1))} dx
• Simplify to L = ∫{1}^{\sqrt{5}} \frac{x}{\sqrt{x^{2} - 1}} dx = [\sqrt{x^{2} - 1}]{1}^{\sqrt{5}} = \sqrt{5 - 1} - 0 = 2.

35) Volume: region 0 ≤ y ≤ sin^{2}(x) for 0 ≤ x ≤ π spun about the y-axis.

• Shells about y-axis: radius = x, height = sin^{2} x,
  $V = 2\pi \int_{0}^{\pi} x \sin^{2} x \, dx.$
• Evaluate using sin^{2} x = (1 - \cos 2x)/2 to obtain a closed form (the integral gives a finite value, not shown here in full). The final form can be written as a combination of constants after integration.

36) Region 0 ≤ y ≤ \sec^{3} x for 0 ≤ x ≤ π/4; rotate about the x-axis.

• Disk method: radius = y = \sec^{3} x;
  V = \pi \int{0}^{\pi/4} [\sec^{3} x]^{2} dx = \pi \int{0}^{\pi/4} \sec^{6} x \, dx.$n
• Use substitution t = \tan x to compute or reduce to standard forms; the evaluated result is
  V = \frac{28}{15}\pi.

Additional worked results (quick recap for study):

• Problem 5: ∫_1^4 (f(x) - x)^{2} dx = 5.
• Problem 6: V = 4π/5.
• Problem 7: V = π^{2}/4.
• Problem 10 (the cable problem) is not evaluated here; setup described above.
• Problem 14: V = π/4.
• Problem 15: V = 16π/5.
• Problem 16: L = π/3.
• Problem 18: V = π.
• Problem 19, 20, 21, 22, and 23 have setups shown above; several yield neat closed forms (e.g., 23 gives 15π − 16π ln 2).
• Problem 28: V = 4π r^{3}/3.
• Problem 31: V = 1152π.
• Problem 32: V = π^{2}/2 + 2π.
• Problem 33: V = 364π.
• Problem 34: Length = 2.
• Problem 35: V expressed as 2π ∫_0^{π} x sin^{2} x dx; can be evaluated to a definite constant.
• Problem 36: V = 28π/15.

Notes on the connections and intent

• These problems reinforce core methods for volumes of solids of revolution: disk/washer and cylindrical shells, with many variants where the cross-sections are different shapes (square, circle, equilateral triangle).
• They also stress when to switch to shells (often around a vertical line) versus disks/washers (around horizontal lines).
• Several problems connect volume information to recover integrals (problem 5), illustrating how different rotation axes encode different integral forms (f^2, x f, and x^2 terms).
• Arc length problems reinforce the standard formula and occasionally reveal geometric shortcuts (e.g., semicircles, circles) when the curve is a part of a circle.
• Practical engineering-type questions (e.g., cable laying, challege problems with lines and curves) are included to emphasize setup and interpretation before computation.

Key formulas to memorize (LaTeX-ready)

• Disk/Washer about x-axis or horizontal axis:
  V = \int{a}^{b} \pi \left(R(x)^{2} - r(x)^{2}\right) dx \text{(Disk: } r(x)=0\text{)} \,\Rightarrow\, V = \int{a}^{b} \pi [R(x)]^{2} dx
• Shells about y-axis or vertical line x=a: V = 2\pi \int_{a}^{b} (\text{radius})\,(\text{height})\,dx
  • For rotation about x=a: radius = |x-a|, height = top(x) − bottom(x).
• Cross-sections perpendicular to the axis:
  • Square: A = s^{2}.
  • Circle (diameter D): A = \frac{\pi}{4} D^{2}.
  • Equilateral triangle (side s): A = \frac{\sqrt{3}}{4} s^{2}.
• Arc length:
  L = \int_{a}^{b} \sqrt{1 + [f'(x)]^{2}} \, dx.$$
• Polar/area-integral shortcuts and common substitutions (e.g., u = tan x for integrals involving sec^{n} x).
• Sphere volume reminder: for a semicircular base rotated about the axis to form a sphere, volume results in V = \frac{4}{3} \pi r^{3}.

End of notes