MA200 Integral Calculus: Area Between Curves, Volumes of Revolution, and Arc Length

Foundations: Area Between Curves and Riemann Sum Generalization
  • Riemann Sum Framework: Definite integrals evaluate net signed area. Pure geometric area between curves is always non-negative.

  • Area Between Curves: For continuous functions with f(x)≥g(x)f(x) \ge g(x) on [a,b][a, b]:   A=∫ab(f(x)−g(x)) dxA = \int_a^b (f(x) - g(x))\,dx

  • Intersecting Regions: Partition the integral at intersection points to maintain (upper curve)−(lower curve)(\text{upper curve}) - (\text{lower curve}).

    • Example: Enclosing y=x3y = x^3 and y=4xy = 4x over [−1,1][-1, 1] splits at x=0x = 0:     A=∫−10(x3−4x) dx+∫01(4x−x3) dx=74+74=72A = \int_{-1}^0 (x^3 - 4x)\,dx + \int_0^1 (4x - x^3)\,dx = \frac{7}{4} + \frac{7}{4} = \frac{7}{2}

  • Standard Evaluations:

    • Enclosed by g(x)=xg(x) = \sqrt{x} and f(x)=xf(x) = x over [0,1][0, 1]: A=∫01(x−x) dx=16A = \int_0^1 (\sqrt{x} - x)\,dx = \frac{1}{6}

    • Enclosed by f(x)=4xf(x) = 4^x, g(x)=2xg(x) = 2^x, and x=1x = 1: A=∫01(4x−2x) dx=12ln⁡(2)A = \int_0^1 (4^x - 2^x)\,dx = \frac{1}{2\ln(2)}

Volume by Slicing and Cross-Sectional Area
  • Slicing Principle: Integrating parallel cross-sectional slices of area A(x)A(x) along the xx-axis yields total volume:   V=∫abA(x) dxV = \int_a^b A(x)\,dx

  • Uniform Cross Section: For a constant 2×22 \times 2 square cross section on [0,10][0, 10], V=∫0104 dx=40 units3V = \int_0^{10} 4\,dx = 40\,\text{units}^3.

  • Square Pyramid: Apex at origin, height h=10h = 10, base side b=6b = 6. Side length scales as s(x)=35xs(x) = \frac{3}{5}x:   A(x)=925x2  ⟹  V=∫010925x2 dx=120 units3A(x) = \frac{9}{25}x^2 \implies V = \int_0^{10} \frac{9}{25}x^2\,dx = 120\,\text{units}^3

  • Right Circular Cone: Apex at origin, height h=10h = 10, base radius r=5r = 5. Radius scales as r(x)=12xr(x) = \frac{1}{2}x:   A(x)=πx24  ⟹  V=∫010πx24 dx=250π3 units3A(x) = \frac{\pi x^2}{4} \implies V = \int_0^{10} \frac{\pi x^2}{4}\,dx = \frac{250\pi}{3}\,\text{units}^3

Solids of Revolution and the Disk Method
  • Disk Method Formula: Rotating y=f(x)y = f(x) around the xx-axis forms cylindrical disks of radius r=f(x)r = f(x):   V=∫abπ(f(x))2 dxV = \int_a^b \pi (f(x))^2\,dx

  • Applications:

    • Cone verification: Revolving y=12xy = \frac{1}{2}x on [0,10][0, 10] gives V=π∫010x24 dx=250π3 units3V = \pi \int_0^{10} \frac{x^2}{4}\,dx = \frac{250\pi}{3}\,\text{units}^3.

    • Rational curve y=1xy = \frac{1}{x} on [1,2][1, 2]: V=π∫12x−2 dx=π2 units3V = \pi \int_1^2 x^{-2}\,dx = \frac{\pi}{2}\,\text{units}^3.

    • Radical curve y=xy = \sqrt{x} on [0,4][0, 4]: V=π∫04x dx=8π units3V = \pi \int_0^4 x\,dx = 8\pi\,\text{units}^3.

Differentiation Rules and Geometric Foundations
  • Calculus Rules:

    • Product Rule: ddx(uv)=u′v+vu′\frac{d}{dx}(uv) = u'v + vu'

    • Quotient Rule: ddx(uv)=u′v−v′uv2\frac{d}{dx}\left(\frac{u}{v}\right) = \frac{u'v - v'u}{v^2}

    • Chain Rule: dydx=dydududx\frac{dy}{dx} = \frac{dy}{du} \frac{du}{dx}

    • Exponentials: ddx(ex)=ex\frac{d}{dx}(e^x) = e^x, ddx(ax)=ln⁡(a)⋅ax\frac{d}{dx}(a^x) = \ln(a) \cdot a^x, and ddx(2−x)=−ln⁡(2)⋅2−x\frac{d}{dx}(2^{-x}) = -\ln(2) \cdot 2^{-x}

  • Distance Relations:

    • Pythagorean Theorem: c=a2+b2c = \sqrt{a^2 + b^2}

    • Distance Formula: d=(x2−x1)2+(y2−y1)2d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}

Arc Length: Derivation and Rigorous Formulation
  • Approximation: Segment sum of secant chords: ∑i=1n(Δxi)2+(Δyi)2\sum_{i=1}^n \sqrt{(\Delta x_i)^2 + (\Delta y_i)^2}.

  • Definite Integral Derivation: Factoring Δxi\Delta x_i forms a Riemann sum converging to:   L=lim⁡n→∞∑i=1n1+(ΔyiΔxi)2Δxi=∫ab1+(f′(x))2 dxL = \lim_{n \rightarrow \infty} \sum_{i=1}^n \sqrt{1 + \left(\frac{\Delta y_i}{\Delta x_i}\right)^2} \Delta x_i = \int_a^b \sqrt{1 + (f'(x))^2}\,dx

  • Condition: f′(x)f'(x) must be continuous on [a,b][a, b].

Arc Length Computations and Analytical Techniques
  • Parabola y=12x2y = \frac{1}{2}x^2 over [0,3][0, 3]:   L=∫031+x2 dx≈5.65L = \int_0^3 \sqrt{1 + x^2}\,dx \approx 5.65

  • Linear function y=2x+1y = 2x + 1 over [0,2][0, 2]:   L=∫021+22 dx=25L = \int_0^2 \sqrt{1 + 2^2}\,dx = 2\sqrt{5}

  • Power function y=x3/2y = x^{3/2} over [0,4][0, 4] (via u=1+94xu = 1 + \frac{9}{4}x):   L=∫041+94x dx=827(103/2−1)≈9.07L = \int_0^4 \sqrt{1 + \frac{9}{4}x}\,dx = \frac{8}{27}(10^{3/2} - 1) \approx 9.07

  • Logarithmic function f(x)=x28−ln⁡(x)f(x) = \frac{x^2}{8} - \ln(x) over [1,2][1, 2]:

    • Radicand forms a perfect square: 1+(f′(x))2=(x4+1x)21 + (f'(x))^2 = \left(\frac{x}{4} + \frac{1}{x}\right)^2

    • Integral evaluation: $$L = \int_1^2 \left(\frac{x}{4} + \frac{1}{x}\right)\,dx = \frac{3}{8} + \ln(2) \approx