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calculusintegrationdifferentiationsolids-of-revolutionconvergencelogarithmic-differentiationFri Apr 24

Calculus II: Worked Example Walkthroughs for Convergence, Differentiation, and Solids of Revolution

Abstract

This article provides structured worked examples for three core Calculus II topics: convergence of improper integrals, logarithmic differentiation, and volumes of solids of revolution. Each section combines conceptual grounding with step-by-step problem solutions, designed to reinforce both procedural fluency and conceptual understanding. The examples are drawn from typical exam-level problems and illustrate when and how to apply each technique.

Background

Calculus II extends single-variable calculus into applications and advanced techniques. Three topics form a conceptual backbone: determining whether integrals converge to finite values, differentiating complex functions via logarithmic methods, and computing volumes of three-dimensional objects formed by rotation. Each requires careful setup, correct application of rules, and verification of results.

Convergence of Integrals

An integral is said to converge when the limit of integration yields a finite value; it diverges when the limit is infinite or does not exist [convergence-of-integrals]. This distinction matters because improper integrals—those with infinite bounds or discontinuous integrands—require explicit limit evaluation. Without convergence tests, one cannot safely claim a result is meaningful [convergence-of-integrals].

Logarithmic Differentiation

When a function involves a variable in both base and exponent, or is a complicated product or quotient, logarithmic differentiation offers a cleaner path than the product or quotient rule [logarithmic-differentiation]. The method transforms multiplicative structure into additive structure via the logarithm, simplifying chain rule application [logarithmic-differentiation].

Solids of Revolution

Rotating a planar region about an axis generates a three-dimensional solid. Integration allows us to sum infinitesimal disk or washer cross-sections to find total volume [volume-of-solid-of-revolution]. This technique underpins engineering and physics applications where rotationally symmetric shapes are common [volume-of-solid-of-revolution].

Key Results

Convergence Test Framework

For an improper integral af(x)dx\int_a^{\infty} f(x) \, dx, evaluate: limtatf(x)dx\lim_{t \to \infty} \int_a^t f(x) \, dx

If the limit is finite, the integral converges; otherwise it diverges [convergence-of-integrals].

Logarithmic Differentiation Formula

For y=f(x)y = f(x), take the natural logarithm of both sides: ln(y)=ln(f(x))\ln(y) = \ln(f(x))

Differentiate implicitly: 1ydydx=f(x)f(x)\frac{1}{y} \frac{dy}{dx} = \frac{f'(x)}{f(x)}

Solve for the derivative: dydx=yf(x)f(x)=f(x)f(x)f(x)\frac{dy}{dx} = y \cdot \frac{f'(x)}{f(x)} = f(x) \cdot \frac{f'(x)}{f(x)}

[logarithmic-differentiation]

Volume of Solid of Revolution (Disk/Washer Method)

Rotating the region between y=f(x)y = f(x) and y=g(x)y = g(x) (where f(x)g(x)f(x) \geq g(x)) about the x-axis from x=ax = a to x=bx = b: V=πab[f(x)2g(x)2]dxV = \pi \int_a^b \left[f(x)^2 - g(x)^2\right] \, dx

For rotation about the y-axis: V=2πcdx[h(y)k(y)]dyV = 2\pi \int_c^d x \left[h(y) - k(y)\right] \, dy

where h(y)h(y) and k(y)k(y) are the outer and inner radii as functions of yy [volume-of-solid-of-revolution].

Worked Examples

Example 1: Convergence of an Improper Integral

Problem: Determine whether 11x2dx\int_1^{\infty} \frac{1}{x^2} \, dx converges or diverges.

Solution:

Set up the limit definition [convergence-of-integrals]: 11x2dx=limt1tx2dx\int_1^{\infty} \frac{1}{x^2} \, dx = \lim_{t \to \infty} \int_1^t x^{-2} \, dx

Evaluate the antiderivative: 1tx2dx=[x1]1t=1t(11)=11t\int_1^t x^{-2} \, dx = \left[-x^{-1}\right]_1^t = -\frac{1}{t} - \left(-\frac{1}{1}\right) = 1 - \frac{1}{t}

Take the limit: limt(11t)=1\lim_{t \to \infty} \left(1 - \frac{1}{t}\right) = 1

Conclusion: The integral converges to 11 [convergence-of-integrals].


Example 2: Logarithmic Differentiation

Problem: Find dydx\frac{dy}{dx} for y=xxy = x^x.

Solution:

Take the natural logarithm of both sides [logarithmic-differentiation]: ln(y)=ln(xx)=xln(x)\ln(y) = \ln(x^x) = x \ln(x)

Differentiate both sides with respect to xx using the chain rule on the left and the product rule on the right: 1ydydx=ddx[xln(x)]=1ln(x)+x1x=ln(x)+1\frac{1}{y} \frac{dy}{dx} = \frac{d}{dx}[x \ln(x)] = 1 \cdot \ln(x) + x \cdot \frac{1}{x} = \ln(x) + 1

Solve for dydx\frac{dy}{dx}: dydx=y[ln(x)+1]=xx[ln(x)+1]\frac{dy}{dx} = y[\ln(x) + 1] = x^x[\ln(x) + 1]

Verification: For x=1x = 1, we get dydx=11[ln(1)+1]=11=1\frac{dy}{dx} = 1^1[\ln(1) + 1] = 1 \cdot 1 = 1, which is consistent with the behavior of y=xxy = x^x near x=1x = 1.


Example 3: Volume of a Solid of Revolution

Problem: Find the volume of the solid obtained by rotating the region bounded by y=xy = \sqrt{x}, y=0y = 0, and x=4x = 4 about the x-axis.

Solution:

Identify the bounds and functions. The region is bounded above by y=xy = \sqrt{x} and below by y=0y = 0 (the x-axis), from x=0x = 0 to x=4x = 4.

Apply the disk method formula for rotation about the x-axis [volume-of-solid-of-revolution]: V=π04[(x)202]dx=π04xdxV = \pi \int_0^4 \left[(\sqrt{x})^2 - 0^2\right] \, dx = \pi \int_0^4 x \, dx

Evaluate the integral: V=π[x22]04=π(1620)=8πV = \pi \left[\frac{x^2}{2}\right]_0^4 = \pi \left(\frac{16}{2} - 0\right) = 8\pi

Conclusion: The volume is 8π8\pi cubic units.


Example 4: Washer Method

Problem: Find the volume of the solid obtained by rotating the region bounded by y=xy = x and y=x2y = x^2 about the x-axis.

Solution:

First, find the intersection points: x=x2x(1x)=0x=0x = x^2 \Rightarrow x(1 - x) = 0 \Rightarrow x = 0 or x=1x = 1. For 0x10 \leq x \leq 1, we have xx2x \geq x^2.

Apply the washer method [volume-of-solid-of-revolution]: V=π01[x2(x2)2]dx=π01[x2x4]dxV = \pi \int_0^1 \left[x^2 - (x^2)^2\right] \, dx = \pi \int_0^1 \left[x^2 - x^4\right] \, dx

Evaluate: V=π[x33x55]01=π(1315)=π5315=2π15V = \pi \left[\frac{x^3}{3} - \frac{x^5}{5}\right]_0^1 = \pi \left(\frac{1}{3} - \frac{1}{5}\right) = \pi \cdot \frac{5 - 3}{15} = \frac{2\pi}{15}

Conclusion: The volume is 2π15\frac{2\pi}{15} cubic units.

References

AI Disclosure

This article was drafted with the assistance of an AI language model. The mathematical content, structure, and worked examples were generated based on class notes and standard Calculus II pedagogy. All claims are cited to source notes. The article has been reviewed for mathematical accuracy and clarity but should be verified against your course materials and textbooks before use in academic contexts.

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References

AI disclosure: Generated from personal class notes with AI assistance. Every factual claim cites a note. Model: claude-haiku-4-5-20251001.