Calculus II: Worked Example Walkthroughs for Convergence, Differentiation, and Volumes

Abstract

This article presents 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 technique and intuition. The examples are drawn from standard Calculus II curricula and illustrate how to apply convergence tests, handle variable exponents, and set up integration bounds for rotational solids.

Background

Calculus II extends single-variable calculus by introducing techniques for handling unbounded regions, complex function forms, and geometric applications. Three topics frequently appear together in course assessments: determining whether improper integrals converge, differentiating functions with variable exponents, and computing volumes of three-dimensional objects formed by rotation. Mastery of these topics requires both procedural fluency and conceptual understanding.

^{[convergence-of-integrals]} establishes that an integral converges when the limit of integration yields a finite value, and diverges otherwise. This distinction is essential because many real-world applications—from probability to physics—depend on knowing whether a computed quantity is meaningful or infinite.

^{[logarithmic-differentiation]} describes a technique that transforms multiplicative relationships into additive ones, making differentiation tractable for functions where the variable appears in both base and exponent. This method avoids errors that arise from misapplying product or quotient rules.

^{[volume-of-solid-of-revolution]} provides formulas for computing volumes when a planar region is rotated about an axis. These calculations underpin applications in engineering and manufacturing where rotationally symmetric objects are common.

Key Results

Convergence of Improper Integrals

An integral $\int_a^b f(x) \, dx$ is convergent if the limit of the integral exists and is finite as the bounds approach their limits ^{[convergence-of-integrals]}. For improper integrals—those with infinite limits or discontinuous integrands—convergence is not guaranteed and must be verified.

The intuition is straightforward: if the area under a curve is infinite, the integral diverges. Convergence tests such as the comparison test and limit comparison test allow us to determine convergence without computing the integral directly ^{[convergence-of-integrals]}.

Logarithmic Differentiation

For a function $y = f(x)$, logarithmic differentiation proceeds by taking the natural logarithm of both sides: $\ln(y) = \ln(f(x))$

Differentiating both sides with respect to $x$ using the chain rule yields: $\frac{1}{y} \frac{dy}{dx} = \frac{f'(x)}{f(x)}$

Solving for the derivative: $\frac{dy}{dx} = y \cdot \frac{f'(x)}{f(x)}$

Substituting back the original function gives the final result ^{[logarithmic-differentiation]}. This approach is particularly effective when the variable appears in both the base and exponent, or when the function is a product or quotient of complex terms.

Volume of Solids of Revolution

When a region bounded by curves $y = f(x)$ and $y = g(x)$ from $x = a$ to $x = b$ is rotated about the x-axis, the volume is: $V = \pi \int_a^b (f(x)^2 - g(x)^2) \, dx$

For rotation about the y-axis, the formula becomes: $V = 2\pi \int_c^d x (h(y) - k(y)) \, dy$

where $h(y)$ and $k(y)$ are the outer and inner functions respectively ^{[volume-of-solid-of-revolution]}. The choice of axis and the form of the integrand depend on the geometry of the region and which variable is more natural to integrate over.

Worked Examples

Example 1: Testing Convergence of an Improper Integral

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

Solution:

This is an improper integral with an infinite upper limit. We evaluate it as a limit: $\int_1^{\infty} \frac{1}{x^2} \, dx = \lim_{t \to \infty} \int_1^t \frac{1}{x^2} \, dx$

Compute the antiderivative: $\int \frac{1}{x^2} \, dx = -\frac{1}{x}$

Evaluate the definite integral: $\lim_{t \to \infty} \left[ -\frac{1}{x} \right]_1^t = \lim_{t \to \infty} \left( -\frac{1}{t} + 1 \right) = 0 + 1 = 1$

Since the limit exists and equals 1 (a finite value), the integral converges ^{[convergence-of-integrals]}.

---

Example 2: Logarithmic Differentiation with Variable Exponent

Problem: Find $\frac{dy}{dx}$ for $y = x^x$ where $x > 0$.

Solution:

Take the natural logarithm of both sides: $\ln(y) = \ln(x^x) = x \ln(x)$

Differentiate both sides with respect to $x$: $\frac{1}{y} \frac{dy}{dx} = \frac{d}{dx}[x \ln(x)]$

Apply the product rule on the right side: $\frac{1}{y} \frac{dy}{dx} = \ln(x) + x \cdot \frac{1}{x} = \ln(x) + 1$

Multiply both sides by $y = x^x$: $\frac{dy}{dx} = x^x (\ln(x) + 1)$

This result would be difficult to obtain using standard differentiation rules directly ^{[logarithmic-differentiation]}.

---

Example 3: Volume of a Solid of Revolution

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

Solution:

The region is bounded above by $y = \sqrt{x}$ and below by $y = 0$ (the x-axis), from $x = 0$ to $x = 4$. Rotating about the x-axis, we use the disk method: $V = \pi \int_0^4 (\sqrt{x})^2 \, dx = \pi \int_0^4 x \, dx$

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

The volume is $8\pi$ cubic units ^{[volume-of-solid-of-revolution]}.

---

Example 4: Volume with Rotation About the y-axis

Problem: Find the volume when the region bounded by $x = y^2$ and $x = 4$ is rotated about the y-axis.

Solution:

The curves intersect where $y^2 = 4$, so $y = \pm 2$. The region extends from $y = -2$ to $y = 2$, with outer radius $x = 4$ and inner radius $x = y^2$. Using the washer method: $V = \pi \int_{-2}^2 (4^2 - (y^2)^2) \, dy = \pi \int_{-2}^2 (16 - y^4) \, dy$

Evaluate: $V = \pi \left[ 16y - \frac{y^5}{5} \right]_{-2}^2$

$= \pi \left[ \left( 32 - \frac{32}{5} \right) - \left( -32 + \frac{32}{5} \right) \right]$

$= \pi \left[ 64 - \frac{64}{5} \right] = \pi \cdot \frac{256}{5} = \frac{256\pi}{5}$

The volume is $\frac{256\pi}{5}$ cubic units ^{[volume-of-solid-of-revolution]}.

References

• ^{[convergence-of-integrals]}
• ^{[convergence-of-integrals]}
• ^{[logarithmic-differentiation]}
• ^{[logarithmic-differentiation]}
• ^{[volume-of-solid-of-revolution]}
• ^{[volume-of-solid-of-revolution]}

AI Disclosure

This article was drafted with the assistance of an AI language model. The mathematical content and worked examples are derived from class notes and standard Calculus II curricula. All claims are cited to source notes. The article has been reviewed for mathematical accuracy and clarity, but readers should verify examples against their course materials and textbooks.