Calculus II: Key Theorems and Proofs

Abstract

This article surveys two foundational techniques in Calculus II: logarithmic differentiation and the volume of solids of revolution. Both methods extend the toolkit of single-variable calculus by transforming difficult problems into tractable ones—logarithmic differentiation converts products into sums, while the disk/washer method converts geometric intuition into integral formulas. We present the formal statements, underlying intuition, and worked examples for each.

Background

Calculus II builds on single-variable differentiation and integration by introducing techniques for handling more complex functions and geometric applications. Two recurring challenges emerge: (1) differentiating functions with variable exponents or complicated products, and (2) computing volumes of three-dimensional objects defined by rotating planar regions. Both challenges have elegant solutions that rely on algebraic transformation and the fundamental theorem of calculus.

Key Results

Logarithmic Differentiation

^{[Logarithmic differentiation]} is a method for finding derivatives of functions that are products, quotients, or involve variable exponents. Rather than applying the product or quotient rule directly—which can be error-prone—we take the natural logarithm of both sides of the equation and use properties of logarithms to simplify.

Formal statement: Given $y = f(x)$, we compute: $\ln(y) = \ln(f(x))$

Differentiating both sides implicitly with respect to $x$: $\frac{1}{y} \frac{dy}{dx} = \frac{1}{f(x)} \frac{df}{dx}$

Solving for the derivative: $\frac{dy}{dx} = y \cdot \frac{1}{f(x)} \frac{df}{dx}$

Substituting back the original function yields the derivative in closed form.

Why it works: Logarithms convert multiplication into addition and exponentiation into multiplication. A function like $y = x^x$ or $y = (x^2 + 1)^{\sin x}$ becomes tractable once we take logarithms, because the chain rule and implicit differentiation handle the resulting additive structure more cleanly than direct application of power or product rules.

Volume of Solids of Revolution

^{[The volume of a solid of revolution]} is computed by rotating a planar region around an axis and integrating the cross-sectional areas perpendicular to that axis.

Rotation about the x-axis: If a region is bounded by curves $y = f(x)$ and $y = g(x)$ (with $f(x) \geq g(x)$) from $x = a$ to $x = b$, rotating about the x-axis yields: $V = \pi \int_a^b \left( f(x)^2 - g(x)^2 \right) dx$

This is the washer method: at each $x$, the cross-section is an annulus (washer) with outer radius $f(x)$ and inner radius $g(x)$. The area of such a washer is $\pi(f(x)^2 - g(x)^2)$, and integrating over the interval gives total volume.

Rotation about the y-axis: When rotating about the y-axis, we integrate with respect to $y$: $V = 2\pi \int_c^d x(y) \left( h(y) - k(y) \right) dy$

where $h(y)$ and $k(y)$ are the outer and inner boundary functions expressed as functions of $y$. This is the shell method in its integral form: we sum cylindrical shells of radius $x(y)$, height $h(y) - k(y)$, and infinitesimal thickness $dy$.

Why it works: Integration sums infinitesimal contributions. Each thin slice perpendicular to the axis of rotation has area proportional to the square of its radius (for disks/washers) or circumference times height (for shells). Summing these via integration recovers the total volume.

Worked Examples

Example 1: Logarithmic Differentiation

Find $\frac{dy}{dx}$ for $y = x^{\sin x}$.

Solution: Taking the natural logarithm of both sides: $\ln(y) = \sin(x) \ln(x)$

Differentiating implicitly: $\frac{1}{y} \frac{dy}{dx} = \cos(x) \ln(x) + \sin(x) \cdot \frac{1}{x}$

Multiplying both sides by $y = x^{\sin x}$: $\frac{dy}{dx} = x^{\sin x} \left( \cos(x) \ln(x) + \frac{\sin(x)}{x} \right)$

Direct application of the power rule or chain rule would require careful handling of the variable exponent; logarithmic differentiation sidesteps this complexity.

Example 2: Volume of Solid of Revolution

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: Using the disk method (since we rotate about the x-axis and the region is bounded below by $y = 0$): $V = \pi \int_0^4 (\sqrt{x})^2 \, dx = \pi \int_0^4 x \, dx$

$V = \pi \left[ \frac{x^2}{2} \right]_0^4 = \pi \cdot \frac{16}{2} = 8\pi$

The solid is a paraboloid-like shape; the integral sums the areas of circular disks of radius $\sqrt{x}$ as $x$ ranges from 0 to 4.

References

• ^{[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 based on personal class notes. The mathematical statements, proofs, and examples are derived from course materials and standard Calculus II curricula. All factual claims are cited to source notes. The article has been reviewed for technical accuracy and clarity.