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Calculus II (Integration & Series) Unit 5: Applications: area, volume, arc length

Integrals measure more than area: volume, arc length, surface area and physical work are all accumulations.

Covers: Volumes (discs, washers, shells), arc length, surface area, work

Study guide

Integrals compute geometric quantities by adding thin slices. Volumes of revolution come in two styles. The disc and washer methods slice perpendicular to the axis of rotation: V = π∫ (R² − r²) dx, with outer radius R and inner radius r (r = 0 gives discs). The shell method slices parallel to the axis: each thin cylindrical shell has volume 2π·radius·height·thickness, so V = 2π∫ r(x)h(x) dx. Shells are often easier when rotating around a vertical line with the region described by functions of x. Radii are always distances from the axis, like x + 1 for the axis x = −1.

Arc length adds up tiny hypotenuses: the length of y = f(x) from a to b is L = ∫ₐᵇ √(1 + (f′(x))²) dx. These integrals are often hard to compute exactly, so many problems only ask you to set them up or approximate.

The surface area of revolution of y = f(x) rotated about the x-axis is S = 2π∫ f(x)√(1 + (f′(x))²) dx: the circumference times the slanted arc length.

Work is force times distance, and when the force varies, W = ∫ F(x) dx. Spring problems use Hooke's law, F = kx. Pumping problems slice the liquid into thin layers, find each layer's weight and the distance it must be lifted, and integrate.

The area between two curves is ∫ₐᵇ (top − bottom) dx, with limits at the intersection points. Split the integral where the curves cross, or integrate (right − left) dy when the region is easier to describe horizontally. Rotating y = f(x) about the y-axis uses radius x: S = 2π∫ x√(1 + (f′(x))²) dx. For polar or parametric curves, use ds = √(r² + (dr/dθ)²) dθ or ds = √((dx/dt)² + (dy/dt)²) dt, and take the distance to the axis as the radius.

Questions often give an integral and ask what it represents. ∫ₐᵇ √(1 + (f′(x))²) dx is the length of the graph from a to b, π∫ f(x)² dx is a disc volume, and ∫ (top − bottom) dx is an area. For pumping, W = ∫ (weight density)·A(y)·(distance lifted) dy, where A(y) is the area of the layer at height y. Water weighs about 9800 N/m³, or 62.4 lb/ft³.

Key ideas

• Shell method: V = 2π∫ (radius)(height) dx, useful when slicing parallel to the axis.
• Disc and washer: V = π∫ (R² − r²) dx, slicing perpendicular to the axis.
• Arc length adds tiny hypotenuses.
• Work = force × distance; when force varies, integrate.

Formulas and rules

Common mistakes

• Using the wrong radius when rotating around a line other than an axis.
• Forgetting the square root in arc length.

Words to know

Shell method volume

V = 2π∫ (radius)(height) dx

Arc length formula

L = ∫ √(1 + (f′(x))²) dx

Work done by a variable force

W = ∫ F(x) dx

Shell vs. washer method

Shells use slices parallel to the axis of rotation; washers use perpendicular slices.

Shell radius

Distance from the slice to the axis of rotation (x + 1 for the axis x = −1).

Surface area of revolution

S = 2π∫ f(x)√(1 + (f′(x))²) dx

Disc method

V = π∫ r² dx

Washer method

V = π∫ (R² − r²) dx

Hooke's law

F = kx for a spring.

Pumping problems

Slice the liquid, find each slice's weight and lift distance, integrate.

Work (variable force)

W = ∫ F(x) dx

Sample practice questions

28 more questions like these, from level 1 to 5, are in the practice set.