Difficulty: Intermediate | Prerequisites: Basic antiderivatives, u-substitution, and the trig integral techniques from the first part of HW 05.
Integration by parts and volumes of revolution are two pillars of a first integral calculus course. Integration by parts extends your toolkit beyond substitution: it handles products of functions where one factor simplifies when differentiated and the other is easy to integrate. Volumes of revolution take the area-under-a-curve idea and spin it into three dimensions, producing solids whose volumes you compute with definite integrals. Both topics are tested extensively and tend to appear together on exams, since the volume integrals often require integration by parts or trig identities to evaluate.
Integration by parts rewrites ∫ u dv as uv - ∫ v du, letting you shift difficulty from one factor to the other. Volumes of revolution use the disk method (V = π ∫ [f(x)]² dx) for single-curve solids and the washer method (V = π ∫ [f(x)² - g(x)²] dx) when a gap exists between two curves.
Integration by parts (IBP)
The formula ∫ u dv = uv - ∫ v du. Think of it as: the product rule for derivatives, run in reverse. You choose u (the part you differentiate) and dv (the part you integrate).
LIATE rule
A mnemonic for choosing u in IBP: Logarithmic, Inverse trig, Algebraic, Trigonometric, Exponential. Pick u from whichever category comes first in this list. In simple terms, choose the function that gets simpler when you differentiate it.
Disk method
V = π ∫ₐᵇ [f(x)]² dx. Used when a region bounded by one curve and an axis is rotated about that axis. The cross-sections are solid disks.
Washer method
V = π ∫ₐᵇ [f(x)² - g(x)²] dx. Used when the region between two curves is rotated about an axis, leaving a hole in the middle. The cross-sections are washers (rings).
Solid of revolution
The three-dimensional shape formed by rotating a two-dimensional region about an axis. In simple terms, imagine the region spinning like a potter's wheel.
The core formula: ∫ u dv = uv - ∫ v du.
The trick is choosing u and dv so that the new integral ∫ v du is simpler than the original.
x ln(x) type (Q11): ∫ 5x ln(x) dx. Let u = ln(x), dv = 5x dx. Then du = (1/x) dx, v = (5/2)x². Result: (5/2)x² ln(x) - (5/4)x² + C.
Polynomial times exponential (Q12): ∫ (x - 2)e^{2x} dx. Let u = (x - 2), dv = e^{2x} dx. Then du = dx, v = (1/2)e^{2x}. After applying IBP: (1/2)(x - 2)e^{2x} - (1/4)e^{2x} + C = (1/4)(2x - 5)e^{2x} + C.
x sin(x) type (Q13): ∫₀^{π/2} x sin(x) dx. Let u = x, dv = sin(x) dx. Then du = dx, v = -cos(x). Result: [-x cos(x) + sin(x)]₀^{π/2} = 1.
When rotating y = f(x) about the x-axis on [a, b], the cross-section at x is a circle with radius f(x).
V = π ∫ₐᵇ [f(x)]² dx.
Example (Q7): Rotate y = 2 sin(x) from x = 0 to x = π/4 about the x-axis. V = π ∫₀^{π/4} 4 sin²(x) dx. Use the half-angle identity: V = 2π ∫₀^{π/4} (1 - cos(2x)) dx = π(π/2 - 1).
Example (Q14): Rotate y = 2/x from x = 2 to x = 3. V = π ∫₂³ 4/x² dx = π[-4/x]₂³ = (2/3)π.
Example (Q15): Rotate y = sec(x) from x = 0 to x = π/3. V = π ∫₀^{π/3} sec²(x) dx = π[tan(x)]₀^{π/3} = π√3.
When the region sits between two curves f(x) (outer) and g(x) (inner), the cross-section is a washer.
V = π ∫ₐᵇ [f(x)² - g(x)²] dx.
Example (Q16): The region between y = 2 and y = 1 + √x, from x = 0 to x = 1, rotated about the x-axis. f(x) = 2, g(x) = 1 + √x. f² - g² = 4 - (1 + √x)² = 3 - 2√x - x. V = π ∫₀¹ (3 - 2√x - x) dx = (7/6)π.
\int u \, dv = uv - \int v \, duV_{\text{disk}} = \pi \int_a^b [f(x)]^2 \, dxV_{\text{washer}} = \pi \int_a^b \bigl[f(x)^2 - g(x)^2\bigr] \, dxVolumes of revolution show up whenever you need the capacity of a rotationally symmetric container: tanks, bottles, turbine housings, or lathe-turned parts. Engineers use the disk and washer methods to calculate material volumes for manufacturing. Integration by parts is essential in probability and statistics for computing expected values when the probability density function involves products of functions.
Students often choose u and dv backwards in IBP. If the new integral is harder than the original, you picked the wrong u. Swap and try again.
Forgetting the constant of integration in indefinite IBP problems. Every indefinite integral answer needs + C.
In volume problems, students sometimes forget to square the function. The formula is π ∫ [f(x)]² dx, not π ∫ f(x) dx. Squaring changes everything.
Using the disk method when the washer method is needed. If the region does not touch the axis of rotation, there is a hole, and you must subtract the inner radius squared.
⚠️ IBP problems almost always appear on the exam. Be ready for x ln(x), x sin(x), x eˣ, and ln(x) alone (where dv = dx).
⚠️ Volume problems frequently combine the disk/washer setup with a trig integral or IBP integral to evaluate. Set up the integral correctly first, then evaluate.
⚠️ Know when to use disk vs washer. If only one curve bounds the region against the axis, use disk. If two curves bound the region, use washer.
⚠️ Read volume problems carefully for the axis of rotation. Rotating about the y-axis requires rewriting in terms of y (or using the shell method, covered later).
True or False: In ∫ x eˣ dx, you should let u = eˣ. (False: let u = x so it simplifies when differentiated.)
Fill in the blank: The disk method formula for rotation about the x-axis is V = π ∫ₐᵇ ______. ([f(x)]² dx)
True or False: The washer method subtracts the inner radius squared, not the inner radius. (True.)
Fill in the blank: ∫ u dv = ______. (uv - ∫ v du)
True or False: When rotating the region between y = 2 and y = 1 + √x about the x-axis, you use the disk method. (False: two curves means washer method.)
Q: Evaluate ∫ 5x ln(x) dx.
A: Let u = ln(x), dv = 5x dx. Then v = (5/2)x², du = (1/x) dx. IBP gives (5/2)x² ln(x) - (5/2) ∫ x dx = (5/2)x² ln(x) - (5/4)x² + C.
Q: Evaluate ∫ (x - 2)e^{2x} dx.
A: Let u = x - 2, dv = e^{2x} dx. v = (1/2)e^{2x}. IBP gives (1/2)(x - 2)e^{2x} - (1/4)e^{2x} + C = (1/4)(2x - 5)e^{2x} + C.
Q: Find the volume of the solid obtained by rotating y = 2 sin(x) from x = 0 to x = π/4 about the x-axis.
A: V = π ∫₀^{π/4} 4 sin²(x) dx = 2π ∫₀^{π/4} (1 - cos(2x)) dx = 2π[x - (1/2) sin(2x)]₀^{π/4} = π(π/2 - 1).
Q: Find the volume obtained by rotating the region between y = 2 and y = 1 + √x (from x = 0 to x = 1) about the x-axis.
A: Washer method. V = π ∫₀¹ (4 - (1 + √x)²) dx = π ∫₀¹ (3 - 2√x - x) dx = π[3x - (4/3)x^{3/2} - (1/2)x²]₀¹ = (7/6)π.
Q: Find the volume obtained by rotating y = sec(x) from x = 0 to x = π/3 about the x-axis.
A: V = π ∫₀^{π/3} sec²(x) dx = π[tan(x)]₀^{π/3} = π√3.
Integration by parts connects to the upcoming topic of partial fractions, where rational functions are decomposed before integrating. The volume techniques here extend naturally to the shell method (cylindrical shells), which handles rotation about the y-axis more elegantly for many problems. Both IBP and volumes also connect to arc length and surface area of revolution, which appear later in the course.
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