Tags: area between curves, volume of revolution, disk method, shell method, work, Hooke's Law, pumping work, arc length, applications of integration, Calculus II, Texas A&M, MATH 172, MATH 2414
This covers the main applications of definite integrals in Calc II: finding areas between curves, computing volumes of solids of revolution (disk and shell methods), calculating work done by variable forces (springs, pumping fluids), and measuring arc length. These problems all follow the same pattern: slice the region, write a formula for each slice, then integrate.
Area between curves
The integral of (top function − bottom function) over the interval where the curves intersect. Can be set up with respect to x or y.
Disk method
Finds volumes of revolution by integrating π[r(x)]² dx, where r(x) is the distance from the curve to the axis of revolution. Used when slicing perpendicular to the axis of revolution.
Shell method
Finds volumes of revolution by integrating 2π(radius)(height) dx. Used when slicing parallel to the axis of revolution. Often simpler than the disk method when revolving around a vertical line and integrating with respect to x.
Work (physics)
Work = Force × Distance. When force varies, W = ∫F(x) dx. Measured in joules (N·m).
Hooke's Law
F = kx, where k is the spring constant and x is displacement from natural length.
Arc length
The length of a curve y = f(x) from a to b: L = ∫ₐᵇ √(1 + [f'(x)]²) dx. For parametric curves: L = ∫ √((dx/dt)² + (dy/dt)²) dt.
The basic idea: integrate the vertical (or horizontal) distance between two curves over their overlapping region.
Integrating with respect to x:
A = ∫ₐᵇ (top − bottom) dx
Find intersection points to determine limits
Integrating with respect to y:
Rewrite curves as x = f(y)
A = ∫ (right − left) dy
Useful when horizontal slices are simpler
Worked example: Area between y = x³ and y = x
Intersections: x³ = x, so x(x² − 1) = 0, giving x = −1, 0, 1
By symmetry, compute 2 × ∫₀¹ (x − x³) dx
= 2[x²/2 − x⁴/4] from 0 to 1
= 2(1/2 − 1/4) = 1/2
Worked example: Area bounded by y = √x, y = x/2, and the y-axis (integrating with respect to y)
Rewrite: x = y² (from y = √x) and x = 2y (from y = x/2)
Intersection: y² = 2y, so y(y − 2) = 0, giving y = 0 and y = 2
In [0, 2], the line x = 2y is to the right of x = y²
A = ∫₀² (2y − y²) dy = [y² − y³/3] from 0 to 2 = 4 − 8/3 = 4/3
When revolving around the x-axis and slicing perpendicular to it, each slice is a disk.
V = ∫ₐᵇ π[r(x)]² dx
Worked example: y = √x, x = 0 to x = 4, revolved around the x-axis
Radius of each disk = √x
V = ∫₀⁴ π(√x)² dx = π∫₀⁴ x dx
= π[x²/2] from 0 to 4 = 8π
When revolving around a vertical line and integrating with respect to x, each slice is a cylindrical shell.
V = ∫ₐᵇ 2π(radius)(height) dx
Radius = distance from x to the axis of revolution
Height = the function value (or difference of functions)
Worked example: y = eˣ, y = 0, x = 0, x = 1, revolved about x = 2
Radius = 2 − x (distance from the slice at x to the axis x = 2)
Height = eˣ
V = 2π∫₀¹ (2 − x)eˣ dx
Integration by parts: let u = 2 − x, dv = eˣ dx, then du = −dx, v = eˣ
(2 − x)eˣ + ∫eˣ dx = (2 − x)eˣ + eˣ = (3 − x)eˣ
V = 2π[(3 − x)eˣ] from 0 to 1 = 2π(2e − 3)
Hooke's Law: F = kx. Find k from the given conditions, then integrate.
W = ∫₀ᵈ kx dx
Worked example: A force of 90 N stretches a spring 1 m. Find the work to stretch it 2 m.
90 = k(1), so k = 90 N/m
W = ∫₀² 90x dx = [45x²] from 0 to 2 = 45(4) = 180 J
The pattern: slice the fluid into thin horizontal layers, compute the weight of each layer, multiply by the distance it must travel, and integrate.
For each slice at height y:
Find the cross-sectional area A(y) using the geometry of the container
Volume of slice: dV = A(y) dy
Weight of slice: ρg · dV
Distance to pump: depends on setup (typically top − y, or just y if measuring from the top)
Work = ∫ ρg · A(y) · (distance) dy
Worked example: Semicircular trough, length 3 m, radius 0.25 m, full of water
At depth y from the top, the width of the semicircular cross-section is 2√(0.25² − y²)
Slice volume: dV = 3 · 2√(0.0625 − y²) dy
Each slice must be lifted a distance y to reach the top
W = ∫₀^(0.25) 1000 · 9.8 · 6√(0.0625 − y²) · y dy
Worked example: Inverted cone, height 6 m, base radius 1.5 m, full of milk (ρ = 1030)
By similar triangles at height h from the bottom: r = h/4
Slice volume at height h: π(h/4)² dh
Distance to pump to the top: (6 − h)
W = ∫₀⁶ 1030 · 9.8 · π(h/4)²(6 − h) dh
Evaluates to approximately 142,855 J
The similar-triangles step is the part most students trip on. Draw the cross-section and label the radius at height h.
L = ∫ₐᵇ √((dx/dt)² + (dy/dt)²) dt
Worked example: x = t², y = t³, for 0 ≤ t ≤ 1
dx/dt = 2t, dy/dt = 3t²
L = ∫₀¹ √(4t² + 9t⁴) dt = ∫₀¹ t√(4 + 9t²) dt
Substitute u = 4 + 9t², du = 18t dt
L = (1/18)∫₄¹³ u^(1/2) du = (1/18)[2u^(3/2)/3] from 4 to 13
= (1/27)(13√13 − 8)
Application | Formula |
|---|---|
Area (wrt x) | ∫(top − bottom) dx |
Area (wrt y) | ∫(right − left) dy |
Disk method | V = ∫π[r(x)]² dx |
Shell method | V = ∫2π(radius)(height) dx |
Work (spring) | W = ∫kx dx |
Work (pumping) | W = ∫ρg · A(y) · distance dy |
Arc length (parametric) | L = ∫√((dx/dt)² + (dy/dt)²) dt |
⚠️ For area between curves, always check which function is on top (or to the right) over the interval. If the curves cross, split the integral at the crossing points.
⚠️ Disk vs shell: choose based on which gives a simpler integral. If revolving around a horizontal axis and the functions are given as y = f(x), disks are usually easier. If revolving around a vertical axis with y = f(x), shells are usually easier.
⚠️ For shell method around x = 2 (not the y-axis), the radius is (2 − x), not just x. The axis of revolution changes the radius expression.
⚠️ Pumping problems: clearly define your coordinate system (measuring from top or bottom) and use similar triangles for tapered containers. The distance each slice travels depends on your coordinate choice.
⚠️ In arc length problems, simplifying the integrand before substituting is essential. Look for common factors you can pull out from under the square root.
Q: Find the area between y = x³ and y = x.
A: 1/2. Intersections at x = −1, 0, 1. By symmetry, 2∫₀¹(x − x³) dx = 1/2.
Q: Find the volume when y = √x (from x = 0 to x = 4) is revolved around the x-axis using the disk method.
A: 8π. V = π∫₀⁴ x dx = 8π.
Q: A spring requires 90 N to stretch 1 m. How much work to stretch it 2 m?
A: 180 J. k = 90, W = ∫₀² 90x dx = 180 J.
Q: Find the volume when y = eˣ (from x = 0 to 1) is revolved about x = 2 using shells.
A: 2π(2e − 3). Shell radius = (2 − x), height = eˣ, then integrate by parts.
Q: Calculate the arc length of x = t², y = t³ for t ∈ [0, 1].
A: (13√13 − 8)/27. Simplify the integrand to t√(4 + 9t²) and use u-substitution.
Q: Find the area in the first quadrant bounded by y = √x, y = x/2, and the y-axis, integrating with respect to y.
A: 4/3. Rewrite as x = y² and x = 2y, integrate ∫₀²(2y − y²) dy.
area between curves, volume of revolution, disk method, washer method, shell method, cylindrical shells, work integral, Hooke's Law, spring work, pumping work, conical tank, trough, similar triangles, arc length, parametric arc length, applications of integrals, Calc II applications, MATH 2414, MATH 172