Difficulty: Intermediate | Prerequisites: Calculus I integration techniques, basic geometry of solids
This material covers how to find the area between curves (vertical and horizontal slicing), how to compute volumes of solids of revolution using the disk/washer and shell methods, and how to set up and evaluate arc length integrals. These are core Exam 2 topics in Calculus II and show up as both computation and setup-only problems.
Area between curves
The integral of the "top function minus bottom function" (vertical slicing) or "right function minus left function" (horizontal slicing) over the interval where they enclose a region.
In simple terms, you are adding up thin rectangular strips between two curves across an interval.
Disk method
A technique for finding the volume of a solid of revolution by integrating pi times the square of the radius function along the axis of rotation: V = pi * integral of [R(x)]^2 dx.
Think of it as stacking circular coins (disks) of varying radius along the axis.
Washer method
A variation of the disk method used when the solid has a hole through its centre. The volume is pi * integral of ([R_outer]^2 - [R_inner]^2) dx.
Think of it as stacking flat rings (washers) rather than solid disks.
Shell method
A technique for finding the volume of a solid of revolution by integrating 2pix*f(x) dx, where x is the distance from the axis of rotation and f(x) is the height of the shell.
Think of it as peeling the solid into nested cylindrical layers, like rings of an onion.
Arc length
The total length of a curve y = f(x) from x = a to x = b, given by the integral of sqrt(1 + [f'(x)]^2) dx over [a, b].
In simple terms, you are measuring the actual distance along the curve, not just the horizontal distance.
Identify the top curve f(x) and the bottom curve g(x) over [a, b].
Area = integral from a to b of [f(x) - g(x)] dx.
Example (Problem 1a): Region bounded by y = sqrt(x) + 2, the y-axis, y = 0, and x = 4.
For 0 <= x <= 4, the curve y = sqrt(x) + 2 sits above y = 0.
A(R) = integral from 0 to 4 of [sqrt(x) + 2] dx = [(2/3)x^(3/2) + 2x] from 0 to 4 = (2/3)(8) + 8 = 16/3 + 8 = 40/3.
Rewrite curves as x = h(y) and x = k(y); the rightmost curve minus the leftmost.
Area = integral from c to d of [h_right(y) - h_left(y)] dy.
Useful when the region is more naturally described by horizontal strips (e.g. Problem 7a(i)).
Find the intersection points first by setting the two functions equal.
Example (Problem 2): y = e^(2x) + 3 and y = 5e^x - 3.
Set e^(2x) + 3 = 5e^x - 3. Let u = e^x: u^2 - 5u + 6 = 0, so u = 2 or u = 3, giving x = ln(2) and x = ln(3).
Check which function is on top between the intersection points.
A(R) = integral from ln(2) to ln(3) of [(5e^x - 3) - (e^(2x) + 3)] dx.
Problem 7: Region bounded by y = sqrt(x), y = x^2, and x = 1/2, for 0 <= x <= 1/2.
Vertical slicing: integral from 0 to 1/2 of [sqrt(x) - x^2] dx (since sqrt(x) >= x^2 on this interval).
Horizontal slicing requires splitting the region into sub-intervals and rewriting x in terms of y.
Revolving around the x-axis: V = pi * integral from a to b of [R(x)]^2 dx (disk) or pi * integral of ([R_outer]^2 - [R_inner]^2) dx (washer).
Revolving around the y-axis: rewrite functions as x = g(y) and integrate with respect to y.
Example (Problem 1b): Revolve the region under y = sqrt(x) + 2 (from x = 0 to x = 4, above y = 0) about the x-axis.
V = pi * integral from 0 to 4 of [sqrt(x) + 2]^2 dx = pi * integral of (x + 4*sqrt(x) + 4) dx.
Evaluate: pi * [x^2/2 + (8/3)x^(3/2) + 4x] from 0 to 4 = pi * [8 + 64/3 + 16] = pi * (136/3).
Example (Problem 8a): Revolve region between y = sqrt(x) and y = x^2 on [0, 1/2] about the x-axis.
V = pi * integral from 0 to 1/2 of [(sqrt(x))^2 - (x^2)^2] dx = pi * integral of [x - x^4] dx.
Used when the axis of rotation is perpendicular to the natural slicing direction.
Revolving about the y-axis with vertical slices: V = 2*pi * integral from a to b of x * f(x) dx.
Revolving about the x-axis with horizontal slices: V = 2*pi * integral from c to d of y * g(y) dy.
Example (Problem 8b): Set up volume of the region between y = sqrt(x) and y = x^2, 0 <= x <= 1/2, about the y-axis.
Shell method: V = 2*pi * integral from 0 to 1/2 of x * [sqrt(x) - x^2] dx.
Disk/washer method: rewrite as x = y^2 and x = sqrt(y), integrate with respect to y.
Problem 8c: Revolve about x = 1/2. The shell radius becomes |x - 1/2| rather than x.
V = 2*pi * integral from 0 to 1/2 of (1/2 - x) * [sqrt(x) - x^2] dx.
The key adjustment: replace x (or y) with the distance from the slice to the line of rotation.
For y = f(x) on [a, b]: L = integral from a to b of sqrt(1 + [f'(x)]^2) dx.
The expression under the square root comes from the Pythagorean theorem applied to an infinitesimal piece of the curve.
Curve: y = ln(cos(x)) on [0, pi/4].
Step 1: Find f'(x) = -sin(x)/cos(x) = -tan(x).
Step 2: Compute 1 + [f'(x)]^2 = 1 + tan^2(x) = sec^2(x).
Step 3: sqrt(sec^2(x)) = |sec(x)| = sec(x) (since sec(x) > 0 on [0, pi/4]).
Step 4: L = integral from 0 to pi/4 of sec(x) dx = ln|sec(x) + tan(x)| evaluated from 0 to pi/4.
Step 5: L = ln(sqrt(2) + 1) - ln(1) = ln(sqrt(2) + 1).
Arc length integrals appear whenever engineers need the actual length of a curved path, for instance when computing the length of cable needed for a suspension bridge or the distance a particle travels along a trajectory.
Students often forget to check which curve is on top before subtracting. If you integrate (bottom - top), you get a negative area. Always sketch or test a point in the interval.
When using the washer method, students sometimes square the entire expression (R_outer - R_inner)^2 instead of computing R_outer^2 - R_inner^2. These are different.
In the shell method, the radius is the distance from the slice to the axis of rotation, not the value of the function. When revolving about x = 1/2, the radius is |x - 1/2|, not x.
For arc length, students occasionally forget to square f'(x) before adding 1. The integrand is sqrt(1 + [f']^2), not sqrt(1 + f').
"Set up only" problems (like 8b and 8c) test whether you understand the method's structure without making you evaluate the integral. Know how to write the integrand, limits, and the correct constant (pi vs. 2*pi).
Problem 7 asks for the same area via both horizontal and vertical slicing. Expect to be tested on translating between the two approaches.
Arc length problems tend to be clean only when 1 + [f']^2 simplifies (as in Problem 4, where it becomes sec^2(x)). If it does not simplify, you may be asked to set up only.
Revolving about a non-standard axis (x = 1/2, y = -1, etc.) is a common exam twist. The only change is the radius expression.
True or False: The disk method always uses pi * integral of [f(x)]^2 dx, even when there is a hole in the solid. (False, that is the washer method.)
Fill in the blank: In the shell method about the y-axis, the volume element is 2*pi * ____ * height * dx. (Answer: x, the radius.)
True or False: To find the area between y = x^2 and y = x^3 on [0, 1], you integrate (x^3 - x^2) dx. (False, x^2 >= x^3 on [0, 1], so it is (x^2 - x^3) dx.)
Fill in the blank: The arc length integrand is sqrt(1 + ____). (Answer: [f'(x)]^2.)
True or False: When revolving about x = 3, the shell radius for a vertical strip at position x is (3 - x) if x < 3. (True.)
Q: Set up (do not evaluate) the integral for the volume obtained by revolving the region between y = sqrt(x) and y = x^2, 0 <= x <= 1/2, about the y-axis using the shell method.
A: V = 2*pi * integral from 0 to 1/2 of x * [sqrt(x) - x^2] dx.
Q: Find the area of the region enclosed by y = e^(2x) + 3 and y = 5e^x - 3.
A: Set e^(2x) + 3 = 5e^x - 3. Let u = e^x, so u^2 - 5u + 6 = 0, giving u = 2, 3 (x = ln 2, ln 3). Between these values, 5e^x - 3 > e^(2x) + 3. Area = integral from ln(2) to ln(3) of [(5e^x - 6) - e^(2x)] dx = [5e^x - 6x - (1/2)e^(2x)] from ln 2 to ln 3 = (15 - 6 ln 3 - 9/2) - (10 - 6 ln 2 - 2) = 3/2 - 6 ln 3 + 6 ln 2 = 3/2 + 6 ln(2/3).
Q: Find the arc length of y = ln(cos x) on [0, pi/4].
A: f'(x) = -tan(x), so 1 + [f']^2 = sec^2(x). L = integral from 0 to pi/4 of sec(x) dx = ln|sec(x) + tan(x)| from 0 to pi/4 = ln(sqrt(2) + 1).
Q: Using the disk method, find the volume generated by revolving the region under y = sqrt(x) + 2, above y = 0, from x = 0 to x = 4, about the x-axis.
A: V = pi * integral from 0 to 4 of (sqrt(x) + 2)^2 dx = pi * integral of (x + 4sqrt(x) + 4) dx = pi * [x^2/2 + (8/3)x^(3/2) + 4x] from 0 to 4 = 136pi/3.
Area between curves, volume of revolution, disk method, washer method, shell method, cylindrical shells, solids of revolution, arc length formula, arc length integral, revolving about x-axis, revolving about y-axis, revolving about a line, horizontal slicing, vertical slicing, cross-sectional area, Calculus II, MA-UY 1124, MA-UY 1424, NYU Courant