MATH 104, Sections 6.3, 6.4, and 6.6 | Source: Rimmer lecture notes
Difficulty: Intermediate | Prerequisites: Definite integrals, derivatives, Volumes by Slicing (Section 6.1)
After learning how to compute volumes of solids, Calculus II extends the same slicing-and-integrating logic to two more geometric measurements: the length of a curve and the surface area of a solid of revolution. Section 6.6 then shifts to physics, using integrals to locate the balance point (centre of mass) of a flat plate or a system of point masses. All three topics rely on the same core skill: setting up a definite integral by approximating a quantity with small pieces and taking a limit.
Arc length measures how long a curve is by integrating sqrt(1 + [f'(x)]^2). Surface area of revolution wraps that arc length around an axis, multiplying by 2 pi times the radius. Centre of mass uses moment integrals (x times mass, y times mass) divided by total mass to find the balance point of a system or a flat region.
Arc length
The total distance measured along a curve between two points. In simple terms, if you laid the curve flat and measured it with a tape measure, that is the arc length.
Surface area of revolution
The area of the surface generated when a curve is rotated around an axis. Think of it as the amount of material you would need to wrap around the resulting 3D shape.
Centre of mass (centroid)
The single point where a system of masses, or a region, would balance perfectly. For a uniform flat plate, the centre of mass is the same as the geometric centroid.
Moment about an axis
A quantity that measures the tendency of a system to rotate about that axis. Moment about the y-axis is the sum of each mass times its x-coordinate. Moment about the x-axis is the sum of each mass times its y-coordinate.
Lamina
A thin, flat plate with uniform density. When the density is constant, it cancels out and the centre of mass depends only on the shape of the region.
The idea is to approximate a curve by a chain of short line segments, compute the length of each segment using the distance formula, then take the limit.
For a segment between consecutive points on the curve, the distance formula gives:
d(P_{k-1}, P_k) = \sqrt{(\Delta x)^2 + (\Delta y)^2}By the Mean Value Theorem, there exists a point x_k* in each sub-interval such that the change in y equals f'(x_k*) times dx. Substituting and factoring out dx:
d(P_{k-1}, P_k) = \sqrt{1 + \bigl[f'(x_k^*)\bigr]^2}\;\Delta xSumming all segments and taking the limit gives the arc length formula:
L = \int_a^b \sqrt{1 + \bigl[f'(x)\bigr]^2}\, dxIf the curve is given as x = g(y) for c <= y <= d, the formula becomes:
L = \int_c^d \sqrt{1 + \bigl[g'(y)\bigr]^2}\, dyThe shorthand ds: The quantity sqrt(1 + [f'(x)]^2) dx is often written as ds, representing an infinitesimal arc length element. This notation reappears in the surface area formula.
When a curve y = f(x) is revolved around an axis, it sweeps out a surface. To find the area of that surface, think of each small arc-length piece ds tracing out a thin band (like a strip of tape wrapped around a cone). The area of each band is 2 pi times the radius times the arc length of the strip.
Rotation about the x-axis:
The radius of each band is the y-value, f(x). The surface area is:
SA = \int_a^b 2\pi\, f(x)\,\sqrt{1 + \bigl[f'(x)\bigr]^2}\, dxIn shorthand: SA = integral of 2 pi f(x) ds.
Rotation about the y-axis:
The radius of each band is the x-value. The surface area is:
SA = \int_a^b 2\pi\, x\,\sqrt{1 + \bigl[f'(x)\bigr]^2}\, dxIf the curve is given as x = g(y), the formula adapts accordingly, with the radius and ds expressed in terms of y.
Important requirement: The function must have a continuous derivative on the interval. If the derivative is undefined at an endpoint (as with y = sqrt(x) at x = 0), treat the integral as improper.
Discrete point masses on a line:
For masses m_1, m_2, ..., m_n at positions x_1, x_2, ..., x_n, the centre of mass is:
\bar{x} = \frac{\sum m_i x_i}{\sum m_i} = \frac{M_y}{\text{total mass}}This is Archimedes' Law of the Lever: the rod balances when m_1 d_1 = m_2 d_2.
Discrete point masses in the plane:
For masses at positions (x_i, y_i), compute two moments:
M_y (moment about the y-axis) = sum of m_i x_i. This gives x-bar.
M_x (moment about the x-axis) = sum of m_i y_i. This gives y-bar.
The centre of mass is the point (x-bar, y-bar) where concentrating all the mass would produce the same moments as the system.
Continuous flat plate (lamina) with uniform density:
For a region bounded above by y = f(x) and below by y = g(x) on [a, b] with constant density rho:
\bar{x} = \frac{1}{A}\int_a^b x\bigl[f(x) - g(x)\bigr]\, dx\bar{y} = \frac{1}{A}\int_a^b \frac{f(x) + g(x)}{2}\bigl[f(x) - g(x)\bigr]\, dxwhere A is the area of the region. The density cancels out when it is constant, so the centroid depends only on the geometry.
The y-bar formula uses the average height [f(x) + g(x)] / 2 because each thin vertical strip has its own centre halfway up.
Topic | Formula | Notes |
|---|---|---|
Arc length (y = f(x)) | L = integral of sqrt(1 + [f'(x)]^2) dx | f' must be continuous on [a, b] |
Arc length (x = g(y)) | L = integral of sqrt(1 + [g'(y)]^2) dy | Same idea, variable switched |
Surface area (rotation about x-axis) | SA = integral of 2 pi f(x) sqrt(1 + [f'(x)]^2) dx | Radius = f(x), length = ds |
Surface area (rotation about y-axis) | SA = integral of 2 pi x sqrt(1 + [f'(x)]^2) dx | Radius = x |
Centre of mass (discrete, 1D) | x-bar = (sum of m_i x_i) / (total mass) | Archimedes' lever law |
Centroid x-bar (lamina) | x-bar = (1/A) integral of x [f(x) minus g(x)] dx | Density cancels for uniform plates |
Centroid y-bar (lamina) | y-bar = (1/A) integral of [(f+g)/2] [f minus g] dx | Uses average height of strip |
Arc length calculations are used in road design and cable engineering: determining how much cable is needed to hang between two poles along a catenary curve. Surface area of revolution is precisely the calculation behind designing pressure vessels, rocket nose cones, and any axially symmetric container where you need to know how much material to fabricate. Centre of mass is fundamental in structural engineering and robotics, telling you where to support a beam or how a robotic arm will balance under load.
Students often forget the square root in the arc length formula. The integrand is sqrt(1 + [f'(x)]^2), not just 1 + [f'(x)]^2.
In the surface area formula, students sometimes use f(x) squared inside the integral, confusing it with the disk method. Surface area uses 2 pi f(x) times ds, with no squaring of the radius.
For centre of mass, students frequently mix up M_x and M_y. Remember: M_y (moment about the y-axis) involves x-coordinates and gives x-bar, while M_x (moment about the x-axis) involves y-coordinates and gives y-bar. The subscript names the axis you are measuring the tendency to rotate about.
In the y-bar formula for a lamina, students forget to use the average height (f + g) / 2 and instead just use f(x).
Arc length integrals are often chosen so that the expression under the square root simplifies neatly. If yours does not simplify, double-check your derivative.
Surface area problems almost always involve rotation about the x-axis. Rotation about the y-axis is less common but does appear.
Centre of mass problems for laminas between two curves are a staple exam question. You need both x-bar and y-bar.
Know the difference between moment about the x-axis and moment about the y-axis cold. Exam questions test whether you can set up the correct moment integral.
True or False: The arc length integrand is 1 + [f'(x)]^2 (no square root).
False. It is sqrt(1 + [f'(x)]^2).
Fill in the blank: In the surface area formula for rotation about the x-axis, the radius of each band is ______.
f(x), the y-value of the curve.
True or False: M_y gives you y-bar.
False. M_y (moment about the y-axis) gives x-bar.
Fill in the blank: For a uniform lamina, the density ______ in the centroid formulas.
cancels out.
Q: Find the arc length of y = (2/3)(1 + x^2)^(3/2) from x = 0 to x = 3.
A: Compute f'(x) = 2x(1 + x^2)^(1/2). Then [f'(x)]^2 = 4x^2(1 + x^2). So 1 + [f'(x)]^2 = 1 + 4x^2 + 4x^4 = (1 + 2x^2)^2. The square root is (1 + 2x^2). Integrate from 0 to 3: integral of (1 + 2x^2) dx = [x + (2/3)x^3] from 0 to 3 = 3 + 18 = 21.
Q: Set up the integral for the surface area obtained by rotating y = sqrt(x), 1 <= x <= 4, about the x-axis.
A: f(x) = sqrt(x), f'(x) = 1/(2 sqrt(x)). SA = integral from 1 to 4 of 2 pi sqrt(x) sqrt(1 + 1/(4x)) dx.
Q: Find the centre of mass of a system with m_1 = 3 at x = 1, m_2 = 5 at x = 4, m_3 = 2 at x = 6.
A: x-bar = (3 times 1 + 5 times 4 + 2 times 6) / (3 + 5 + 2) = (3 + 20 + 12) / 10 = 35/10 = 3.5.
Arc length and surface area build directly on the slicing logic from Sections 6.1 and 6.2 (volumes). The ds element that appears in both formulas will resurface if you study line integrals in multivariable calculus. Centre of mass connects to moment of inertia and to the theorem of Pappus, which says that the volume of a solid of revolution equals 2 pi times the distance the centroid travels times the area of the region.
Arc length formula, curve length, surface area of revolution, surface integral, centre of mass, centroid, moment about an axis, moment of a system, lamina, uniform density, Pappus theorem, ds element, MATH 104, Calculus II applications, Rimmer, Penn State Math 104