June 10, 2026 - Calculus 2 - Calculus Applications: Surface Area, Mass Density, and Work (Concise)

The surface area of an object rotated around the x-axis is calculated using the formula: $S = ∣_a^b 2\pi f(x) \sqrt{1 + (f'(x))^2}\,dx$
Rotating around the y-axis requires bounds and functions to be expressed in terms of $y$ .
For $f(x) = \sqrt{x}$ on the interval $[1, 4]$ , the integral simplifies by combining the square roots: $2\pi \int_1^4 \sqrt{x + \frac{1}{4}}\,dx$
For $f(x) = \sqrt[3]{x}$ on the $y$ -interval $[0, 2]$ , bounds must be converted to $x$ -bounds. Here, $x = 0$ and $x = 8$ , with the derivative $f'(x) = \frac{1}{3}x^{-2/3}$ .

The mass ( $M$ ) of a thin wire or rod with variable density is the integral of its density function $\rho(x)$ : $M = \int_a^b \rho(x)\,dx$
Linear density ( $\rho(x)$ ) in physics describes the density of a one-dimensional object (like a line), which differs from the mathematical definition of a linear function (degree-one polynomial).
Example: A rod from $0$ to $2\pi$ with density $\rho(x) = \sin(x)$ has mass calculated as: $M = \int_0^{2\pi} \sin(x)\,dx$

For a thin plate where density $\rho(x)$ depends on the radius $x$ , mass is calculated by integrating the density over the circumference: $M = \int_0^R 2\pi x \rho(x)\,dx$
If $\rho(x) = \sqrt{x}$ and the radius is $4$ , the mass integral includes the product of circumference and density: $M = \int_0^4 2\pi x \cdot x^{1/2}\,dx = \int_0^4 2\pi x^{3/2}\,dx$
If $\rho(x) = (x+1)e^{-x}$ for a disc of radius $10$ , integration by parts ( $u-dv$ substitution) is required to evaluate: $M = 2\pi \int_0^{10} x(x+1)e^{-x}\,dx$

Force ( $F$ ) is defined as the push or pull on an object. Constant force work is $W = F \times d$ .
Variable force work over an interval $[a, b]$ is calculated as: $W = \int_a^b f(x)\,dx$
Units for work include Joules ( $J$ ), which are Newton-meters ( $N \cdot m$ ), and foot-pounds ( $ft \cdot lb$ ).

Hooke's Law states the force required to stretch or compress a spring $x$ units from equilibrium ( $x=0$ ) is: $F(x) = kx$
The value of the spring constant $k$ must be determined from provided data before calculating work.
Example: If $10\,N$ of force is used to compress a spring by $0.2\,m$ : $10 = k(0.2) \rightarrow k = 50$
Work done to stretch this spring from $0$ to $0.5\,m$ : $W = \int_0^{0.5} 50x\,dx$

Calculating the work to pump water involves lifting thin horizontal slices (disks) of water out of a tank.
The volume of a disk is the area times thickness ( $A(x)dx$ ). The force involves gravity and the density of water.
Common water weight/force constants:
- Metric: $9800\,N/m^3$ (lifting a kilogram weight against gravity).
- Imperial: $62.4\,lb/ft^3$ (lifting a pound weight).
Distance moved depends on the tank depth and the height of the lift to the exit point.

Question: How do physicists and mathematicians define "linear" differently?
Answer: Mathematicians refer to a degree-one polynomial (a straight line), while physicists use it to describe an object of one dimension (like a wire), though the density function itself could be exponential.
Question: Should functions be simplified before putting them into a calculator?
Answer: While the calculator can handle complex inputs, simplifying reduces the chance of user input errors.
Question: Is a negative force value possible?
Answer: In the context of springs, negative force indicates the direction of displacement relative to the frame of reference, not a gain of energy.