Exam Cram: Applications of Integration

Topic Weighting

[!IMPORTANT] This module typically accounts for 20–25% of a standard Calculus II exam. It is highly cumulative, requiring mastery of substitution ( $u$ -substitution) and fundamental integration rules.

Topic	Frequency	Difficulty
Volumes of Revolution (Disk/Washer/Shell)	High	★★★★☆
Area Between Curves	High	★★☆☆☆
Work and Physical Applications	Medium	★★★☆☆
Arc Length & Surface Area	Low/Medium	★★★☆☆
Moments and Centers of Mass	Medium	★★★★☆

Key Concepts Summary

Area Between Curves: The integral of the "top" function minus the "bottom" function. If functions intersect, you must split the integral at the intersection points.
Volumes by Slicing:
- Disk Method: Used when there is no "hole" in the solid.
- Washer Method: Used when the region is bounded by two functions, creating a hollow center.
- Cylindrical Shells: Often easier when revolving around an axis parallel to the dependent variable's axis.
Physical Applications:
- Work: The accumulation of force over a distance ( $W = \int F(x) dx$ ). Common for springs (Hooke's Law) and pumping liquids.
- Hydrostatic Force: Force exerted by a fluid on a submerged plate; depends on depth and area.
Centroids: The geometric center of a region. For a thin plate of constant density, it is the point $(\bar{x}, \bar{y})$ where the plate would balance.

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Common Pitfalls

Incorrect Radius in Washers: Students often use $(R - r)^2 instead of the correct R^2 - r^2$ . Don't subtract radii before squaring.
Integration Limits: Forgetting to change $x$ -limits to $y-limits when integrating with respect to y$ .
Shell vs. Washer Confusion: Using $2\pi$ for washers or $\pi for shells. Remember: **Shells = 2\pi$ (circumference), Washers = $\pi$ (area).**
Units of Work: Confusing mass and weight in US Customary units. Weight is a force ( $lb$ ), but mass in SI ( $kg) must be multiplied by g = 9.8$ to get Newtons ( $N$ ).

Mnemonics / Memory Triggers

Area: "Top Minus Bottom" (TMB) or "Right Minus Left" (RML).
Disk/Washer: $\pi \int (R^2 - r^2) \, d(\text{axis})$ . Think of it as "Pie on the Plate" (Area of a circle).
Shells: $2\pi \int rh \, d(\text{radius})$ . Think of it as "Two Pies in a Shell" (Circumference of a circle).
Arc Length Formula: Look for the "1" and the "Prime". $L = \int \sqrt{1 + (f')^2}$ .

Formula / Equation Sheet

Application	Formula (x-axis / dx)	Notes
Area	$A = \int_{a}^{b} [f(x) - g(x)] \, dx$	$f(x) \ge g(x)$
Disk Volume	$V = \pi \int_{a}^{b} [R(x)]^2 \, dx$	No inner radius
Washer Volume	$V = \pi \int_{a}^{b} ([R(x)]^2 - [r(x)]^2) \, dx$	$R$ is outer, $r$ is inner
Shell Volume	$V = 2\pi \int_{a}^{b} x f(x) \, dx$	Rotation about y-axis
Arc Length	$L = \int_{a}^{b} \sqrt{1 + [f'(x)]^2} \, dx$	Function must be smooth
Work (Spring)	$W = \int_{a}^{b} kx \, dx$	$k$ = spring constant
Moment ( $M_y$ )	$M_y = \rho \int_{a}^{b} x [f(x) - g(x)] \, dx$	Distance to y-axis is $x$

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Practice Set

Compound Region: Find the area bounded by $y = x^2$ $y = x^{2}$ and $y = 2x - x^2$ $y = 2 x - x^{2}$ .
- Tip: Set them equal to find bounds ( $x=0, x=1$ ).
Washer Method: Revolve the region bounded by $y = \sqrt{x}$ $y = x$ and $y = x^2$ $y = x^{2}$ about the x-axis.
- Answer Setup: $V = \pi \int_{0}^{1} ((\sqrt{x})^2 - (x^2)^2) \, dx$ .
Shell Method: Revolve the region bounded by $y = e^{-x^2}, y=0, x=0, x=1$ $y = e^{- x^{2}}, y = 0, x = 0, x = 1$ about the y-axis.
- Tip: This requires $u-substitution after setting up the 2\pi x f(x)$ integral.
Work (Pumping): A rectangular tank (10ft long, 5ft wide, 6ft deep) is full of water ($62.4 , lb/ft^3$). Find the work to pump all water over the top edge.
- Recall: $$W = \int (Weight Density) \cdot (Area) \cdot (Distance to lift) $\, dy$ .
Centroid: Find the center of mass of a semicircular plate of radius $r$ $r$ centered at the origin.
- Symmetry Tip: By symmetry, $\bar{x} = 0$. You only need to solve for $\bar{y}$ .