2.3 Geometric Applications: Areas, Volumes, and Arc Length

2.3 Geometric Applications: Areas, Volumes, and Arc Length

Geometric applications of definite integrals on the GRE Mathematics Subject Test test the ability to set up and evaluate integrals representing physical and spatial quantities. Problems require seamless translation between Cartesian, parametric, and polar coordinates, as well as choosing the most algebraically advantageous integration method for solids of revolution.

Areas in Cartesian and Polar Coordinates

Cartesian Coordinates

The area of a region bounded above by $y = f(x)$ and below by $y = g(x)$ on $[a, b]$ is: $A = \int_a^b (f(x) - g(x))\,dx$ When curves are expressed as functions of $y$, partitioning horizontally with respect to $y$ is often more convenient: $A = \int_c^d (x_{\text{right}}(y) - x_{\text{left}}(y))\,dy$

Polar Coordinates

A polar curve $r = f(\theta)$ sweeps out differential triangular sectors of area $dA = \frac{1}{2} r^2,d\theta$. The area bounded by $r = f(\theta)$ from $\theta = \alpha$ to $\theta = \beta$ is: $A = \frac{1}{2} \int_\alpha^\beta [f(\theta)]^2\,d\theta$ The area between two polar curves $r_{\text{outer}}(\theta) \ge r_{\text{inner}}(\theta) \ge 0$ is: $A = \frac{1}{2} \int_\alpha^\beta \left( [r_{\text{outer}}(\theta)]^2 - [r_{\text{inner}}(\theta)]^2 \right)\,d\theta$

Rose Curves and Petals

For polar rose curves $r = a\cos(k\theta)$ or $r = a\sin(k\theta)$:

• If $k$ is odd, the curve has $k$ petals, traced fully as $\theta$ ranges from $0$ to $\pi$.
• If $k$ is even, the curve has $2k$ petals, traced fully as $\theta$ ranges from $0$ to $2\pi$.
• A single petal boundary is found by setting $r = 0$ to identify consecutive radial zero crossings.

Volumes of Solids of Revolution

Choosing between the Washer Method (slicing perpendicular to the rotation axis) and the Cylindrical Shells Method (slicing parallel to the rotation axis) depends on which variable isolates the boundary functions more simply.

Method Comparison for Revolution Around Coordinate and Shifted Axes

In the washer formulation, $R$ is the outer radius (distance from the axis to the outer boundary) and $r$ is the inner radius (distance from the axis to the inner boundary). In the shell formulation, $|x - h|$ is the radius and $[y_{\text{top}} - y_{\text{bottom}}]$ is the cylinder height.

Arc Length Formulations

The infinitesimal arc length element $ds = \sqrt{dx^2 + dy^2}$ adapts across coordinate frameworks:

• Cartesian: For $y = f(x)$ on $[a, b]$: $L = \int_a^b \sqrt{1 + [f'(x)]^2}\,dx$
• Parametric: For $x = x(t), y = y(t)$ on $[t_1, t_2]$: $L = \int_{t_1}^{t_2} \sqrt{[x'(t)]^2 + [y'(t)]^2}\,dt$
• Polar: For $r = f(\theta)$ on $[\alpha, \beta]$, using $x = r\cos\theta, y = r\sin\theta$: $L = \int_\alpha^\beta \sqrt{r^2 + \left(\frac{dr}{d\theta}\right)^2}\,d\theta$

GRE arc length problems are engineered so the quantity under the radical simplifies to a perfect square (e.g., $1 + \sinh^2 x = \cosh^2 x$, or algebraic terms matching $(u + 1/u)^2$).

Surface Area of Revolution

Revolving an arc length element $ds$ around an axis generates a circular band of radius $\rho$ and area $dS = 2\pi \rho,ds$:

• Rotation about $x$-axis: $\rho = |y|$, yielding $S = 2\pi \int |y|,ds$.
• Rotation about $y$-axis: $\rho = |x|$, yielding $S = 2\pi \int |x|,ds$.

Centroids and Pappus's Theorems

Let a planar region have area $A$ with centroid coordinates $(\bar{x}, \bar{y})$: $\bar{x} = \frac{1}{A} \int_a^b x (f(x) - g(x))\,dx, \quad \bar{y} = \frac{1}{2A} \int_a^b \left( [f(x)]^2 - [g(x)]^2 \right)\,dx$

First Theorem of Pappus (Surface Area)

The surface area $S$ generated by revolving a plane curve of length $L$ about an external non-intersecting coplanar axis is equal to the product of $L$ and the distance traveled by the curve's centroid: $S = 2\pi d L$

Second Theorem of Pappus (Volume)

The volume $V$ of a solid generated by revolving a plane area $A$ about an external non-intersecting coplanar axis equals the product of $A$ and the distance traveled by the region's centroid: $V = 2\pi d A$ For instance, revolving a circle of radius $r$ centered at distance $R > r$ from an axis produces a solid torus with volume $V = (2\pi R)(\pi r^2) = 2\pi^2 R r^2$ and surface area $S = (2\pi R)(2\pi r) = 4\pi^2 R r$.

Worked Problem: Shells vs. Washers on Shifted Axis

Problem: Find the volume generated by revolving the region bounded by $y = x^2$ and $y = 4$ about the line $y = 4$. Solution: The region lies between $x = -2$ and $x = 2$, bounded below by $y = x^2$ and above by $y = 4$. Slicing perpendicular to the horizontal axis of rotation $y = 4$ produces solid disks (no inner hole). The radius of each disk is $R(x) = 4 - x^2$. $V = \pi \int_{-2}^2 [R(x)]^2\,dx = \pi \int_{-2}^2 (4 - x^2)^2\,dx = 2\pi \int_0^2 (16 - 8x^2 + x^4)\,dx$ $V = 2\pi \left[ 16x - \frac{8}{3}x^3 + \frac{1}{5}x^5 \right]_0^2 = 2\pi \left[ 32 - \frac{64}{3} + \frac{32}{5} \right] = 2\pi \cdot 32 \left( 1 - \frac{2}{3} + \frac{1}{5} \right) = 64\pi \left( \frac{8}{15} \right) = \frac{512\pi}{15}$

Common GRE Traps & Pitfalls

• Squaring the difference instead of difference of squares: In the washer method, writing $(R - r)^2$ instead of $R^2 - r^2$ is a frequent calculation mistake.
• Omitting the factor of $1/2$ in polar area: Evaluating $\int_\alpha^\beta r^2,d\theta$ without the $\frac{1}{2}$ pre-factor leads to an answer twice the true area.
• Double-counting polar petals: Integrating $\theta$ from $0$ to $2\pi$ for an odd-petal rose curve traces each petal twice because negative $r$ values redraw existing regions.
• Negative radius in cylindrical shells: When the rotation axis is to the right of the region ($x = h$ with $h \ge b$), the shell radius is $h - x$, not $x - h$.