Applications of Integration

The area between $y = f(x)$ (top) and $y = g(x)$ (bottom) from $x = a$ to $x = b$ is $A = \int_a^b [f(x) - g(x)]\,dx$.

The integrand is always (top minus bottom) or (right minus left). If the curves cross, split the integral at the crossing points and take care with which function is on top in each sub-interval.

To find crossing points, solve $f(x) = g(x)$. These are the limits of integration.

For curves given as $x = h(y)$, integrate with respect to $y$: $A = \int_c^d [h_{\text{right}}(y) - h_{\text{left}}(y)]\,dy$. Sometimes integrating in $y$ is simpler because it avoids splitting at crossings.

When you rotate a region around an axis, it sweeps out a solid. To find its volume, slice perpendicular to the axis of rotation.

Disk method (no hole): rotating $y = f(x)$ around the $x$-axis from $a$ to $b$: $V = \int_a^b \pi [f(x)]^2\,dx$. Each cross-section is a disk with radius $f(x)$.

Washer method (with a hole): if the region is between $y = f(x)$ (outer) and $y = g(x)$ (inner), $V = \int_a^b \pi\left([f(x)]^2 - [g(x)]^2\right)dx$. Each cross-section is a washer (annulus).

For rotation around the $y$-axis, express $x$ as a function of $y$ and integrate with respect to $y$.

For rotation around a line $y = c$ (not the $x$-axis), adjust the radii: outer radius is $|f(x) - c|$ and inner radius is $|g(x) - c|$.

The shell method computes volumes of revolution by wrapping thin cylindrical shells around the axis of rotation, rather than slicing perpendicular to it.

For rotation around the $y$-axis: $V = \int_a^b 2\pi x \cdot f(x)\,dx$. Each shell is a thin hollow cylinder with radius $x$ (distance from the axis), height $f(x)$, and thickness $dx$. Unrolling a shell gives a rectangle with area $2\pi x \cdot f(x)$.

For rotation around the $x$-axis: $V = \int_c^d 2\pi y \cdot h(y)\,dy$, where $h(y)$ is the horizontal width of the region at height $y$.

For rotation around a vertical line $x = c$: the radius becomes $|x - c|$ instead of $x$.

When to use shells vs. washers: use whichever makes the integral simpler. Shells are ideal when: (a) the function is easier to express as $y = f(x)$ but you're rotating around a vertical axis, or (b) the washer method would require splitting into multiple integrals but shells give a single clean integral.

General formula: $V = \int 2\pi \cdot (\text{radius from axis}) \cdot (\text{height of shell}) \cdot d(\text{variable})$. Always identify the three components before writing the integral.

The length of a curve $y = f(x)$ from $x = a$ to $x = b$ is $L = \int_a^b \sqrt{1 + [f'(x)]^2}\,dx$.

Derivation: on a tiny interval $[x, x+dx]$, the curve covers a horizontal distance $dx$ and a vertical distance $dy = f'(x)\,dx$. By Pythagoras, the actual distance along the curve is $\sqrt{(dx)^2 + (dy)^2} = \sqrt{1 + [f'(x)]^2}\,dx$.

Warning: most arc length integrals are hard or impossible to evaluate in closed form. The formula is simple; the computation rarely is. Numerical methods are often used in practice.

For parametric curves $x = x(t)$, $y = y(t)$: $L = \int_\alpha^\beta \sqrt{[x'(t)]^2 + [y'(t)]^2}\,dt$.

A parametric curve is defined by $x = x(t)$ and $y = y(t)$ as a parameter $t$ varies. The point $(x(t), y(t))$ traces out a path in the plane.

Example: $x = \cos t$, $y = \sin t$ for $t \in [0, 2\pi]$ traces the unit circle.

The slope of the tangent line is $\frac{dy}{dx} = \frac{dy/dt}{dx/dt} = \frac{y'(t)}{x'(t)}$, provided $x'(t) \neq 0$.

The second derivative: $\frac{d^2y}{dx^2} = \frac{\frac{d}{dt}(dy/dx)}{dx/dt}$.

Arc length: $L = \int_\alpha^\beta \sqrt{[x'(t)]^2 + [y'(t)]^2}\,dt$.

Area under a parametric curve: $A = \int_\alpha^\beta y(t) \cdot x'(t)\,dt$ (careful with orientation).

Parametric curves are essential for describing motion: $t$ is time, $(x(t), y(t))$ is position, $(x'(t), y'(t))$ is the velocity vector.

When rotating $y = f(x)$ around the $x$-axis, each tiny piece of the curve sweeps out a band (frustum) on the surface. The surface area of that band is the circumference $2\pi f(x)$ times the arc length element $ds = \sqrt{1 + [f'(x)]^2}\,dx$.

Full formula: $S = \int_a^b 2\pi f(x) \sqrt{1 + [f'(x)]^2}\,dx$. This is the integral of $2\pi r \cdot ds$ where $r = f(x)$ is the radius (distance from the axis) and $ds$ is the length element along the curve.

For rotation around the $y$-axis: the radius is $x$ instead of $f(x)$, giving $S = \int_a^b 2\pi x \sqrt{1 + [f'(x)]^2}\,dx$.

For parametric curves: $S = \int_\alpha^\beta 2\pi y(t) \sqrt{[x'(t)]^2 + [y'(t)]^2}\,dt$ (rotation around the $x$-axis).

Warning: surface area integrals are typically harder than volume integrals. The $\sqrt{1+[f']^2}$ factor rarely simplifies to something nice. Most surface area problems either have exact answers engineered to work out (like cones and spheres) or require numerical methods.

In physics, work = force × distance. When the force varies over the displacement, you slice the motion into tiny steps where the force is approximately constant and integrate: $W = \int_a^b F(x)\,dx$.

Spring problems (Hooke's Law): the force to stretch a spring $x$ units beyond its natural length is $F(x) = kx$. The work to stretch from $x = a$ to $x = b$ is $W = \int_a^b kx\,dx = \frac{k}{2}(b^2 - a^2)$. Note: $x$ measures displacement from natural length, not the total length of the spring.

Pumping problems: to pump liquid out of a tank, slice the liquid into thin horizontal layers. Each layer at height $y$ has volume $A(y)\,dy$ (where $A(y)$ is the cross-sectional area), weighs $\rho g \cdot A(y)\,dy$, and must be lifted a distance $d(y)$ to the top. Total work: $W = \int_0^h \rho g \cdot A(y) \cdot d(y)\,dy$. The tricky part is expressing $A(y)$ and $d(y)$ correctly for the tank's shape.

Hydrostatic force: the pressure at depth $d$ below the surface of a fluid is $P = \rho g d$. The total force on a submerged vertical plate is $F = \int_a^b \rho g \cdot d(y) \cdot w(y)\,dy$, where $w(y)$ is the width of the plate at depth $d(y)$.

The unifying idea: in every application, you identify the small piece, write its contribution ($dW$, $dF$, etc.), and integrate. The calculus is the same; only the physical setup changes.