$$V = \pi \int^b_a f(x)^2 - g(x)^2 \> dx$$
where $f(x) > g(x)$
+## Length of a curve
+
+$$L = \int^b_a \sqrt{1 + ({dy \over dx})^2} \> dx \quad \text{(Cartesian)}$$
+
+$$L = \int^b_a \sqrt{{dx \over dt} + ({dy \over dt})^2} \> dt \quad \text{(parametric)}$$
+
+Evaluate on CAS. Or use Interactive $\rightarrow$ Calculation $\rightarrow$ Line $\rightarrow$ `arcLen`.
+
## Rates
### Related rates
$${da \over db} \quad \text{(change in } a \text{ with respect to } b)$$
-#### Gradient at a point on parametric curve
+### Gradient at a point on parametric curve
$${dy \over dx} = {{dy \over dt} \div {dx \over dt}} \> \vert \> {dx \over dt} \ne 0$$
$$\int^b_a f(x) \> dx = F(b) - F(a)$$
where $F$ is any antiderivative of $f$
+
+## Differential equations
+
+One or more derivatives
+
+**Order** - highest power inside derivative
+**Degree** - highest power of highest derivative
+e.g. ${\left(dy^2 \over d^2 x\right)}^3$: order 2, degree 3
+
+### Verifying solutions
+
+Start with $y=\dots$, and differentiate. Substitute into original equation.
+
+### Function of the dependent variable
+
+If ${dy \over dx}=g(y)$, then ${dx \over dy} = 1 \div {dy \over dx} = {1 \over g(y)}$. Integrate both sides to solve equation. Only add $c$ on one side. Express $e^c$ as $A$.