Quick questions on Arc length and surface area explained: H2 Further Mathematics

A short piece of curve has length $\mathrm{d}s = \sqrt{\mathrm{d}x^2 + \mathrm{d}y^2}$. Dividing inside the root by $\mathrm{d}x^2$ gives the Cartesian arc length:

For a parametric curve, divide inside the root by $\mathrm{d}t^2$ instead:

Rotating an arc through $2\pi$ generates a surface. Each band has area $2\pi(\text{radius})\,\mathrm{d}s$, where the radius is the distance from the axis. About the $x$-axis the radius is $y$:

These integrals are often awkward unless the expression under the root simplifies. Many exam curves are designed so that $1 + (\mathrm{d}y/\mathrm{d}x)^2$ becomes a perfect square, removing the root cleanly. Always expand and look for that before reaching for a substitution.

About the $x$-axis the radius is $y$; about the $y$-axis it is $x$. Using the wrong one gives the wrong surface.

$\sqrt{(\cdots)^2}$ is the positive value on the interval; check the sign of the simplified expression.

Write the Cartesian arc-length formula for $y = \mathrm{f}(x)$ between $x = a$ and $x = b$. [1 mark]

State the surface-area integral for an arc rotated about the $x$-axis. [1 mark]

For the parametric curve $x = t$, $y = t^2$, write the integrand $\sqrt{(\dot{x})^2 + (\dot{y})^2}$. [1 mark]