Arc length and area, across every coordinate system
The same two questions — how long, how much area — answered differently in rectangular, parametric, and polar.
Match the formula to how the curve is GIVEN, not to the shape it draws. A curve given as y=f(x) uses the rectangular arc-length integrand \sqrt{1+[f'(x)]^2}. A curve given as x(t),y(t) uses the parametric speed integrand \sqrt{[x'(t)]^2+[y'(t)]^2} — which reduces to the rectangular form when x(t)=t. A curve given as r(\theta) uses the polar-specific integrand \sqrt{[r(\theta)]^2+[r'(\theta)]^2}, derived from the SAME parametric formula applied to x=r\cos\theta,\ y=r\sin\theta.
Arc length, by how the curve is given
Formula
Arc length of a curve
For a smooth function f with continuous derivative on [a,b], the length of the curve y=f(x) from x=a to x=b is L = \int_a^b \sqrt{1+\big[f'(x)\big]^2}\,dx.
Show proof
Approximate the curve by a polygonal path connecting points at x_0,x_1,\ldots,x_n. Between consecutive points, the path rises \Delta y_i over a run of \Delta x, so that segment's length is \sqrt{(\Delta x)^2+(\Delta y_i)^2} = \sqrt{1+(\Delta y_i/\Delta x)^2}\,\Delta x. By the Mean Value Theorem, \Delta y_i/\Delta x = f'(x_i^*) for some point x_i^* in that subinterval, so the segment's length is \sqrt{1+[f'(x_i^*)]^2}\,\Delta x. Summing over all n segments gives a Riemann sum for \sqrt{1+[f'(x)]^2}, and letting n\to\infty turns that sum into the stated integral, with the polygonal path's total length converging to the true arc length.
For x=x(t), y=y(t) with continuous derivatives on [a,b], and the curve traced exactly once as t runs from a to b, the arc length is L = \int_a^b \sqrt{\big[x'(t)\big]^2+\big[y'(t)\big]^2}\,dt.
Show proof
Approximate the curve by a polygonal path through the points at t_0,t_1,\ldots,t_n. The segment from t_{i-1} to t_i runs \Delta x_i across and \Delta y_i up, so its length is \sqrt{(\Delta x_i)^2+(\Delta y_i)^2}. Factoring out \Delta t gives \sqrt{(\Delta x_i/\Delta t)^2+(\Delta y_i/\Delta t)^2}\,\Delta t, and the Mean Value Theorem replaces each difference quotient with x'(t_i^*) and y'(t_i^*) at (possibly different) points in the subinterval. Summing over all n segments produces a Riemann sum for \sqrt{[x'(t)]^2+[y'(t)]^2}, and letting n\to\infty turns that sum into the stated integral, with the polygonal path's length converging to the curve's true arc length.
For a curve r=r(\theta) with continuous derivative on [\alpha,\beta], traced exactly once, the arc length is L = \int_\alpha^\beta \sqrt{\big[r(\theta)\big]^2 + \left[\frac{dr}{d\theta}\right]^2}\;d\theta.
Show proof
Write the curve parametrically as x=r(\theta)\cos\theta, y=r(\theta)\sin\theta, and apply the parametric arc-length integral L=\int_\alpha^\beta\sqrt{(dx/d\theta)^2+(dy/d\theta)^2}\,d\theta. Using \dfrac{dx}{d\theta}=r'\cos\theta-r\sin\theta and \dfrac{dy}{d\theta}=r'\sin\theta+r\cos\theta (product rule on each, writing r' for dr/d\theta), \left(\frac{dx}{d\theta}\right)^2+\left(\frac{dy}{d\theta}\right)^2 = (r'\cos\theta-r\sin\theta)^2 + (r'\sin\theta+r\cos\theta)^2. Expanding both squares, the cross terms -2rr'\sin\theta\cos\theta and +2rr'\sin\theta\cos\theta cancel exactly, leaving (r')^2\cos^2\theta + r^2\sin^2\theta + (r')^2\sin^2\theta + r^2\cos^2\theta = (r')^2\big(\cos^2\theta+\sin^2\theta\big) + r^2\big(\sin^2\theta+\cos^2\theta\big) = (r')^2 + r^2, by the Pythagorean identity. Substituting this simplified sum under the square root gives the stated formula.
Area, by how the region is given
If x=x(t), y=y(t) trace a curve exactly once, with y(t)\ge 0, as t runs from \alpha to \beta with x(\alpha)\le x(\beta), the area between the curve and the x-axis is A = \int_\alpha^\beta y(t)\,x'(t)\,dt.
Show proof
The ordinary area formula A=\int_a^b y\,dx still applies once x is recognized as a function of t: substituting x=x(t) turns dx into x'(t)\,dt, and the bounds x=a,\,x=b become t=\alpha,\,t=\beta through a=x(\alpha), b=x(\beta). No new geometry is involved — this is u-substitution run on the area integral itself, with the parameter t playing the role of u.
For a curve r=r(\theta) with r(\theta)\ge0 and continuous on [\alpha,\beta], where the ray at \theta=\alpha to the ray at \theta=\beta sweeps out the region exactly once, the enclosed area is A = \frac{1}{2}\int_\alpha^\beta \big[r(\theta)\big]^2\,d\theta.
Example
Show proof
Partition [\alpha,\beta] into n subintervals of width \Delta\theta. On each subinterval, approximate the thin polar region by a circular SECTOR of radius r(\theta_i^*) (for some sample angle \theta_i^* in that subinterval) and central angle \Delta\theta. By the sector area formula, that one sector has area \tfrac12\big[r(\theta_i^*)\big]^2\Delta\theta. Summing all n sectors gives \sum_{i=1}^n \frac12\big[r(\theta_i^*)\big]^2\,\Delta\theta, a Riemann sum for \tfrac12\big[r(\theta)\big]^2. As n\to\infty and \Delta\theta\to0, the sector approximation's error shrinks to zero and the sum converges to the stated integral.
For a circle of radius r, the area of a sector with central angle \theta (in radians) is A = \frac{1}{2} r^2 \theta, \qquad \theta \text{ in radians.}
Show proof
A sector with angle \theta is one particular fraction of the whole circle — specifically, the fraction \frac{\theta}{2\pi}, since 2\pi is the angle of the complete circle. The whole circle has area \pi r^2, so the sector's area is that same fraction of the total: A = \frac{\theta}{2\pi} \cdot \pi r^2 = \frac{\theta \pi r^2}{2\pi} = \frac{1}{2}r^2\theta. The \pi in the numerator and denominator cancel, leaving a formula with no \pi in it at all — it's already baked into \theta being in radians.
Surfaces of revolution
For a smooth, non-negative function f with continuous derivative on [a,b], revolving y=f(x) about the x-axis sweeps out a surface of area S = 2\pi\int_a^b f(x)\sqrt{1+\big[f'(x)\big]^2}\,dx.
Show proof
Approximate the curve by short straight segments, as in the arc length formula. Revolving one segment of length ds=\sqrt{1+[f'(x_i^*)]^2}\,\Delta x about the x-axis, where the segment sits at height \approx f(x_i^*), sweeps out a thin frustum (a truncated cone). A frustum's lateral surface area is 2\pi\,(\text{average radius})\times(\text{slant length}), and for a short enough segment the average radius is well approximated by f(x_i^*) itself, giving area \approx 2\pi f(x_i^*)\,ds. Summing over all segments gives a Riemann sum for 2\pi f(x)\sqrt{1+[f'(x)]^2}, and the limit as the segments shrink turns that sum into the stated integral.
For x=x(t), y=y(t)\ge 0 with continuous derivatives on [a,b], traced exactly once, revolving the curve about the x-axis sweeps out a surface of area S = 2\pi\int_a^b y(t)\,\sqrt{\big[x'(t)\big]^2+\big[y'(t)\big]^2}\,dt.
Show proof
Approximate the curve by short segments, as in the parametric arc length derivation. Revolving one segment about the x-axis, where the segment sits at height \approx y(t_i^*) and has length ds\approx\sqrt{[x'(t_i^*)]^2+[y'(t_i^*)]^2}\,\Delta t, sweeps out a thin frustum of lateral area \approx 2\pi\,y(t_i^*)\,ds. Summing over all segments gives a Riemann sum for 2\pi\,y(t)\sqrt{[x'(t)]^2+[y'(t)]^2}, and the limit as the segments shrink turns that sum into the stated integral.