Why do internal boundaries cancel out when summing line integrals over small rectangles in the proof of Green's Theorem?

The ordinary integral $\int_{-\infty}^{\infty} g(t)e^{-2\pi i ft} dt$ is well-defined if $g(t)$ is absolutely integrable (i.e., $\int |g(t)| dt < \infty$). However, ideal sinusoids sustained indefinitely are not absolutely integrable because their energy spreads over infinite time.

Conditions: Considering the limit as the time window $T \to \infty$.; Analyzing either decaying transient signals or sustained periodic signals.

The definite integral can be evaluated by recognizing that the integrand $y = \sqrt{9-x^2}$ graphs as an upper semicircle of radius 3. Because the function is continuous and nonnegative on $[-3, 3]$, the integral equals the ordinary geometric area of this region.

Conditions: The integrand is recognized as the upper semicircle of $x^2+y^2=9$.; The function is continuous and nonnegative on the interval $[-3, 3]$.; Use the real geometric area formula for a circle.

The equation $x^2 + y^2 = 9$ describes a full circle centered at the origin with radius 3, which includes both upper and lower branches. However, the original function is defined using the principal square root, $y = \sqrt{9-x^2}$.

Conditions: Working over the real numbers.; Using the principal (nonnegative) square root convention.; The underlying relation is the circle $x^2 + y^2 = 9$.

To evaluate the definite integral $\int_{-3}^{3} \sqrt{9-x^2}\,dx$ geometrically, recognize that the integrand $y = \sqrt{9-x^2}$ represents the upper semicircle of a circle centered at the origin with radius 3. Because the function is nonnegative and continuous on the interval $[-3, 3]$, the definite integral equals the ordinary geometric area of this shaded region.

Conditions: The integrand is $\sqrt{9-x^2}$ and the limits of integration are -3 and 3.; The square root denotes the principal (nonnegative) root, restricting the graph to $y \ge 0$.; The function is continuous and nonnegative on the closed interval $[-3, 3]$, ensuring the definite integral equals the ordinary area under the curve.

The calculation averages complex points sampled uniformly in *time*, not uniformly along the *arc length* of the trajectory. Because the signal $g(t)$ modulates the radius and the rotation speed varies with frequency, equal time intervals do not correspond to equal distances traveled along the curve.

Conditions: Sampling is done at uniform time intervals $t_k$.; The path is defined by $z_f(t) = g(t)e^{-2\pi i ft}$.