C.5 · Improper integrals and geometric applications
GRE Subject Test · GRE Mathematics Topic 16 2:56 English narration · English + 中文 subtitles burned in
Chapters
Transcript
This unit defines improper integrals by limits at each singular boundary, selects area, volume and arc-length formulas from the geometry, and separates convergence of integrals from signed cancellation.
本单元用每个奇异边界处的极限来定义反常积分，根据几何情形选择面积、体积和弧长公式，并把积分的收敛与带符号的抵消区分开。
Replace an infinite endpoint by a finite bound and a limit.
把无穷端点换成有限上限再取极限。
At a singular point inside the interval, split and require both sides to converge separately: for 1 over x across zero, symmetric cutoffs cancel, but both sides diverge.
区间内部有奇点时要拆开，并要求两边各自收敛：对跨越零点的 1/x，对称截断可以抵消，但两边都发散。
x to the minus p converges from 1 to infinity exactly when p exceeds 1, and from 0 to 1 exactly when p is below 1.
x^{−p} 在 1 到 ∞ 上收敛当且仅当 p > 1，在 0 到 1 上收敛当且仅当 p < 1。
For an integral of a maximum, solve the branch crossings and check which expression dominates on each interval.
对最大值函数的积分，先求分支的交点，再看每段上哪个表达式更大。
For the maximum of the semicircle and x plus 1, the interior crossing is zero: use the semicircle on minus 1, 0, and the line on 0, 1.
对半圆与 x + 1 的最大值，内部交点是零：在 上用半圆，在 上用直线。
The area is pi over 4 plus 3 over 2.
面积是 π/4 + 3/2。
For rotation around the x-axis, washers integrate pi times R squared minus r squared.
绕 x 轴旋转时，垫圈法积分 π(R² − r²)。
Shells use 2 pi times radius times height.
柱壳法用 2π × 半径 × 高。
Arc length integrates the square root of 1 plus f prime squared, and surface area adds the factor 2 pi f.
弧长积分 √(1 + f′²)，旋转面面积再乘上 2πf。
These are different quantities, and an unbounded interval still needs a convergence test.
这些是不同的量，区间无界时仍需检验收敛。
Rotate y equals 1 over x, for x at least 1, around the x-axis.
把 y = 1/x（x ≥ 1）绕 x 轴旋转。
The disk volume is pi times the integral of 1 over x squared: exactly pi.
圆盘法体积是 π 乘以 1/x² 的积分：恰好是 π。
The surface integrand is at least 2 pi over x, whose integral diverges.
表面积的被积函数至少为 2π/x，其积分发散。
Finite volume does not imply finite surface area.
体积有限并不意味着表面积有限。
Never count a principal value as a convergent improper integral, or infer one geometric quantity's finiteness from another.
不要把柯西主值当作收敛的反常积分，也不要由一个几何量有限推出另一个几何量有限。
Split at 1.
在 1 处拆开。
Near zero, convergence requires p below 1; near infinity, p above 1.
在零附近收敛要求 p < 1；在无穷远处要求 p > 1。
No real p meets both.
没有实数 p 同时满足。
For 1 over x across zero, the two sides diverge separately, so the ordinary integral diverges; symmetric cutoffs give only a principal value.
对跨越零点的 1/x，两边各自发散，所以普通积分发散；对称截断只给出主值。
Disks give pi times the limit of 1 minus 1 over b, which is pi.
圆盘法给出 π 乘以 (1 − 1/b) 的极限，即 π。
For the surface, y prime is minus 1 over x squared, and the nonnegative integrand is at least 2 pi over x, whose integral diverges.
对表面积，y′ = −1/x²，非负的被积函数至少为 2π/x，其积分发散。
Comparison proves infinite lateral area.
比较法证明侧面积无穷大。
Check yourself.
自我检测。
From 1 to infinity, x to the minus p needs p above 1; from 0 to 1, p below 1.
从 1 到 ∞，x^{−p} 需要 p > 1；从 0 到 1，需要 p < 1。
1 over x across zero diverges.
跨越零点的 1/x 发散。
And the horn has volume pi but infinite surface area.
号角的体积是 π，但表面积无穷大。