Skip to content
CalcGospel 微積分福音
返回

108 台灣大學微積分(B) 第 20 題

考題 / 轉學考微積分 / 台灣大學 / 微積分B

108學年度 · 108台大微積分B · 第 20 題

題目

Problem

PART 1: Fill in the blanks.

  • Work need not be shown, only answers will be graded.

  • 4 points for each blank and 100 points total.

  • Each answer need to be clearly labeled on the answer sheet.

  • Use of any device with computer algebra system during the exam will result in zero points.

  1. The flux of the vector field F⃗=(sin⁡y+x3,3yz2+ez,3zy2)\vec{F}=(\sin y+x^3,3yz^2+e^z,3zy^2) through the surface x2+y2+z2=1, z≥0x^2+y^2+z^2=1,\ z\ge 0, where the surface is equipped with the upward normal, is (25)‾\underline{\quad(25)\quad}.

解答

解法一

思路

展開

補上底圓盤形成封閉上半球,再用散度定理;底面通量另外檢查。

答題過程

展開

以底圓盤 z=0z=0 補成上半球。散度為

∇⋅F⃗=3x2+3y2+3z2=3r2.\nabla\cdot\vec F=3x^2+3y^2+3z^2=3r^2.

底圓盤的外法向量向下,而該處 Fz=0F_z=0,故底面通量為零。由散度定理,所求通量為

∭上半球3r2 dV=3⋅2π5=6π5.\iiint_{\text{上半球}}3r^2\,\mathrm{d}V =3\cdot\frac{2\pi}{5} =\boxed{\frac{6\pi}{5}}.