Ngô Quốc Anh

January 31, 2008

GT4 – Bài Tập 05

Filed under: Giải Tích 4 — Ngô Quốc Anh @ 21:43

Lấy bài tập ở đây: gt4-05.pdf

January 19, 2008

Steinmetz Solid

Filed under: Các Bài Tập Nhỏ, Giải Tích 5 — Ngô Quốc Anh @ 4:52

The solid common to two (or three) right circular cylinders of equal radii intersecting at right angles is called the Steinmetz solid. Two cylinders intersecting at right angles are called a bicylinder, and three intersecting cylinders a tricylinder. Half of a bicylinder is called a vault.

Steinmetz solid for two cylinders

For two cylinders of radius r oriented long the z– and x-axes gives the equations


which can be solved for x and y gives the parametric equations of the edges of the solid,

x = +/-z
y = +/-sqrt(r^2-z^2).

The surface area can be found as intxds, where

ds = sqrt(1+((dy)/(dz))^2)dz
= r/(sqrt(r^2-z^2))dz.

Taking the range of integration as a quarter or one face and then multiplying by 16 gives


The volume common to two cylinders was known to Archimedes (Heath 1953, Gardner 1962) and the Chinese mathematician Tsu Ch’ung-Chih (Kiang 1972), and does not require calculus to derive. Using calculus provides a simple derivation, however. Noting that the solid has a square cross section of side-half-length sqrt(r^2-z^2), the volume is given by


(Moore 1974). The volume can also be found using cylindrical algebraic decomposition, which reduces the inequalities

{x^2+y^2<1; -L<z<L; y^2+z^2<1; -L<x<L


{-1<x<1; -sqrt(1-x^2)<y<sqrt(1-x^2); -sqrt(1-y^2)<z<sqrt(1-y^2),

giving the integral


If the two right cylinders are of different radii a and b with a>b, then the volume common to them is


where K(k) is the complete elliptic integral of the first kind, E(k) is the complete elliptic integral of the second kind, and k=b/a is the elliptic modulus.

Steinmetz curve

The curves of intersection of two cylinders of radii a and b, shown above, are given by the parametric equations

x(t) = bcost
y(t) = bsint
z(t) = +/-sqrt(a^2-b^2sin^2t)

(Gray 1997, p. 204).

The volume common to two elliptic cylinders

(x^2)/(a^2)+(z^2)/(c^2)==1    (y^2)/(b^2)+(z^2)/(c^('2))==1

with c<c^' is


where k==c/c^' (Bowman 1961, p. 34).


For three cylinders of radii r intersecting at right angles, The resulting solid has 12 curved faces. If tangent planes are drawn where the faces meet, the result is a rhombic dodecahedron (Wells 1991). The volume of intersection can be computed in a number of different ways,

V_3(r,r,r) = 16r^3int_0^(pi/4)int_0^1ssqrt(1-s^2cos^2t)dsdt
= (sqrt(2)r)^3+6int_(r/sqrt(2))^r(2sqrt(r^2-z^2))^2dz
= 8(2-sqrt(2))r^3

(Moore 1974). According to the protagonist Christopher in the novel The Curious Incident of the Dog in the Night-Time, “…People go on holidays to see new things and relax, but it wouldn’t make me relaxed and you can see new things by looking at earth under a microscope or drawing the shape of the solid made when 3 circular rods of equal thickness intersect at right angles” (Haddon 2003, p. 178), which is of course precisely the Steinmetz solid formed by three symmetrically placed cylinders.

Steinmetz tetrahedra

Four cylinders can also be placed with axes along the lines joining the vertices of a tetrahedron with the centers on the opposite sides. The resulting solid of intersection has volume


and 24 curved faces analogous to a cube-octahedron compound (Moore 1974, Wells 1991).


Six cylinders can be placed with axes parallel to the face diagonals of a cube. The resulting solid of intersection has volume


and 36 curved faces, 24 of which are kite-shaped and 12 of which are rhombic (Moore 1974).


January 7, 2008

Tính compact của toán tử tuyến tính và đạo hàm của nó

Filed under: Các Bài Tập Nhỏ, Linh Tinh — Ngô Quốc Anh @ 20:58

Let G be an open set in Banach space X. Assume that G \to X is a compact operator and x_0 \in G. If f'(x_0) exists, then f'(x_0) is also a compact operator.

\displaystyle \boxed{ f : G \to X \text{ is compact}} \xrightarrow{{f'\left( {x_0 } \right)\text{ exists}}} \boxed{f'(x_0) : X \to X \text{ is compact}}.

GT4 – Bài Tập 06

Filed under: Giải Tích 4 — Ngô Quốc Anh @ 0:17

 Lấy bài tập ở đây: gt4-06.pdf

January 6, 2008

GT4 – Bài Tập 04

Filed under: Giải Tích 4 — Ngô Quốc Anh @ 23:39

Lấy bài tập ở đây: gt4-04.pdf

GT4 – Bài Tập 03

Filed under: Giải Tích 4 — Ngô Quốc Anh @ 21:31

Lấy bài tập ở đây: gt4-03.pdf

January 5, 2008

GT4 – Bài Tập 02

Filed under: Giải Tích 4 — Ngô Quốc Anh @ 23:44

Lấy bài tập ở đây: gt4-02.pdf

GT4 – Bài Tập 01

Filed under: Giải Tích 4 — Ngô Quốc Anh @ 21:04

Lấy bài tập ở đây: gt4-01.pdf

GT3 – Bài Tập 08

Filed under: Giải Tích 3 — Ngô Quốc Anh @ 20:57

Lấy bài tập ở đây: gt3-08.pdf

Luận Văn Thạc Sĩ

Filed under: Luận Văn — Ngô Quốc Anh @ 6:56

Tính giải được của một lớp hệ phương trình elliptic không tuyến tính




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