求體積分(橢圓球體)

A = πab for an ellipseA(z) = π*x(z)*y(z) for many variable ellipseswhere x(z) = a√(c^2-z^2)/cy(z) = b√(c^2-z^2)/cdV(z) = A(z)*dz = π(c^2-z^2)ab*dz/c^2 E = (2左右/8象限)*∫∫∫xyz*dV ;; z = 0~c= 1/4*∫(c^2-z^2)z(ab/c^2)^2*π(c^2-z^2)dz= π(ab/c^2)^2/4*∫z(c^2-z^2)^2*dz= π(ab/c^2)^2/8*∫(c^2-z^2)d(z^2)= -π(ab/c^2)^2/8*∫(c^2-z^2)d(c^2-z^2)= -π(ab/c^2)^2/16 * (c^2-z^2)^2= -π(ab/c^2)^2/16 * [(c^2-c^2)^2 - (c^2-0)^2]= π(abc^4/c^2)^2/16= π(abc^2)^2/16= Answer


2015-06-17 06:41:43 補充：
漏打平方.修正如下:

E = -π(ab/c^2)^2/16*∫(c^2-z^2)^2*d(c^2-z^2)

= -π(ab/c^2)^2/16 * (c^2-z^2)^3/3

= -π(ab/c^2)^2/48 * [(c^2-c^2)^3 - (c^2-0)^3]

= π(ab)^2*c^6/48c^4

= π(abc)^2/48

2015-06-17 07:03:24 補充：
修改係數:

E = -π(ab/c^2)^2/8*∫(c^2-z^2)^2*d(c^2-z^2)

= π(abc)^2/24

2015-06-24 17:34:05 補充：
(2) dV 與 A dz 不相同?

Ans: dV = A(z)*dz 成立


Ex1.直圓錐

dV = A(z)*dz

= πx(z)^2*dz

= π[r*(h-z)/h]^2

V = π(r/h)^2∫(0~h)(h^2-2hz+z^2)dz

= π(r/h)^2(zh^2 - hz^2 + z^3/3)

= π(r/h)^2(h^3 - h^3 + h^3/3)

= πh*r^2/3


Ex2.半球體

dV = A(z)*dz

= πx(z)^2*dz

2015-06-24 17:35:01 補充：
V = π∫(r^2 - z^2)dz ;; z=0~r

= π(zr^2 - z^3/3)

= π(r^3 - r^3/3)

= 2πr^3/3


以上兩例證明 dV = A(z)*dz 成立

the integral = (abc)^2 / 48

2015-06-21 23:21:36 補充：
dV 與 A dz 不相同

2015-06-24 20:19:21 補充：
∫∫∫ f(z) dV = ∫ f(z)A(z) dz是對的
但∫∫∫ f(x,y,z) dV 就不對了
∫∫∫ f(x,y,z) dV 其中點(x,y,z)在立體區域內, (x,y)不是常數, 也不是限定為立體區域的邊界
懂嗎?

2015-06-24 20:41:52 補充：
(x,y)不是z的函數