5. 다중적분

5.1 서론

5.2 이중, 삼중적분

2-1.png
(1)
\begin{align} \int_a^b f(x) dx = \lim_{ \Delta x \rightarrow 0 } \sum_{x=1}^N f \left( a + {{(b-a)} \over N} \right) \Delta x \end{align}
2-2.png
(2)
\begin{align} \iint f(x, y) dx dy \iff 표면\ z=f(x, y)의\ 아래공간\ 부피 \end{align}

반복적분 — 예제 1:

2-3.png
(3)
\begin{align} V & = \iint\limits_A f(x, y) dx dy = \iint\limits_A z dx dy \end{align}
2-4.png
(4)
\begin{align} \int_0^{2-2x} dy\ z(x, y) & = \int_0^{2-2x} dy\ (1 + y) = \left [ y + {1 \over 2} y^2 \right]_0^{2-2x} \\ & = 4 - 6x + 2x^2 \\ V = \int dV & = \int_0^1 (4 - 6x + 2 x^2 ) dx = {5 \over 3} \end{align}

한편 $y$를 고정하고 $x = [0, 1-y/2 ]$까지 적분할 경우,

(5)
\begin{align} \int_{y=0}^2 \left( \int_{x=0}^{1-y/2} (1+y) dx \right) dy & = \int_{y=0}^2 \left. (1+y) x \right\rvert_{x=0}^{1-y/2} dy \\ & = \int_{y=0}^2 (1+y)(1-y/2)\ dy \\ & = \int_{y=0}^2 (1 + {y \over 2} - {y^2 \over 2} )\ dy = {5 \over 3} \end{align}

으로 그 결과는 같다.

위 결과를 요약하면

(6)
\begin{align} V & = \int_{x=0}^1 dx \int_{y=0}^{2-2x} dy\ (1+y) \\ & = \int_{y=0}^2 dy \int_{x=0}^{1-y/2} dx\ (1+y) \end{align}

예제 3:

(7)
\begin{align} \int dx\ dy\ dz = \int_{x=0}^1 dx\ \int_{y=0}^{2-2x} dy\ \int_{z=0}^{i+y} dz \end{align}

이것을 일반적으로 쓰면

dlfqksghk.png
(8)
\begin{align} A & = \int_a^b dx\ \int_{y_1 (x)}^{y_2 (x)}\ dy\ z \\ & = \int_c^d dy\ \int_{x_1(y)}^{y_2 (y)}\ dx\ z \end{align}

5.3 단일적분, 다중적분

예제 1:

3-1.png 3-2.png

  • 면적
(9)
\begin{align} A = \int_{x=0}^1 dx\ x^2 = \left. {x^3 \over 3} \right\rvert_0^1 {1 \over 3} \end{align}
  • 질량:
(10)
\begin{align} M & = \int_{x=0}^1 \int_{y=0}^{x^2} xy\ dy\ dx = \int_0^1 x\ dx \left[ {y^2 \over 2} \right]_0^{x^2} \\ & = \int_0^1 {x^5 \over 2} dx = {1 \over 12} \end{align}
  • 호의 길이
(11)
\begin{align} ds^2 & = dx^2 + dy^2 \\ ds & = \sqrt{ dx^2 + dy^2 } = dx\ \sqrt{1 + \left( {{dy} \over {dx}} \right)^2 } = dy\ \sqrt{ 1 + \left( {{dx} \over {dy}} \right)^2 } \end{align}
(12)
\begin{align} \int_{x=0}^{x=1} ds & = \int_0^1 dx\ \sqrt{ 1 + \left( {{dy} \over {dx}} \right)^2 } \\ & = \int_0^1 \sqrt{1 + 4x^2} dx = {{2 \sqrt{5} + \ln (2+ \sqrt{5} )} \over 4} \end{align}
  • 면적의 질량중심
(13)
\begin{align} \int \bar{x}\ dM = \int x\ dM, \qquad \int \bar{y}\ dM = \int y\ dM, \qquad \int \bar{z}\ dM = \int z\ dM \end{align}
(14)
\begin{align} \iint \bar{x}\ dA = \iint x\ dA, \qquad \iint \bar{y}\ dy = \iint y\ dA \end{align}
(15)
\begin{align} \iint x_M dM = \iint x dM = \iint x \rho\ dA \end{align}
  • 곡선의 질량중심
(16)
\begin{align} \int \bar{x}\ ds & = \int x\ ds = \int_0^1 x\ dx\ \sqrt{ 1 + \left( {{dy} \over {dx}} \right)^2 } = \int_0^1 dx\ x\ \sqrt{1 + 4x^2} \\ \int \bar{y}\ ds & = \int y\ ds = \int_0^1 y\ dx\ \sqrt{1 + 4x^2 } = \int_0^1 x^2 dx\ \sqrt{1 + 4x^2 } \end{align}
  • 관성모멘트
(17)
\begin{align} I \equiv ml^2 \end{align}
3-3.png
(18)
\begin{align} I_x & = \int y^2\ dM = \int dx\ dy xy^3 = \int_0^1 dx\ \int_0^{x^2} dy\ xy^3 = {1 \over 40}, \\ I_y & = \int x^2\ dM = \int dx\ dy\ xy = {1 \over 16}, \\ I_z & = \int (x^2 + y^2 ) dM = I_x + I_y = {y \over 80} \end{align}

예제 2:

3-4.png3-5.png

  • 부피
(19)
\begin{align} V = \int_0^1 \pi y^2\ dx = \int_0^1 \pi x^4\ dx = { \pi \over 5} & = \iiint dx\ dy\ dz \\ & = \int_{x=0}^1 dx\ \int_{y=-x^2}^{x^2} dy\ \int_{z= - \sqrt{x^4 - y^2} }^{\sqrt{x^4 - y^2}} dz = {\pi \over 5} \end{align}
  • 관성모멘트
(20)
\begin{equation} \end{equation}

[[/math]]

5.4 적분의 변수변환, 야코비 행렬식

극좌표계:

4-1.png
(21)
\begin{align} x = r \cos \theta, \qquad y = r \sin \theta \end{align}
(22)
\begin{align} dA = dr \cdot r\ d \theta = r\ dr\ d \theta \end{align}
(23)
\begin{align} ds^2 & = dr^2 + r^2\ d \theta^2 \\ ds & = \sqrt{ \left( {{dr} \over {d \theta}} \right)^2 + r^2 }\ d \theta = \sqrt{ 1 + r^2\ \left( {{d \theta} \over {dr}} \right)^2 }\ dr \end{align}

원통좌표계:

4-4.png
(24)
\begin{align} x = r \cos \theta, \qquad y = r \sin \theta, \qquad z = z \end{align}
(25)
\begin{align} dV & = r\ dr\ d \theta\ dz \\ ds^2 & = dr^2 + r^2\ d \theta^2 + dz^2 \\ dA & = a\ d \theta\ dz \end{align}

구면좌표계:

4-5.png
(26)
\begin{align} x = r \sin \theta \cos \phi, \qquad y = r \sin \theta \sin \phi, \qquad z = r \cos \theta \end{align}
(27)
\begin{align} dV & = dr\ r\ d \theta \cdot r \sin \theta d \phi = r^2 \sin \theta d \theta d \phi \\ ds^2 & = dr^2 + r^2\ d \theta^2 + r^2 \sin^2 \theta d \phi^2 \\ dA & = a^2 \sin \theta\ d \theta\ d \phi \end{align}

야코비 행렬식:

(28)
\begin{align} x(r, s) \qquad y(r, s) \\ (x, y) \longrightarrow (r, s) \end{align}
(29)
\begin{align} J = {{ \partial (x, y) } \over {\partial (s, t)}} = \begin{vmatrix} {{\partial x} \over {\partial s}} & {{\partial x} \over {\partial t}} \\ {{\partial y} \over {\partial s}} & {{\partial y} \over {\partial t}} \end{vmatrix} \end{align}
(30)
\begin{align} A = \iint dx\ dy = \iint \left\lvert J \right\rvert dr\ ds \end{align}

예컨대 극좌표의 경우,

(31)
\begin{align} J = \begin{vmatrix} \cos \theta & - r \sin \theta \\ \sin \theta & r \cos \theta \end{vmatrix} = r \end{align}
(32)
\begin{align} \implies \int dx\ dy = \int r\ dx\ dy \end{align}

삼차원-삼중적분에 대한 일반화:

(33)
\begin{align} \iiint dx\ dy\ dz = \iiint \left\lvert J \right\rvert\ du\ dv\ dw \end{align}
(34)
\begin{align} J = \begin{vmatrix} {{\partial x} \over {\partial u}} & {{\partial x} \over {\partial v}} & {{\partial x} \over {\partial w}} \\ {{\partial y} \over {\partial u}} & {{\partial y} \over {\partial v}} & {{\partial y} \over {\partial w}} \\ {{\partial z} \over {\partial u}} & {{\partial z} \over {\partial v}} & {{\partial z} \over {\partial w}} \end{vmatrix} \end{align}

구면좌표계의 경우:

(35)
\begin{align} J = {{ \partial (x, y, z) } \over { \partial (r, \theta, \phi ) }} & = \begin{vmatrix} \sin \theta \cos \phi & r \cos \theta \cos \phi & - r \sin \theta \sin \phi \\ \sin \theta \sin \phi & r \cos \theta \sin \phi & r \sin \theta \cos \phi \\ \cos \theta & - r \sin \theta & 0 \end{vmatrix} \\ & = \cos \theta \left[ r^2 \sin \theta \cos \theta \sin^2 \phi + r^2 \sin \theta \cos \theta \cos^2 \phi \right] \\ & \qquad + r \sin \theta \left[ r \sin^2 \theta \cos^2 \phi + r \sin^2 \theta \sin^2 \phi \right] \\ & = r^2 \sin \theta \end{align}

예제 2:

4-6.png
(36)
\begin{align} V = \iiint dx\ dy\ dz = \iiint J\ dr\ d \theta\ d \phi = \iiint r^2 \sin \theta\ dr\ d \theta\ d \phi \end{align}
(37)
\begin{align} M & = \iiint \rho\ dV = \iiint \rho\ r\ dr\ d \theta\ d z \\ & = \rho \int_0^h dz\ \int_0^z dr\ \int_0^{2 \pi} d \theta\ r = { \pi \over 3} \rho h^3 \end{align}
(38)
\begin{align} \bar{z} {\pi \over 3} h^3 = \iiint \bar{z}\ dV & = \iiint z\ dV = \int_0^h dz\ \int_0^z dr\ \int_0^{2 \pi} d \theta\ r z = {1 \over 4} \pi h^4 \\ & \implies \bar{z} = {3 \over 4} h \end{align}
(39)
\begin{align} I_z = \iiint r^2 \rho\ dV = { \pi \over 10} \rho h^5 = { \pi \over 10} h^5 \times {{M} \over { {\pi \over 3} h^3 }} = {3 \over 10} h^3 M \end{align}

예제 3: 반지름 $a$인 고체공의 지름에 대한 관성모멘트

(40)
\begin{align} M = \iiint \rho\ dV = {4 \over 3} \pi a^3 \rho \end{align}
(41)
\begin{align} I_z & = \iiint ( x^2 + y^2 ) \rho\ dV = \rho \iiint r^2 \sin^2 \theta\ r^2 \sin \theta\ d \theta\ d \phi \\ & = \rho \int_0^{2 \pi} d \phi \int_0^\pi d \theta \int_0^a dr\ r^4 \sin^4 \theta \\ & = \rho \cdot {a^5 \over 5} \cdot {4 \over 3} \cdot 2 \pi = {{ 8 \pi a^5 \rho } \over 15} = {2 \over 5} M a^2 \end{align}

예제 4: 타원체의 z축에 대한 회전관성

(42)
\begin{align} {x^2 \over a^2} + {y^2 \over b^2} + {z^2 \over c^2} =1 \end{align}

변수 변환 - $x = a x' , y = b y', z = c z' \implies x'^2 + y'^2 + z'^2 = 1$

(43)
\begin{align} M = \iiint \rho\ dx\ dy\ dz = \rho abc \iiint dx' dy' dz' = {4 \over 3} \pi \rho abc \end{align}
(44)
\begin{align} I_z = \iiint (x^2 + y^2) \rho\ dx\ dy\ dz = \rho abc \iiint dx' dy' dz' (a^2 x'^2 + b^2 y'^2 ) d V' \end{align}
(45)
\begin{equation} \end{equation}

곡면좌표계에서 호의 길이 구하기:

(46)
\begin{align} ds^2 & = dx^2 + dy^2 + dy^2 + dz^2 \qquad (직교) \\ & = dr^2 + r^2\ d \theta^2 + dz^2 \qquad (원통) \\ & = rd^2 + (r\ d \theta )^2 + ( r \sin \theta\ d \phi )^2 \qquad (구면) \end{align}

5.5 면적분

5-1.png
  • $dA$: 면적요소
  • $\gamma$: 면적요소와 $x-y$ 평면 사이의 예각
(47)
\begin{align} dA & = dx\ dy = {1 \over {\cos \gamma}}\ dx\ dy \\ \iint dA & = \iint {1 \over {\cos \gamma}}\ dx\ dy \end{align}

면의 식 $\phi (x, y, z) = \mathrm{const.}$라고 하면,

(48)
\begin{align} \operatorname{grad} \phi = \hat{x} {{\partial \phi } \over {\partial x}} + \hat{y} {{\partial \phi} \over {\partial y}} + \hat{z} {{\partial \phi } \over {\partial z}} \end{align}

이 벡터는 면 $\phi$에 수직이다. 그러면 그 방향의 단위법선벡터는

(49)
\begin{align} \hat{n} & = {{ \vec{\nabla} \phi} \over { \left\lvert \vec{\nabla} \phi \right\rvert }} \\ \hat{n} \cdot \hat{z} & = \cos \gamma = {{ \vec{\nabla} \phi \cdot \hat{z} } \over { \left\lvert \vec{\nabla} \phi \right\rvert }} = {{ \partial \phi / \partial z} \over { \left\lvert \vec{\nabla} \phi \right\rvert }} \\ \cos \gamma & = {{ \vec{ \nabla} \phi \cdot \hat{z} } \over { \left\lvert \vec{\nabla} \phi \right\rvert }} = {{ \left( {{\partial \phi } \over {\partial z}} \right) } \over \sqrt{\left( {{\partial \phi} \over {\partial x}} \right)^2 + \left( {{\partial \phi} \over {\partial y}} \right)^2 + \left( {{\partial \phi} \over {\partial z}} \right)^2 }} \end{align}

예제 1:
공껍질

(50)
\begin{align} \phi (x, y, z) = x^2 + y^2 + z^2 & = 1 \end{align}

이 원기둥

(51)
\begin{align} x^2 + y^2 - y = x^2 + \left( y- {1 \over 2} \right)^2 - {1 \over 4} = 0 \end{align}

에 의해 잘린 면적 구하기.

5-2.png
(52)
\begin{align} \vec{\nabla} \phi & = (2x, 2y, 2z), \\ \hat{n} & = {{(2x, 2y, 2z) } \over \sqrt{ 4x^2 + 4y^2 + 4z^2 }} \\ \hat{n} \cdot \hat{z} & = {{ 2z } \over \sqrt{4 x^2 + 4 y^2 + 4z^2}} \end{align}
(53)
\begin{align} A = & \iint {1 \over { \cos \gamma}}\ dx\ dy = \iint { \sqrt{4x^2 + 4y^2 + 4z^2} \over {2z}}\ dx\ dy = \iint {1 \over z}\ dx\ dy \\ & = \int_0^1 dy\ \int_{ - \sqrt{y-y^2} }^{\sqrt{y-y^2}} dx\ {1 \over \sqrt{1 - x^2 - y^2 }} = 2 \int_0^1 dy\ \int_0^\sqrt{y-y^2} dx\ {1 \over \sqrt{1 - x^2 - y^2}} \\ & = 2 \int d \theta\ \int dr\ r {1 \over \sqrt{1 - r^2}} \qquad \longleftarrow \begin{cases} (x, y) = (r \cos \theta , r \sin \theta) \\ x^2 + y^2 - y = 0 = r^2 - r \sin \theta \\ r = \sin \theta \end{cases} \\ & = 2 \int_0^{\pi \over 2} d \theta\ \int_0^{ \cos \theta} (- dz) \qquad \longleftarrow \begin{cases} z & = \sqrt{1 - r^2} \\ dz & = {{ -r\ dr} \over \sqrt{ 1 - r^2} } \end{cases} \end{align}