高数 07.08 二重积分的计算

§第七章第八节二重积分的计算 \color{blue}{\S 第七章第八节二重积分的计算}

一、利用直角坐标计算二重积分
二、利用极坐标计算二重积分

一、利用直角坐标计算二重积分 \color{blue}{一、利用直角坐标计算二重积分}

由曲顶柱体体积的计算可知,当被积函数f(x,y)≥0且在D上连续时,若D为X−型区域由曲顶柱体体积的计算可知,当被积函数f(x, y) \geq 0\\ 且在D上连续时,若D为X-型区域
D:{φ1(x)≤y≤φ2(x)a≤x≤b D:\left \{ \begin{array}{l}\varphi_1(x) \leq y \leq \varphi_2(x) \\ a \leq x \leq b \end{array} \right.
则∬Df(x,y)dxdy=∫ba[∫φ2(x)φ1(x)f(x,y)dy]dx 则\iint_D{f(x, y)dxdy} = \int_a^b[{\int_{\varphi_1(x)}^{\varphi_2(x)}{f(x, y)dy}}]dx
若D为Y−型区域若D为Y-型区域
D:{ψ1(y)≤x≤ψ2(x)c≤y≤d D:\left \{ \begin{array}{l}\psi_1(y) \leq x \leq \psi_2(x) \\ c \leq y \leq d \end{array} \right.
则∬Df(x,y)dxdy=∫dc[∫ψ2(y)ψ1(y)f(x,y)dx]dy 则\iint_D{f(x, y)dxdy} = \int_c^d[{\int_{\psi_1(y)}^{\psi_2(y)}{f(x, y)dx}}]dy

说明:(1)若积分区域即是X−型区域又是Y−型区域,则有：∬Df(x,y)dxdy=∫ba[∫φ2(x)φ1(x)f(x,y)dy]dx=∫dc[∫ψ2(y)ψ1(y)f(x,y)dx]dy为计算方便，可选择积分序,必要时还可以交换积分序.(2)若积分域复杂,可将它分成若干个X−型域或Y−型域. 说明:(1)若积分区域即是X-型区域又是Y-型区域,则有：\\ \iint_D{f(x, y)dxdy} \\ = \int_a^b[{\int_{\varphi_1(x)}^{\varphi_2(x)}{f(x, y)dy}}]dx \\ = \int_c^d[{\int_{\psi_1(y)}^{\psi_2(y)}{f(x, y)dx}}]dy \\ 为计算方便，可选择积分序,必要时还可以交换积分序.\\ (2)若积分域复杂,可将它分成若干个X-型域或Y-型域.

例1.计算I=∬Dxydσ,其中D是直线y=1,x=2及y=x所围成的区域. 例1.计算I = \iint_D{xyd\sigma},其中D是直线y = 1,x = 2及y = x所围成的区域.
解:解法1.将D看作X−型区域,则D:{1≤y≤x1≤x≤2 解:解法1.将D看作X-型区域, 则D:\left \{ \begin{array}{l}1 \leq y \leq x \\1 \leq x \leq 2 \end{array} \right.
I=∫21dx∫x1xydy=∫21[12xy2]x1=∫21[12x3−12x]dx=[18x4−14x2]21=98 I = \int_1^2 dx \int_1^x{xydy} \\ = \int_1^2{[\dfrac{1}{2}xy^2]_1^x} \\ = \int_1^2[\dfrac{1}{2}x^3 - \dfrac{1}{2}x]dx \\ = [\dfrac{1}{8}x^4 - \dfrac{1}{4}x^2]_1^2 \\ = \dfrac{9}{8}
解法1.将D看作Y−型区域,则D:{y≤x≤21≤y≤2 解法1.将D看作Y-型区域, 则D:\left \{ \begin{array}{l}y \leq x \leq 2 \\1 \leq y \leq 2 \end{array} \right.
I=∫21dy∫2yxydx=∫21[12x2y|2y]dy=∫21[2y−12y3]dy=[y2−18y4]21=98 I = \int_1^2 dy \int_y^2{xydx} \\ = \int_1^2 [\dfrac{1}{2}x^2y|_y^2]dy \\ = \int_1^2 [2y - \dfrac{1}{2}y^3]dy \\ = [y^2 - \dfrac{1}{8}y^4]_1^2 \\ = \dfrac{9}{8}

例2.计算∬Dxydσ,其中D是抛物线y2=x及直线y=x−2所围成的闭区域. 例2.计算\iint_Dxyd\sigma,其中D是抛物线y^2 = x及直线y = x - 2所围成的闭区域.
解:为计算简便,先对x后对y积分,则解:为计算简便,先对x后对y积分,则
D:{y2≤x≤y+2−1≤y≤2 D:\left \{ \begin{array}{l}y^2 \leq x \leq y + 2 \\ -1 \leq y \leq 2 \end{array} \right.
∬Dxydσ=∫21dy∫y2y+2xydx=∫2−1[12x2y]y+2y2dy=∫2−1[12(y+2)2y−12y5]dy=∫2−1[12(−y5+y3+4y2+4y)]dy=12[−16y6+14y4+43y3+2y2]2−1=458 \iint_Dxyd\sigma = \int_1^2dy \int_{y + 2}^{y^2}{xydx} \\ =\int_{-1}^2[\dfrac{1}{2}x^2y]_{y^2}^{y + 2}dy \\ =\int_{-1}^2[\dfrac{1}{2}(y + 2)^2y -\dfrac{1}{2}y^5]dy \\ =\int_{-1}^2[\dfrac{1}{2}(-y^5 + y^3 + 4y^2 + 4y)]dy \\ =\dfrac{1}{2}[-\dfrac{1}{6}y^6 + \dfrac{1}{4}y^4 + \dfrac{4}{3}y^3 + 2y^2]_{-1}^2 \\ = \dfrac{45}{8}

例3.计算∬Dsinxxdxdy,其中D是直线y=x,y=0,x=π所围成的闭区域. 例3.计算\iint_D\dfrac{\sin{x}}{x}dxdy,其中D是直线y = x, \\ y = 0, x = \pi所围成的闭区域.
解: 解:
D:{0≤x≤π0≤y≤x D:\left \{ \begin{array}{l} 0 \leq x \leq \pi \\ 0 \leq y \leq x \end{array} \right.
∬Dsinxxdxdy=∫π0dx∫x0sinxxdy=∫π0sinxx[y]x0dx=∫π0sinxxxdx=∫π0sinxdx=−[cosx]π0=2 \iint_D\dfrac{\sin{x}}{x}dxdy = \int_0^{\pi}dx \int_0^x \dfrac{\sin{x}}{x}dy\\ = \int_0^{\pi} \dfrac{\sin{x}}{x} [y]_0^x dx \\ = \int_0^{\pi} \dfrac{\sin{x}}{x} x dx \\ = \int_0^{\pi} \sin{x} dx \\ = -[\cos{x}]_0^{\pi} \\ = 2

例4.交换下列积分顺序I=∫20dx∫x220f(x,y)dy+∫22√2dx∫8−x2√0f(x,y)dy 例4.交换下列积分顺序\\ I = \int_0^2 dx \int_0^{\frac{x^2}{2}}f(x, y) dy + \int_2^{2\sqrt{2}}dx\int_0^{\sqrt{8 -x^2}}f(x,y) dy
解:y=1x2x2+y2=8交点(2,2) 解:\\ y = \dfrac{1}{x^2} \\ x^2 + y^2 = 8 \\ 交点(2, 2)
原区域D1:⎧⎩⎨0≤x≤20≤y≤12x2 原区域 D_1:\left \{ \begin{array}{l} 0 \leq x \leq 2 \\ 0 \leq y \leq \dfrac{1}{2}x^2 \end{array} \right.
原区域D2:{2≤x≤22√0≤y≤8−x2−−−−−√ 原区域 D_2:\left \{ \begin{array}{l} 2 \leq x \leq 2\sqrt{2} \\ 0 \leq y \leq \sqrt{8 - x^2} \end{array} \right.
交换积分顺序，变为先对x积分，再对y积分交换积分顺序，变为先对x积分，再对y积分
原区域D:{0≤y≤22y−−√≤x≤8−y2−−−−−√ 原区域 D:\left \{ \begin{array}{l} 0 \leq y \leq 2 \\ \sqrt{2y} \leq x \leq \sqrt{8 - y^2} \end{array} \right.
I=∫20dy∫8−y2√2y√f(x,y)dx I = \int_0^2 dy \int_{\sqrt{2y}}^{\sqrt{8 - y^2}}f(x,y) dx

二、利用极坐标计算二重积分 \color{blue}{二、利用极坐标计算二重积分}

在极坐标系下,用同心圆r=常数及射线θ=常数,分划区域D为Δσk(k=1,2,⋯,n)则除包含边界点的小区域外,小区域的面积Δσk=12(rk+Δrk)2⋅Δθk−12r2k⋅Δθk=12[rk+(rk+Δrk)]Δrk⋅Δθk=rk¯Δrk⋅Δθk在Δσk内取点(rk¯,θk¯),对应有ξk=rk¯cosθk¯+rk¯sinθk¯limλ→0∑nk=1f(ξk,ηk)Δσk=limλ→0∑nk=1f(rk¯cosθk¯,rk¯sinθk¯)rk¯ΔrkΔθ即∬Df(x,y)dσ=∬Df(rcosθ,rsinθ)rdrdθ 在极坐标系下,用同心圆r = 常数及射线\theta = 常数,分划区域D为\\ \Delta{\sigma_k} (k = 1, 2, \cdots, n) \\ 则除包含边界点的小区域外,小区域的面积\\ \Delta{\sigma_k} = \dfrac{1}{2}(r_k + \Delta{r_k})^2 \cdot \Delta{\theta_k} - \dfrac{1}{2}r_k^2 \cdot \Delta{\theta_k} \\ = \dfrac{1}{2}[r_k + (r_k + \Delta{r_k})]\Delta{r_k} \cdot \Delta{\theta_k}\\ =\bar{r_k} \Delta{r_k} \cdot \Delta{\theta_k}\\ 在\Delta{\sigma_k}内取点(\bar{r_k}, \bar{\theta_k}),对应有\\ \xi_k = \bar{r_k}\cos{\bar{\theta_k}} + \bar{r_k}\sin{\bar{\theta_k}} \\ \lim_{\lambda \rightarrow 0}{\sum_{k = 1}^{n}{f(\xi_k, \eta_k)\Delta{\sigma_k}}} \\ = \lim_{\lambda \rightarrow 0}{\sum_{k = 1}^{n}{f(\bar{r_k}\cos{\bar{\theta_k}}, \bar{r_k}\sin{\bar{\theta_k}})\bar{r_k}\Delta{r_k}\Delta{\theta}}} \\ 即\iint_Df(x, y)d\sigma = \iint_D f(r\cos{\theta}, r\sin{\theta})rdr d\theta
设D:{φ1(θ)≤r≤φ2(θ)α≤θ≤β则设D:\left \{ \begin{array}{l} \varphi_1(\theta) \leq r \leq \varphi_2(\theta) \\ \alpha \leq \theta \leq \beta \end{array} \right. 则
∬Df(rcosθ,rsinθ)rdrdθ=∫βαdθ∫φ2(θ)φ1(θ)f(rcosθ,rsinθ)rdr \iint_D f(r\cos{\theta}, r\sin{\theta})rdrd\theta \\ = \int_{\alpha}^{\beta} d\theta \int_{\varphi_1(\theta)}^{\varphi_2(\theta)} f(r\cos{\theta}, r\sin{\theta})rdr

特别,对D:{φ1(θ)≤r≤φ2(θ)0≤θ≤2π则特别,对D:\left \{ \begin{array}{l} \varphi_1(\theta) \leq r \leq \varphi_2(\theta) \\ 0 \leq \theta \leq 2\pi \end{array} \right. 则
∬Df(rcosθ,rsinθ)rdrdθ=∫2π0dθ∫φ(θ)0f(rcosθ,rsinθ)rdr \iint_D f(r\cos{\theta}, r\sin{\theta})rdrd\theta \\ = \int_{0}^{2\pi} d\theta \int_{0}^{\varphi(\theta)} f(r\cos{\theta}, r\sin{\theta})rdr

若f≡1,则可求得D的面积若f \equiv 1,则可求得D的面积
σ=∬Ddσ=12∫2π0φ2(θ)dθ \sigma = \iint_D d\sigma = \dfrac{1}{2}\int_0^{2\pi} \varphi^2(\theta) d\theta

思考:下列个图中域D分别与x,y轴相切于原点,试问θ的变化范围是什么? 思考:下列个图中域D分别与x,y轴相切于原点,试问 \theta 的变化范围是什么?
(1)r=φ(θ)在一、二象限与x轴相切,θ∈[0,π] (1) r = \varphi(\theta)在一、二象限与x轴相切, \theta \in [0, \pi]
(2)r=φ(θ)在一、四象限与y轴相切,θ∈[−π2,π2] (2) r = \varphi(\theta)在一、四象限与y轴相切, \theta \in [-\dfrac{\pi}{2}, \dfrac{\pi}{2}]

例5.计算∬De−x2−y2dxdy,其中D:x2+y2≤a2. 例5.计算\iint_D e^{-x^2-y^2}dxdy,其中D:x^2 + y^2 \leq a^2.
由于e−x2−y2的原函数不是初等函数,故无法用直角坐标系计算由于e^{-x^2 - y^2}的原函数不是初等函数,故无法用直角坐标系计算
解:在极坐标系下D:{0≤r≤a0≤θ≤2π 解:在极坐标系下D:\left \{ \begin{array}{l}0 \leq r \leq a \\ 0 \leq \theta \leq 2\pi \end{array} \right.
∬De−x2−y2dxdy=∬De−r2rdrdθ=∫2π0dθ∫a0e−r2rdr=∫2π0[−12e−r2]a0dθ=∫2π01−e−a22dθ=π(1−e−a2) \iint_D e^{-x^2-y^2}dxdy\\ = \iint_D e^{-r^2} rdr d\theta \\ = \int_0^{2\pi} d\theta \int_0^a e^{-r^2}rdr \\ = \int_0^{2\pi} [-\dfrac{1}{2}e^{-r^2}]_0^a d\theta \\ = \int_0^{2\pi} \dfrac{1 - e^{-a^2}}{2} d\theta \\ = \pi ( 1 - e^{-a^2})

例6.求球体x2+y2+z2≤4a2被圆柱体x2+y2=2ax(a>0)所截得的(含在柱体内的)立体的体积. 例6.求球体x^2 + y^2 + z^2 \leq 4a^2被圆柱体x^2 + y^2 = 2ax(a > 0)所截得的(含在柱体内的)立体的体积.
解:设D:0≤r≤2acosθ,0≤θ≤π2 解:设D:0 \leq r \leq 2a\cos{\theta}, 0 \leq \theta \leq \dfrac{\pi}{2}
由对称性可知由对称性可知
V=4∬D4a2−r2−−−−−−−√rdrdθ=4∫π20dθ∫2acosθ04a2−r2−−−−−−−√rdr=323a3∫π20(1−sin3θ)dθ=323a3(π2−23) V=4\iint_D \sqrt{4a^2 - r^2} rdrd\theta \\ =4\int_0^{\frac{\pi}{2}} d\theta \int_0^{2a\cos{\theta}} \sqrt{4a^2 - r^2} rdr \\ = \dfrac{32}{3}a^3 \int_0^{\frac{\pi}{2}}(1 - \sin^3{\theta})d\theta \\ = \dfrac{32}{3}a^3(\dfrac{\pi}{2} - \dfrac{2}{3})

内容小结
(1)二重积分化为累次积分的方法
直角坐标系情形:
若积分区域为:
D={(x,y)|a≤x≤b,y1(x)≤y≤y2(x)} D = \lbrace (x, y) | a \leq x \leq b, y_1(x) \leq y \leq y_2(x) \rbrace
则∬Df(x,y)dσ=∫badx∫y2(x)y1(x)f(x,y)dy 则 \iint_D f(x, y) d\sigma = \int_a^b dx \int_{y_1(x)}^{y_2(x)} f(x, y)dy

若积分区域为:
D={(x,y)|c≤y≤d,x1(y)≤x≤x2(y)} D = \lbrace (x, y) | c \leq y \leq d, x_1(y) \leq x \leq x_2(y) \rbrace
则∬Df(x,y)dσ=∫dcdy∫x2(y)x1(y)f(x,y)dx 则 \iint_D f(x, y) d\sigma = \int_c^d dy \int_{x_1(y)}^{x_2(y)} f(x, y)dx

极坐标系情形：
若积分区域为
D={(r,θ)|α≤θ≤β,φ1(θ)≤r≤φ2(θ)} D = \lbrace(r, \theta) | \alpha \leq \theta \leq \beta, \varphi_1(\theta) \leq r \leq \varphi_2(\theta) \rbrace
则∬Df(x,y)dσ=∬Df(rcosθ,rsinθ)rdrdθ=∫βαdθ∫φ2(θ)φ1(θ)f(rcosθ,rsinθ)rdr 则\iint_D f(x, y)d\sigma = \iint_D f(r\cos{\theta}, r\sin{\theta})rdrd\theta \\ = \int_{\alpha}^{\beta} d\theta \int_{\varphi_1(\theta)}^{\varphi_2(\theta)} f(r\cos{\theta}, r\sin{\theta})rdr

(2)计算步骤及注意事项
先画出积分域
选择坐标系
确定积分序
写出积分限
计算要简便，充分利用对称性

练习
1.设f(x)∈C[0,1],且∫10f(x)dx=A,求I=∫10dx∫1xf(x)f(y)dy 1.设f(x) \in C[0,1], 且\int_0^1 f(x) dx = A, 求I = \int_0^1dx \int_x^1 f(x)f(y)dy
解:交换积分顺序后,x,y互换I=∫10dx∫1xf(x)f(y)dy=∫10dx∫x0f(x)f(y)dy∴2I=∫10dx∫1xf(x)f(y)dy+∫10dy∫x0f(x)f(y)dy=∫10dx∫10f(x)f(y)dy=∫10f(x)dx∫10f(y)dy=A2I=A22 解:\\ 交换积分顺序后,x, y互换 I = \int_0^1dx \int_x^1 f(x)f(y)dy = \int_0^1 dx \int_0^x f(x)f(y)dy \\ \therefore 2I = \int_0^1dx \int_x^1 f(x)f(y)dy + \int_0^1dy \int_0^x f(x)f(y)dy \\ = \int_0^1 dx\int_0^1 f(x)f(y)dy \\ = \int_0^1 f(x)dx \int_0^1 f(y)dy \\ = A^2 \\ I = \dfrac{A^2}{2}

2.给定I=∫2a0dx∫2ax√2ax−x2√f(x,y)dy(a>0,改变积分次序 2.给定I = \int_0^{2a} dx \int_{\sqrt{2ax - x^2}}^{\sqrt{2ax}}f(x, y)dy (a > 0,改变积分次序
解:y=2ax−−−√⟹x=y22ay=2ax−x2−−−−−−−√⟹x=a±a2−y2−−−−−−√I=∫2a0dx∫2ax√2ax−x2√f(x,y)dy=∫10dy∫a−a2−y2√y22f(x,y)dx+∫a0dy∫2aa+a2−y2√f(x,y)dx+∫2aa∫2ay22af(x,y)dx 解:\\ y = \sqrt{2ax} \Longrightarrow x = \dfrac{y^2}{2a}\\ y = \sqrt{2ax - x^2} \Longrightarrow x = a \pm \sqrt{a^2 - y^2}\\ I = \int_0^{2a} dx \int_{\sqrt{2ax - x^2}}^{\sqrt{2ax}}f(x, y)dy \\ = \int_0^1 dy \int_{\frac{y^2}{2}}^{a - \sqrt{a^2 - y^2}}f(x,y)dx + \int_0^a dy\int_{a + \sqrt{a^2 - y^2}}^{2a} f(x, y) dx + \int_a^{2a} \int_{\frac{y^2}{2a}}^{2a} f(x, y) dx