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二、二重积分的计算

小题​

（一）直角坐标法：便于定限的积分次序​

  1. **【1991-4-5 分】**计算 I=∬Dy dx dy\displaystyle I=\iint_{D} y ~d x ~d y，其中 D\displaystyle D 是由 xa+yb=1\displaystyle \sqrt{\frac{x}{a}}+\sqrt{\frac{y}{b}}=1、x=0\displaystyle x=0、y=0\displaystyle y=0 围成，a>0,b>0\displaystyle a>0,b>0。

  2. **【1995-2-5 分】**计算∬Dx2y dx dy\displaystyle \iint_{D} x^{2} y ~d x ~d y，其中 D\displaystyle D 由双曲线x2−y2=1\displaystyle x^{2}-y^{2}=1及直线y=0,y=1\displaystyle y=0,y=1围成。

  3. 【2003-34-4 分】a>0\displaystyle a>0，

f(x)=g(x)={a,0≤x≤1,0,其他,f(x)=g(x)= \begin{cases} a, &0 \leq x \leq 1,\\ 0, &\text{其他}, \end{cases}

D\displaystyle D为全平面，求I=∬Df(x)g(y−x)dx dy\displaystyle I=\iint_{D} f(x) g(y-x) d x ~d y。

  1. 【2008-4-4 分】 ∫12dx∫01xyln⁡xdy=\int_{1}^{2} d x \int_{0}^{1} x^{y} \ln x d y=

  2. 【2022-3-5 分】

f(x)={ex,0≤x≤1,0,其他,f(x)= \begin{cases} e^{x}, &0 \leq x \leq 1,\\ 0, &\text{其他}, \end{cases}

求∫−∞+∞dx∫−∞+∞f(x)f(y−x)dy\displaystyle \int_{-\infty}^{+\infty} d x \int_{-\infty}^{+\infty} f(x) f(y-x) d y。

（二）直角坐标法：第一步积分较简单​

  1. **【1987-45-5 分】**计算∬Dex2dxdy\displaystyle \iint_{D} e^{x^{2}} d x d y，D\displaystyle D为第一象限y=x,y=x3\displaystyle y=x,y=x^3围成区域。

  2. **【1990-45-5 分】**计算∬Dxe−y2dxdy\displaystyle \iint_{D} x e^{-y^{2}} d x d y，D\displaystyle D是y=4x2,y=9x2\displaystyle y=4x^2,y=9x^2在第一象限围成区域。

  3. 【1995-4-5 分】 ∫−∞+∞∫−∞+∞min⁡{x,y}e−(x2+y2)dx dy\int_{-\infty}^{+\infty} \int_{-\infty}^{+\infty} \min \{x, y\} e^{-(x^{2}+y^{2})} d x ~d y

  4. 【2016-3-4 分】D={(x,y)∣∣x∣≤y≤1,−1≤x≤1}\displaystyle D=\{(x, y)|| x | \leq y \leq 1,-1 \leq x \leq 1\}，计算∬Dx2e−y2dxdy\displaystyle \iint_{D} x^{2} e^{-y^{2}} d x d y。

（三）交换积分次序​

  1. 【1988-45-4 分】 ∫0π6dy∫yπ6cos⁡xxdx\int_{0}^{\frac{\pi}{6}} d y \int_{y}^{\frac{\pi}{6}} \dfrac{\cos x}{x} d x

  2. 【1990-12-3 分】 ∫02dx∫x2e−y2dy\int_{0}^{2} d x \int_{x}^{2} e^{-y^{2}} d y

  3. **【1992-4-3 分】**交换次序： ∫01 dy∫y2−y2f(x,y)dx=\int_{0}^{1} ~d y \int_{\sqrt{y}}^{\sqrt{2-y^{2}}} f(x, y) d x=

  4. **【2001-1-3 分】**交换次序： ∫−10 dy∫21−yf(x,y)dx=\int_{-1}^{0} ~d y \int_{2}^{1-y} f(x, y) d x=

  5. 【2002-3-3 分】

∫014 dy∫yyf(x,y)dx+∫1412 dy∫y12f(x,y)dx=\int_{0}^{\frac{1}{4}} ~d y \int_{y}^{\sqrt{y}} f(x, y) d x+\int_{\frac{1}{4}}^{\frac{1}{2}} ~d y \int_{y}^{\frac{1}{2}} f(x, y) d x=

  1. 【2007-234-4 分】 ∫π2πdx∫sin⁡x1f(x,y)dy=\int_{\frac{\pi}{2}}^{\pi} d x \int_{\sin x}^{1} f(x, y) d y= A. ∫01dy∫π+arcsin⁡yπf(x,y)dx\displaystyle \int_{0}^{1} d y \int_{\pi+\arcsin y}^{\pi} f(x, y) d x B. ∫01dy∫π−arcsin⁡yπf(x,y)dx\displaystyle \int_{0}^{1} d y \int_{\pi-\arcsin y}^{\pi} f(x, y) d x C. ∫01dy∫π2π+arcsin⁡yf(x,y)dx\displaystyle \int_{0}^{1} d y \int_{\frac{\pi}{2}}^{\pi+\arcsin y} f(x, y) d x D. ∫01dy∫π2π−arcsin⁡yf(x,y)dx\displaystyle \int_{0}^{1} d y \int_{\frac{\pi}{2}}^{\pi-\arcsin y} f(x, y) d x

  2. 【2009-2-4 分】

∫12dx∫x2f(x,y)dy+∫12dy∫y4−yf(x,y)dx=\int_{1}^{2} d x \int_{x}^{2} f(x, y) d y+\int_{1}^{2} d y \int_{y}^{4-y} f(x, y) d x=

A. ∫12dx∫14−xf(x,y)dy\displaystyle \int_{1}^{2} d x \int_{1}^{4-x} f(x, y) d y B. ∫12dx∫x4−xf(x,y)dy\displaystyle \int_{1}^{2} d x \int_{x}^{4-x} f(x, y) d y C. ∫12dy∫14−yf(x,y)dx\displaystyle \int_{1}^{2} d y \int_{1}^{4-y} f(x, y) d x D. ∫12dy∫y2f(x,y)dx\displaystyle \int_{1}^{2} d y \int_{y}^{2} f(x, y) d x

  1. 【2014-3-4 分】 ∫01dy∫y1(ex2x−ey2)dx\int_{0}^{1} d y \int_{y}^{1}\left(\dfrac{e^{x^{2}}}{x}-e^{y^{2}}\right) d x

  2. 【2017-2-4 分】 ∫01dy∫y1tan⁡xxdx\int_{0}^{1} d y \int_{y}^{1} \dfrac{\tan x}{x} d x

  3. 【2020-2-4 分】 ∫01dy∫y1x3+1dx\int_{0}^{1} d y \int_{\sqrt{y}}^{1} \sqrt{x^{3}+1} d x

  4. 【2021-2-5 分】 f(t)=∫1t2dx∫xtsin⁡xydyf(t)=\int_{1}^{t^{2}} d x \int_{\sqrt{x}}^{t} \sin \dfrac{x}{y} d y 求f′(π2)\displaystyle f'(\dfrac{\pi}{2})。

  5. 【2022-2-5 分】 ∫02dy∫y2y1+x3dx=\int_{0}^{2} d y \int_{y}^{2} \dfrac{y}{\sqrt{1+x^{3}}} d x= A. 26\displaystyle \dfrac{\sqrt2}{6}    B. 13\displaystyle \dfrac13 C. 23\displaystyle \dfrac{\sqrt2}{3}    D. 23\displaystyle \dfrac23

  6. 【2024-23-5 分】 ∫π6π2dx∫sin⁡x1f(x,y)dy=\int_{\frac{\pi}{6}}^{\frac{\pi}{2}} d x \int_{\sin x}^{1} f(x, y) d y= A. ∫121dy∫π6arcsin⁡yf(x,y)dx\displaystyle \int_{\dfrac12}^{1} d y \int_{\frac{\pi}{6}}^{\arcsin y} f(x,y)dx B. ∫121dy∫arcsin⁡yπ2f(x,y)dx\displaystyle \int_{\dfrac12}^{1} d y \int_{\arcsin y}^{\frac{\pi}{2}} f(x,y)dx C. ∫012dy∫π6arcsin⁡yf(x,y)dx\displaystyle \int_{0}^{\dfrac12} d y \int_{\frac{\pi}{6}}^{\arcsin y} f(x,y)dx D. ∫012dy∫arcsin⁡yπ2f(x,y)dx\displaystyle \int_{0}^{\dfrac12} d y \int_{\arcsin y}^{\frac{\pi}{2}} f(x,y)dx

  7. 【2025-3-5 分】 ∫01dy∫0yf(x)dx=\int_{0}^{1} d y \int_{0}^{y} f(x) d x= A. ∫01xf(x)dx\displaystyle \int_{0}^{1} x f(x) d x B. ∫01(1+x)f(x)dx\displaystyle \int_{0}^{1}(1+x) f(x) d x C. ∫01(x−1)f(x)dx\displaystyle \int_{0}^{1}(x-1) f(x) d x D. ∫01(1−x)f(x)dx\displaystyle \int_{0}^{1}(1-x) f(x) d x

  8. 【2025-12-5 分】 ∫−22dx∫4−x24f(x,y)dy=\int_{-2}^{2} d x \int_{4-x^{2}}^{4} f(x, y) d y= A. ∫04[∫−2−4−yf(x,y)dx+∫4−y2f(x,y)dx]dy\displaystyle \int_{0}^{4}\left[\int_{-2}^{-\sqrt{4-y}} f(x, y) d x+\int_{\sqrt{4-y}}^{2} f(x, y) d x\right] d y B. ∫04[∫−24−yf(x,y)dx+∫4−y2f(x,y)dx]dy\displaystyle \int_{0}^{4}\left[\int_{-2}^{\sqrt{4-y}} f(x, y) d x+\int_{\sqrt{4-y}}^{2} f(x, y) d x\right] d y C. ∫04[∫−2−4−yf(x,y)dx+∫24−yf(x,y)dx]dy\displaystyle \int_{0}^{4}\left[\int_{-2}^{-\sqrt{4-y}} f(x, y) d x+\int_{2}^{\sqrt{4-y}} f(x, y) d x\right] d y D. 2∫04dy∫4−y2f(x,y)dx\displaystyle 2 \int_{0}^{4} d y \int_{\sqrt{4-y}}^{2} f(x, y) d x

（四）极坐标法​

  1. 【1989-5-5 分】D\displaystyle D第一象限，x2+y2=1,x=0,y=0\displaystyle x^2+y^2=1,x=0,y=0围成，计算 ∬D1−x2−y21+x2+y2dxdy\iint_{D} \dfrac{1-x^{2}-y^{2}}{1+x^{2}+y^{2}} d x d y

  2. 【2011-2-4 分】D\displaystyle D由y=x,x2+y2=2y,y\displaystyle y=x,x^2+y^2=2y,y轴围成，∬Dxydσ=\displaystyle \iint_{D} x y d \sigma=

  3. 【2015-3-4 分】D={(x,y)∣x2+y2≤2x,x2+y2≤2y}\displaystyle D=\{(x,y)|x^2+y^2\le2x,x^2+y^2\le2y\}， A. ∫0π4dθ∫02cos⁡θf(rcos⁡θ,rsin⁡θ)rdr+∫π4π2dθ∫02sin⁡θf(rcos⁡θ,rsin⁡θ)rdr\displaystyle \int_{0}^{\frac{\pi}{4}} d \theta \int_{0}^{2 \cos \theta} f(r \cos \theta, r \sin \theta) r d r+\int_{\frac{\pi}{4}}^{\frac{\pi}{2}} d \theta \int_{0}^{2 \sin \theta} f(r \cos \theta, r \sin \theta) r d r B. ∫0π4dθ∫02sin⁡θf(rcos⁡θ,rsin⁡θ)rdr+∫π4π2dθ∫02cos⁡θf(rcos⁡θ,rsin⁡θ)rdr\displaystyle \int_{0}^{\frac{\pi}{4}} d \theta \int_{0}^{2 \sin \theta} f(r \cos \theta, r \sin \theta) r d r+\int_{\frac{\pi}{4}}^{\frac{\pi}{2}} d \theta \int_{0}^{2 \cos \theta} f(r \cos \theta, r \sin \theta) r d r C. 2∫01dx∫1−1−x2xf(x,y)dy\displaystyle 2 \int_{0}^{1} d x \int_{1-\sqrt{1-x^{2}}}^{x} f(x, y) d y D. 2∫01dx∫x2x−x2f(x,y)dy\displaystyle 2 \int_{0}^{1} d x \int_{x}^{\sqrt{2 x-x^{2}}} f(x, y) d y

（五）坐标系的转换​

  1. 【1996-4-3 分】 ∫0π2 dθ∫0cos⁡θf(rcos⁡θ,rsin⁡θ)r dr=\int_{0}^{\frac{\pi}{2}} ~d \theta \int_{0}^{\cos \theta} f(r \cos \theta, r \sin \theta) r ~d r= A. ∫01dy∫0y−y2f(x,y)dx\displaystyle \int_{0}^{1} d y \int_{0}^{\sqrt{y-y^{2}}} f(x, y) d x B. ∫01dy∫01−y2f(x,y)dx\displaystyle \int_{0}^{1} d y \int_{0}^{\sqrt{1-y^{2}}} f(x, y) d x C. ∫01dx∫01f(x,y)dy\displaystyle \int_{0}^{1} d x \int_{0}^{1} f(x, y) d y D. ∫01dx∫0x−x2f(x,y)dy\displaystyle \int_{0}^{1} d x \int_{0}^{\sqrt{x-x^{2}}} f(x, y) d y

  2. 【2004-2-4 分】D={(x,y)∣x2+y2≤2y}\displaystyle D=\{(x,y)|x^2+y^2\le2y\}，∬Df(xy)dx dy=\displaystyle \iint_{D} f(x y) d x ~d y= A. ∫−11dx∫−1−x21−x2f(xy)dy\displaystyle \int_{-1}^{1} d x \int_{-\sqrt{1-x^{2}}}^{\sqrt{1-x^{2}}} f(x y) d y B. 2∫02dy∫02y−y2f(xy)dx\displaystyle 2 \int_{0}^{2} d y \int_{0}^{\sqrt{2 y-y^{2}}} f(x y) d x C. ∫0πdθ∫02sin⁡θf(r2sin⁡θcos⁡θ)dr\displaystyle \int_{0}^{\pi} d \theta \int_{0}^{2 \sin \theta} f\left(r^{2} \sin \theta \cos \theta\right) d r D. ∫0πdθ∫02sin⁡θf(r2sin⁡θcos⁡θ)rdr\displaystyle \int_{0}^{\pi} d \theta \int_{0}^{2 \sin \theta} f\left(r^{2} \sin \theta \cos \theta\right) r d r

  3. 【2006-12-4 分】 ∫0π4dθ∫01f(rcos⁡θ,rsin⁡θ)rdr=\int_{0}^{\frac{\pi}{4}} d \theta \int_{0}^{1} f(r \cos \theta, r \sin \theta) r d r= A. ∫022dx∫x1−x2f(x,y)dy\displaystyle \int_{0}^{\frac{\sqrt2}{2}} d x \int_{x}^{\sqrt{1-x^{2}}} f(x, y) d y B. ∫022dx∫01−x2f(x,y)dy\displaystyle \int_{0}^{\frac{\sqrt2}{2}} d x \int_{0}^{\sqrt{1-x^{2}}} f(x, y) d y C. ∫022dy∫y1−y2f(x,y)dx\displaystyle \int_{0}^{\frac{\sqrt2}{2}} d y \int_{y}^{\sqrt{1-y^{2}}} f(x, y) d x D. ∫022dy∫01−y2f(x,y)dx\displaystyle \int_{0}^{\frac{\sqrt2}{2}} d y \int_{0}^{\sqrt{1-y^{2}}} f(x, y) d x

  4. 【2012-3-4 分】 ∫0π2dθ∫2cos⁡θ2f(r2)rdr=\int_{0}^{\frac{\pi}{2}} d \theta \int_{2 \cos \theta}^{2} f(r^{2}) r d r= A. ∫02dx∫2x−x24−x2x2+y2f(x2+y2)dy\displaystyle \int_{0}^{2} d x \int_{\sqrt{2 x-x^{2}}}^{\sqrt{4-x^{2}}} \sqrt{x^{2}+y^{2}} f\left(x^{2}+y^{2}\right) d y B. ∫02dx∫2x−x24−x2f(x2+y2)dy\displaystyle \int_{0}^{2} d x \int_{\sqrt{2 x-x^{2}}}^{\sqrt{4-x^{2}}} f\left(x^{2}+y^{2}\right) d y C. ∫02dy∫1+1−y24−y2x2+y2f(x2+y2)dx\displaystyle \int_{0}^{2} d y \int_{1+\sqrt{1-y^{2}}}^{\sqrt{4-y^{2}}} \sqrt{x^{2}+y^{2}} f\left(x^{2}+y^{2}\right) d x D. ∫02dy∫1+1−y24−y2f(x2+y2)dx\displaystyle \int_{0}^{2} d y \int_{1+\sqrt{1-y^{2}}}^{\sqrt{4-y^{2}}} f\left(x^{2}+y^{2}\right) d x

  5. 【2014-1-4 分】 ∫01dy∫−1−y21−yf(x,y)dx=\int_{0}^{1} d y \int_{-\sqrt{1-y^{2}}}^{1-y} f(x, y) d x= A. ∫01dx∫0x−1f(x,y)dy+∫−10dx∫01−x2f(x,y)dy\displaystyle \int_{0}^{1} d x \int_{0}^{x-1} f(x, y) d y+\int_{-1}^{0} d x \int_{0}^{\sqrt{1-x^{2}}} f(x, y) d y B. ∫01dx∫01−xf(x,y)dy+∫−10dx∫−1−x20f(x,y)dy\displaystyle \int_{0}^{1} d x \int_{0}^{1-x} f(x, y) d y+\int_{-1}^{0} d x \int_{-\sqrt{1-x^{2}}}^{0} f(x, y) d y C. ∫0π2dθ∫01cos⁡θ+sin⁡θf(rcos⁡θ,rsin⁡θ)dr+∫π2πdθ∫01f(rcos⁡θ,rsin⁡θ)dr\displaystyle \int_{0}^{\frac{\pi}{2}} d \theta \int_{0}^{\frac{1}{\cos \theta+\sin \theta}} f(r \cos \theta, r \sin \theta) d r+\int_{\frac{\pi}{2}}^{\pi} d \theta \int_{0}^{1} f(r \cos \theta, r \sin \theta) d r D. ∫0π2dθ∫01cos⁡θ+sin⁡θf(rcos⁡θ,rsin⁡θ)rdr+∫π2πdθ∫01f(rcos⁡θ,rsin⁡θ)rdr\displaystyle \int_{0}^{\frac{\pi}{2}} d \theta \int_{0}^{\frac{1}{\cos \theta+\sin \theta}} f(r \cos \theta, r \sin \theta) r d r+\int_{\frac{\pi}{2}}^{\pi} d \theta \int_{0}^{1} f(r \cos \theta, r \sin \theta) r d r

  6. 【2015-12-4 分】D\displaystyle D第一象限，2xy=1,4xy=1,y=x,y=3x\displaystyle 2xy=1,4xy=1,y=x,y=\sqrt3x围成， A. ∫π4π3dθ∫12sin⁡2θ1sin⁡2θf(rcos⁡θ,rsin⁡θ)rdr\displaystyle \int_{\frac{\pi}{4}}^{\frac{\pi}{3}} d \theta \int_{\frac{1}{2 \sin 2 \theta}}^{\frac{1}{\sin 2 \theta}} f(r \cos \theta, r \sin \theta) r d r B. ∫π4π3dθ∫12sin⁡2θ1sin⁡2θf(rcos⁡θ,rsin⁡θ)rdr\displaystyle \int_{\frac{\pi}{4}}^{\frac{\pi}{3}} d \theta \int_{\frac{1}{\sqrt{2 \sin 2 \theta}}}^{\frac{1}{\sqrt{\sin 2 \theta}}} f(r \cos \theta, r \sin \theta) r d r C. ∫π4π3dθ∫12sin⁡2θ1sin⁡2θf(rcos⁡θ,rsin⁡θ)dr\displaystyle \int_{\frac{\pi}{4}}^{\frac{\pi}{3}} d \theta \int_{\frac{1}{2 \sin 2 \theta}}^{\frac{1}{\sin 2 \theta}} f(r \cos \theta, r \sin \theta) d r D. ∫π4π3dθ∫12sin⁡2θ1sin⁡2θf(rcos⁡θ,rsin⁡θ)dr\displaystyle \int_{\frac{\pi}{4}}^{\frac{\pi}{3}} d \theta \int_{\frac{1}{\sqrt{2 \sin 2 \theta}}}^{\frac{1}{\sqrt{\sin 2 \theta}}} f(r \cos \theta, r \sin \theta) d r

（六）对称性的应用：奇偶性​

  1. 【1991-123-3 分】D\displaystyle D顶点(1,1),(−1,1),(−1,−1)\displaystyle (1,1),(-1,1),(-1,-1)，D1\displaystyle D_1第一象限部分，则∬D(xy+cos⁡xsin⁡y)dx dy=\displaystyle \iint_{D}(x y+\cos x \sin y) d x ~d y= A. 2∬D1cos⁡xsin⁡ydxdy\displaystyle 2 \iint_{D_{1}} \cos x \sin y d x d y B. 2∬D1xydxdy\displaystyle 2 \iint_{D_{1}} x y d x d y C. 4∬D1(xy+cos⁡xsin⁡y)dxdy\displaystyle 4 \iint_{D_{1}}(x y+\cos x \sin y) d x d y D. 0\displaystyle 0

  2. 【2008-4-4 分】f\displaystyle f奇，g\displaystyle g偶，D={(x,y)∣0≤x≤1,−x≤y≤x}\displaystyle D=\{(x,y)|0\le x\le1,-\sqrt x\le y\le\sqrt x\}，则( ) A. ∬Df(y)g(x)dxdy=0\displaystyle \iint_{D} f(y) g(x) d x d y=0 B. ∬Df(x)g(y)dxdy=0\displaystyle \iint_{D} f(x) g(y) d x d y=0 C. ∬D[f(x)+g(y)]dxdy=0\displaystyle \iint_{D}[f(x)+g(y)] d x d y=0 D. ∬D[f(y)+g(x)]dxdy=0\displaystyle \iint_{D}[f(y)+g(x)] d x d y=0

  3. 【2012-2-4 分】D:y=sin⁡x,x=±π2,y=1\displaystyle D:y=\sin x,x=\pm\dfrac\pi2,y=1围成，∬D(xy5−1)dxdy=\displaystyle \iint_{D}(x y^{5}-1) d x d y= A. π\displaystyle \pi    B. 2\displaystyle 2    C. −2\displaystyle -2    D. −π\displaystyle -\pi

  4. 【2018-2-4 分】 ∫−10dx∫−x2−x2(1−xy)dy+∫01dx∫x2−x2(1−xy)dy=\int_{-1}^{0} d x \int_{-x}^{2-x^{2}}(1-x y) d y+\int_{0}^{1} d x \int_{x}^{2-x^{2}}(1-x y) d y= A. 53\displaystyle \dfrac53    B. 56\displaystyle \dfrac56 C. 73\displaystyle \dfrac73    D. 76\displaystyle \dfrac76

（七）对称性的应用：轮换对称性​

  1. 【1994-12-3 分】D:x2+y2≤R2\displaystyle D:x^2+y^2\le R^2，∬D(x2a2+y2b2)dx dy=\displaystyle \iint_{D}(\dfrac{x^{2}}{a^{2}}+\dfrac{y^{2}}{b^{2}}) d x ~d y=

  2. 【1995-1-5 分】f(x)\displaystyle f(x)在[0,1]\displaystyle [0,1]连续，∫01f(x)dx=A\displaystyle \int_{0}^{1} f(x) d x=A，求∫01 dx∫x1f(x)f(y)dy\displaystyle \int_{0}^{1} ~d x \int_{x}^{1} f(x) f(y) d y。

  3. 【2005-2-4 分】D={(x,y)∣x2+y2≤4,x≥0,y≥0}\displaystyle D=\{(x,y)|x^2+y^2\le4,x\ge0,y\ge0\}， ∬Daf(x)+bf(y)f(x)+f(y)dσ=\iint_{D} \dfrac{a \sqrt{f(x)}+b \sqrt{f(y)}}{\sqrt{f(x)}+\sqrt{f(y)}} d \sigma= A. abπ\displaystyle ab\pi    B. ab2π\displaystyle \dfrac{ab}{2}\pi C. (a+b)π\displaystyle (a+b)\pi    D. a+b2π\displaystyle \dfrac{a+b}{2}\pi

大题​

（一）直角坐标法：便于定限的积分次序​

  1. **【1997-4-7 分】**设 D\displaystyle D 是以O(0,0),A(1,2),B(2,1)\displaystyle O(0,0),A(1,2),B(2,1)为顶点的三角形区域，求 ∬Dx dx dy\displaystyle \iint_{D} x ~d x ~d y。

  2. 【2000-4-6 分】

f(x,y)={x2y,1≤x≤2,0≤y≤x,0,其他,f(x, y)= \begin{cases} x^{2} y, &1 \leq x \leq 2,0 \leq y \leq x,\\ 0, &\text{其他}, \end{cases}

D={(x,y)∣x2+y2≥2x}\displaystyle D=\{(x, y) | x^{2}+y^{2} \geq 2 x\}，求 ∬Df(x,y)dx dy\displaystyle \iint_{D} f(x, y) d x ~d y。

  1. **【2008-23-11 分】**计算 ∬Dmax⁡{xy,1}dxdy\displaystyle \iint_{D} \max \{x y, 1\} d x d y，其中 D={(x,y)∣0≤x≤2,0≤y≤2}\displaystyle D=\{(x, y) | 0 \leq x \leq 2,0 \leq y \leq 2\}。

  2. **【2012-3-10 分】**计算∬Dexxydxdy\displaystyle \iint_{D} e^{x} x y d x d y，D\displaystyle D是以y=x, y=1x\displaystyle y=\sqrt{x},\ y=\dfrac1{\sqrt{x}}、y\displaystyle y轴为边界无界区域。

  3. 【2018-3-10 分】D\displaystyle D由y=3(1−x2), y=3x\displaystyle y=\sqrt{3(1-x^{2})},\ y=\sqrt3 x、y\displaystyle y轴围成，计算∬Dx2dxdy\displaystyle \iint_{D} x^{2} d x d y。

  4. 【2024-23-12 分】D\displaystyle D在第一象限，由xy=13,xy=3,y=13x,y=3x\displaystyle xy=\dfrac13,xy=3,y=\dfrac13 x,y=3x围成，计算 I=∬D(1+x−y)dxdyI=\iint_{D}(1+x-y) d x d y

  5. 【2025-3-12 分】D={(x,y)∣y2≤x,x2≤y}\displaystyle D=\{(x, y) | y^{2} \leq x, x^{2} \leq y\}，计算∬D(x−y+1)2dxdy\displaystyle \iint_{D}(x-y+1)^{2} d x d y。

（二）直角坐标法：第一步积分较简单​

  1. 【2002-1-7 分】D={(x,y)∣0≤x≤1,0≤y≤1}\displaystyle D=\{(x, y) | 0 \leq x \leq 1,0 \leq y \leq 1\}，计算∬Demax⁡(x2,y2)dx dy\displaystyle \iint_{D} e^{\max (x^{2}, y^{2})} d x ~d y。

  2. 【2006-3-7 分】D\displaystyle D由y=x,y=1,x=0\displaystyle y=x,y=1,x=0围成，计算∬Dy2−xydxdy\displaystyle \iint_{D} \sqrt{y^{2}-x y} d x d y。

  3. 【2013-23-10 分】D\displaystyle D由x=3y,y=3x,x+y=8\displaystyle x=3y,y=3x,x+y=8围成，计算∬Dx2dxdy\displaystyle \iint_{D} x^{2} d x d y。

  4. 【2017-3-10 分】D\displaystyle D第一象限，边界y=x\displaystyle y=\sqrt{x}、y\displaystyle y轴，计算∬Dy3(1+x2+y4)2dxdy\displaystyle \iint_{D} \dfrac{y^{3}}{(1+x^{2}+y^{4})^{2}} d x d y。

（三）交换积分次序​

  1. 【1988-2-6 分】
∫12 dx∫xxsin⁡πx2y dy+∫24 dx∫x2sin⁡πx2y dy\int_{1}^{2} ~d x \int_{\sqrt{x}}^{x} \sin \dfrac{\pi x}{2 y} ~d y+\int_{2}^{4} ~d x \int_{\sqrt{x}}^{2} \sin \dfrac{\pi x}{2 y} ~d y
  1. 【1992-2-6 分】
∫1412 dy∫12yeyx dx+∫121 dy∫yyeyx dx\int_{\frac{1}{4}}^{\frac{1}{2}} ~d y \int_{\frac{1}{2}}^{\sqrt{y}} e^{\frac{y}{x}} ~d x+\int_{\frac{1}{2}}^{1} ~d y \int_{y}^{\sqrt{y}} e^{\frac{y}{x}} ~d x

（四）极坐标法​

  1. 【1994-4-6 分】D={(x,y)∣x2+y2≤x+y+1}\displaystyle D=\{(x, y) | x^{2}+y^{2} \leq x+y+1\}，计算∬D(x+y)dx dy\displaystyle \iint_{D}(x+y) d x ~d y。

  2. 【1996-2-6 分】D={(x,y)∣0≤y≤x,x2+y2≤2x}\displaystyle D=\{(x, y) | 0 \leq y \leq x,x^{2}+y^{2} \leq 2x\}，计算∬Dx2+y2dxdy\displaystyle \iint_{D} \sqrt{x^{2}+y^{2}} d x d y。

  3. 【1998-34-6 分】D={(x,y)∣x2+y2≤x}\displaystyle D=\{(x,y)|x^2+y^2\le x\}，计算∬Dx dx dy\displaystyle \iint_{D} \sqrt{x} ~d x ~d y。

  4. 【1999-34-7 分】D\displaystyle D由x=−2,y=0,y=2,x=−2y−y2\displaystyle x=-2,y=0,y=2,x=-\sqrt{2y-y^2}围成，计算∬Dy dx dy\displaystyle \iint_{D} y ~d x ~d y。

  5. 【2000-3-6 分】D\displaystyle D由y=−a+a2−x2,y=−x(a>0)\displaystyle y=-a+\sqrt{a^2-x^2},y=-x(a>0)围成，计算 ∬Dx2+y24a2−x2−y2 dσ\iint_{D} \dfrac{\sqrt{x^{2}+y^{2}}}{\sqrt{4 a^{2}-x^{2}-y^{2}}} ~d \sigma

  6. 【2002-4-7 分】D:x2+y2≤y,x≥0\displaystyle D:x^{2}+y^{2} \leq y,x \geq 0， f(x,y)=1−x2−y2−8π∬Df(u,v)du dvf(x, y)=\sqrt{1-x^{2}-y^{2}}-\dfrac{8}{\pi} \iint_{D} f(u, v) d u ~d v 求f(x,y)\displaystyle f(x,y)。

  7. 【2003-34-8 分】D={(x,y)∣x2+y2≤π}\displaystyle D=\{(x, y) | x^{2}+y^{2} \leq \pi\}， I=∬De−(x2+y2−π)sin⁡(x2+y2)dx dyI=\iint_{D} e^{-(x^{2}+y^{2}-\pi)} \sin (x^{2}+y^{2}) d x ~d y

  8. 【2005-234-9 分】D={(x,y)∣0≤x≤1,0≤y≤1}\displaystyle D=\{(x, y) | 0 \leq x \leq 1,0 \leq y \leq 1\}，计算∬D∣x2+y2−1∣dσ\displaystyle \iint_{D}|x^{2}+y^{2}-1| d \sigma。

  9. 【2005-1-11 分】D={(x,y)∣x2+y2≤2,x≥0,y≥0}\displaystyle D=\{(x,y)|x^2+y^2\le\sqrt2,x\ge0,y\ge0\}，[t]\displaystyle [t]为取整，计算∬Dxy[1+x2+y2]dxdy\displaystyle \iint_{D} x y[1+x^{2}+y^{2}] d x d y。

  10. 【2009-23-10 分】D={(x,y)∣(x−1)2+(y−1)2≤2,y≥x}\displaystyle D=\{(x,y)|(x-1)^2+(y-1)^2\le2,y\ge x\}，计算∬D(x−y)dxdy\displaystyle \iint_{D}(x-y) d x d y。

  11. 【2012-2-10 分】D\displaystyle D由r=1+cos⁡θ(0≤θ≤π)\displaystyle r=1+\cos\theta(0\le\theta\le\pi)与极轴围成，计算∬Dxydσ\displaystyle \iint_{D} x y d \sigma。

  12. 【2016-1-10 分】D={(r,θ)∣2≤r≤2(1+cos⁡θ),−π2≤θ≤π2}\displaystyle D=\{(r,\theta)|2\le r\le2(1+\cos\theta),-\dfrac\pi2\le\theta\le\dfrac\pi2\}，计算∬Dxdxdy\displaystyle \iint_{D} x d x d y。

  13. 【2019-2-10 分】D={(x,y)∣∣x∣≤y,(x2+y2)3≤y4}\displaystyle D=\{(x,y)||x|\le y,(x^2+y^2)^3\le y^4\}，计算∬Dx+yx2+y2 dx dy\displaystyle \iint_{D} \dfrac{x+y}{\sqrt{x^{2}+y^{2}}} ~d x ~d y。

  14. 【2020-2-10 分】D\displaystyle D由x=1,x=2,y=x,x\displaystyle x=1,x=2,y=x,x轴围成，I=∬Dx2+y2xdσ\displaystyle I=\iint_{D} \dfrac{\sqrt{x^{2}+y^{2}}}{x} d \sigma。

  15. 【2021-2-12 分】D\displaystyle D:(x2+y2)2=x2−y2(x≥0,y≥0)\displaystyle (x^2+y^2)^2=x^2-y^2(x\ge0,y\ge0)与x\displaystyle x轴围成，∬Dxydxdy\displaystyle \iint_{D} x y d x d y。

  16. 【2021-3-12 分】D:x2+y2=1,y=x,x\displaystyle D:x^2+y^2=1,y=x,x轴第一象限，∬De(x+y)2⋅(x2−y2)dxdy\displaystyle \iint_{D} e^{(x+y)^{2}} \cdot(x^{2}-y^{2}) d x d y。

  17. 【2022-123-12 分】D={(x,y)∣−2+y≤x≤4−y2,0≤y≤2}\displaystyle D=\{(x,y)|-2+y\le x\le\sqrt{4-y^2},0\le y\le2\}， I=∬D(x−y)2x2+y2dxdyI=\iint_{D} \dfrac{(x-y)^{2}}{x^{2}+y^{2}} d x d y

  18. 【2023-2-12 分】D\displaystyle D第一象限，x2+y2−xy=1,x2+y2−xy=2,y=3x,y=0\displaystyle x^2+y^2-xy=1,x^2+y^2-xy=2,y=\sqrt3 x,y=0围成，∬D13x2+y2dxdy\displaystyle \iint_{D} \dfrac{1}{3 x^{2}+y^{2}} d x d y。

  19. 【2023-3-12 分】D={(x,y)∣(x−1)2+y2≤1}\displaystyle D=\{(x,y)|(x-1)^2+y^2\le1\}，∬D∣x2+y2−1∣dx\displaystyle \iint_{D}|\sqrt{x^{2}+y^{2}}-1| d x。

  20. 【2024-1-12 分】D={(x,y)∣1−y2≤x≤1,−1≤y≤1}\displaystyle D=\{(x,y)|\sqrt{1-y^2}\le x\le1,-1\le y\le1\}，I=∬Dxx2+y2dxdy\displaystyle I=\iint_{D} \dfrac{x}{\sqrt{x^{2}+y^{2}}} d x d y。

  21. 【2025-2-12 分】D={(x,y)∣x2+y2≤4x,x2+y2≤4y}\displaystyle D=\{(x,y)|x^2+y^2\le4x,x^2+y^2\le4y\}，∬D(x−y)dxdy\displaystyle \iint_{D}(x-y)dxdy。

（五）坐标系的转换​

  1. 【2010-2-10 分】D={(r,θ)∣0≤r≤sec⁡θ,0≤θ≤π4}\displaystyle D=\{(r,\theta)|0\le r\le\sec\theta,0\le\theta\le\dfrac\pi4\}， I=∬Dr2sin⁡θ1−r2cos⁡2θdrdθI=\iint_{D} r^{2} \sin \theta \sqrt{1-r^{2} \cos 2 \theta} d r d \theta

（六）对称性的应用：奇偶性​

  1. 【2001-34-6 分】D:y=x,y=−1,x=1\displaystyle D:y=x,y=-1,x=1围成，计算∬Dy[1+xe12(x2+y2)]dxdy\displaystyle \iint_{D} y\left[1+x e^{\dfrac12(x^2+y^2)}\right]dxdy。

  2. 【2004-34-8 分】D\displaystyle D由x2+y2=4,(x+1)2+y2=1\displaystyle x^2+y^2=4,(x+1)^2+y^2=1围成，∬D(x2+y2+y)dσ\displaystyle \iint_{D}(\sqrt{x^{2}+y^{2}}+y) d \sigma。

  3. 【2006-12-10 分】D={(x,y)∣x2+y2≤1,x≥0}\displaystyle D=\{(x,y)|x^2+y^2\le1,x\ge0\}，I=∬D1+xy1+x2+y2dxdy\displaystyle I=\iint_{D} \dfrac{1+x y}{1+x^{2}+y^{2}} d x d y。

  4. 【2007-234-11 分】

f(x,y)={x2,∣x∣+∣y∣≤1,1x2+y2,1<∣x∣+∣y∣≤2,f(x,y)= \begin{cases} x^{2},&|x|+|y|\le1,\\ \dfrac{1}{\sqrt{x^{2}+y^{2}}},&1<|x|+|y|\le2, \end{cases}

D={(x,y)∣∣x∣+∣y∣≤2}\displaystyle D=\{(x,y)||x|+|y|\le2\}，求∬Df(x,y)dσ\displaystyle \iint_{D} f(x,y) d \sigma。

  1. 【2010-3-10 分】D\displaystyle D由x=1+y2,x±2y=0\displaystyle x=\sqrt{1+y^2},x\pm\sqrt2 y=0围成，∬D(x+y)3dxdy\displaystyle \iint_{D}(x+y)^3dxdy。

  2. 【2015-23-10 分】D={(x,y)∣x2+y2≤2,y≥x2}\displaystyle D=\{(x,y)|x^2+y^2\le2,y\ge x^2\}，∬Dx(x+y)dxdy\displaystyle \iint_{D} x(x+y) d x d y。

  3. 【2016-2-10 分】D:y=1,y=x,y=−x\displaystyle D:y=1,y=x,y=-x围成，∬Dx2−xy−y2x2+y2dxdy\displaystyle \iint_{D} \dfrac{x^{2}-x y-y^{2}}{x^{2}+y^{2}} d x d y。

  4. 【2017-2-10 分】D={(x,y)∣x2+y2≤2y}\displaystyle D=\{(x,y)|x^2+y^2\le2y\}，∬D(x+1)2dxdy\displaystyle \iint_{D}(x+1)^{2} d x d y。

（七）对称性的应用：轮换对称性​

  1. 【2014-23-10 分】D={(x,y)∣1≤x2+y2≤4,x≥0,y≥0}\displaystyle D=\{(x,y)|1\le x^2+y^2\le4,x\ge0,y\ge0\}， ∬Dxsin⁡(πx2+y2)x+ydxdy\iint_{D} \dfrac{x \sin \left(\pi \sqrt{x^{2}+y^{2}}\right)}{x+y} d x d y