25.1 第一型曲面积分

例$25.1.1$

设 $S$ 为 $z=\sqrt{x^2+y^2}$ 被 $x^2+y^2=2ax$ 割下的部分，求

I = iint_(S) (x² y² + y² z² + z² x²) d S.$$I={\iint }_S\left(x^2{y}^2+y^2{z}^2+z^2{x}^2\right)\mathrm{d}S.$$

解$1$

见图 $25.1$，在直角坐标系中计算

c zₓ = x/z, z_(y) = y/z, sqrt1 + zₓ² + z_(y)² = √(2), I = iintₓ² + y² ≤slant 2 a x [ x² y² + (x² + y²)² ] √(2) d x d y.$$\begin{array}{c}{z}_x=\frac{x}{z},z_y=\frac{y}{z},\\ \sqrt{1+z_x^2+z_y^2}=\sqrt{2},\\ I={\iint }_{x^2+y^2⩽2ax}[x^2{y}^2+(x^2+y^2{)}^2]\sqrt{2}\mathrm{d}x\mathrm{d}y.\end{array}$$

图25.125.1

用极坐标变换求上述二重积分，则

r l I & = √(2) ∫_(- π / 2)^(π / 2) d θ ∫₀² a cos θ (r⁴ cos² θ sin² θ + r⁴) r d r & = √(2) ∫_(- π / 2)^(π / 2) (cos² θ sin² θ + 1) · (1/6 r⁶ |₀² a cos θ) d θ & = (√(2))/(6) (2 a)⁶ ∫_(- π / 2)^(π / 2) cos⁶ θ (cos² θ sin² θ + 1) d θ = (29)/(8) √(2) π a⁶.$$\begin{array}{rl}I& =\sqrt{2}{\int }_{-\pi /2}^{\pi /2}\mathrm{d}\theta {\int }_0^{2a\cos \theta }(r^4{\cos }^2\theta {\sin }^2\theta +r^4)r\mathrm{d}r\\ & =\sqrt{2}{\int }_{-\pi /2}^{\pi /2}({\cos }^2\theta {\sin }^2\theta +1)\cdot \left(\frac{1}{6}{r}^6{|}_0^{2a\cos \theta }\right)\mathrm{d}\theta \\ & =\frac{\sqrt{2}}{6}(2a{)}^6{\int }_{-\pi /2}^{\pi /2}{\cos }^6\theta ({\cos }^2\theta {\sin }^2\theta +1)\mathrm{d}\theta =\frac{29}{8}\sqrt{2}\pi a^6.\end{array}$$

解$2$

用参数式计算。 $z=\sqrt{x^2+y^2}$ 在球坐标系中的方程为 $\phi =\frac{\pi }{4}$，因此 $S$ 的参数方程为

x = (1)/(√(2)) r cos θ , y = (1)/(√(2)) r sin θ , z = (1)/(√(2)) r, (r, θ) ∈ D.$$x=\frac{1}{\sqrt{2}}r\cos \theta ,y=\frac{1}{\sqrt{2}}r\sin \theta ,z=\frac{1}{\sqrt{2}}r,(r,\theta )\in D.$$

又 $S$ 的边界线

z = sqrtx² + y², x² + y² = 2 a x$$z=\sqrt{x^2+y^2},x^2+y^2=2ax$$

的球坐标表示为

φ = (π)/(4), r² sin² φ = 2 a r sin φ cos θ .$$\phi =\frac{\pi }{4},r^2{\sin }^2\phi =2ar\sin \phi \cos \theta .$$

于是

D = {(r, θ) | - (π)/(2) ≤slant θ ≤slant (π)/(2), 0 ≤slant r ≤slant 2 √(2) a cos θ }.$$D=\left\{(r,\theta )\mid -\frac{\pi }{2}⩽\theta ⩽\frac{\pi }{2},0⩽r⩽2\sqrt{2}a\cos \theta \right\}.$$

计算得

E = fracr²2, F = 0, G = 1,$$E=\frac{r^2}{2},F=0,G=1,$$

最后得到

l I = ∫_(- π / 2)^(π / 2) d θ ∫₀² √(2) a cos θ (1/4 r⁴ cos² θ sin² θ + 1/4 r⁴) (r)/(√(2)) d r = (29)/(8) √(2) π a⁶.$$\begin{array}{l}I={\int }_{-\pi /2}^{\pi /2}\mathrm{d}\theta {\int }_0^{2\sqrt{2}a\cos \theta }\left(\frac{1}{4}{r}^4{\cos }^2\theta {\sin }^2\theta +\frac{1}{4}{r}^4\right)\frac{r}{\sqrt{2}}\mathrm{d}r\\ =\frac{29}{8}\sqrt{2}\pi a^6.\end{array}$$

例$25.1.2$

求上半球面 $z=\sqrt{a^2-x^2-y^2}$ 被 $x^2+y^2=ax$ 截取部分的面积与质心坐标，其中 $a>0$.

解

见图 $22.11$，这就是 $Viviani$ 体的上表面。 由于

zₓ = frac- xsqrta² - x² - y², z_(y) = frac- ysqrta² - x² - y²,$$z_x=\frac{-x}{\sqrt{a^2-x^2-y^2}},z_y=\frac{-y}{\sqrt{a^2-x^2-y^2}},$$

故所求面积

r l S & = iintₓ² + y² ≤slant a x fracasqrta² - x² - y² d x d y & = ∫_(- π / 2)^(π / 2) d θ ∫₀^(a cos θ) fraca rsqrta² - r² d r & = (π - 2) a².$$\begin{array}{rl}S& ={\iint }_{x^2+y^2⩽ax}\frac{a}{\sqrt{a^2-x^2-y^2}}\mathrm{d}x\mathrm{d}y\\ & ={\int }_{-\pi /2}^{\pi /2}\mathrm{d}\theta {\int }_0^{a\cos \theta }\frac{ar}{\sqrt{a^2-r^2}}\mathrm{d}r\\ & =(\pi -2)a^2.\end{array}$$

下面求质心坐标，由对称性知 $y_0=0$，且

x₀ & = frac1(π - 2)a²iint limits_(S)xdS & = frac2(π - 2)a²iint limits_(substackx² + y²≤slant ax y≥slant 0)fracaxsqrta² - x² - y²dxdy & = (2)/((π - 2)a)∫₀^(π /2)cos θ dθ ∫₀^(acos θ)fracrsqrta² - r² rdr (r = acos t) & = (2)/((π - 2)a)∫₀^(π /2)cos θ dθ ∫_(π /2)^(θ)fraca²cos²tasin t (-asin t)dt & = (a)/(π - 2)∫₀^(π /2)cos θ ((π)/(2) -θ -sin θ cos θ)dθ & = (2a)/(3(π - 2)); z₀ & = frac1(π - 2)a²iint limits_(S)zdS & = frac1(π - 2)a²iint limitsₓ² + y²≤slant axfracasqrta² - x² - y²sqrta² - x² - y²dxdy & = (π a)/(4(π - 2)).$$\begin{array}{rl}{x}_0& =\frac{1}{(\pi -2)a^2}\underset{S}{\iint }x\mathrm{d}S\\ & =\frac{2}{(\pi -2)a^2}\underset{\begin{array}{c}{x}^2+y^2⩽ax\\ y⩾0\end{array}}{\iint }\frac{ax}{\sqrt{a^2-x^2-y^2}}\mathrm{d}x\mathrm{d}y\\ & =\frac{2}{(\pi -2)a}{\int }_0^{\pi /2}\cos \theta \mathrm{d}\theta {\int }_0^{a\cos \theta }\frac{r}{\sqrt{a^2-r^2}}r\mathrm{d}r(r=a\cos t)\\ & =\frac{2}{(\pi -2)a}{\int }_0^{\pi /2}\cos \theta \mathrm{d}\theta {\int }_{\pi /2}^{\theta }\frac{a^2{\cos }^{2}t}{a\sin t}(-a\sin t)\mathrm{d}t\\ & =\frac{a}{\pi -2}{\int }_0^{\pi /2}\cos \theta \left(\frac{\pi }{2}-\theta -\sin \theta \cos \theta \right)\mathrm{d}\theta \\ & =\frac{2a}{3(\pi -2)};\\ z_0& =\frac{1}{(\pi -2)a^2}\underset{S}{\iint }z\mathrm{d}S\\ & =\frac{1}{(\pi -2)a^2}\underset{x^2+y^2⩽ax}{\iint }\frac{a}{\sqrt{a^2-x^2-y^2}}\sqrt{a^2-x^2-y^2}\mathrm{d}x\mathrm{d}y\\ & =\frac{\pi a}{4(\pi -2)}.\end{array}$$