其他情况下的体积的计算都不是特别的麻烦。只需计算 椭圆与直线相交构成的面积的表达即可。
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至于表面积的计算，个人感觉理论意义不是很大，因为最简单的椭球的表面积以及 椭圆的周长 都不能用初等函数表达。更何况计算截体的表面积了
对于7#计算结果：
\(V_1=\frac{bc}{8a}(\pi a^2+4(2x-a)\sqrt{x(a-x)}-2a^2\arcsin{\frac{a-2x}{a}})\)
\(V_2=\frac{\pi cdx^2(3a-2x)}{6a^2}\)
当\(x=a\)时
\(V_1=\frac{\pi abc}{8}\)
\(V_2=\frac{\pi acd}{6}\)
然而当\(x=a\)时
\(V_1=\pi\frac{a}{2}\frac{c}{2}b=\frac{\pi abc}{4}\) 椭圆柱的体积公式
\(V_2= \frac{4\pi}{3}\frac{a}{2}\frac{c}{2}d=\frac{\pi acd}{3}\)椭球的体积公式
为了将问题更数学表述：
可以得到左边封头半椭球方程：
\(\frac{(x+\frac{a}{2})^2}{(\frac{a}{2})^2}+\frac{(y+\frac{b}{2})^2}{d^2}+\frac{z^2}{(\frac{c}{2})^2}=1\) ..................................(1)
椭圆柱方程：
\(\frac{x^2}{(\frac{a}{2})^2}+\frac{z^2}{(\frac{c}{2})^2}=1\) .................................................................................................(2)
\(-\frac{b}{2}\leq y \leq\frac{b}{2}\)......................................................................................................................................(3)
可以得到右边封头半椭球方程：
\(\frac{(x-\frac{a}{2})^2}{(\frac{a}{2})^2}+\frac{(y-\frac{b}{2})^2}{d^2}+\frac{z^2}{(\frac{c}{2})^2}=1\) .....................................(4)
A类: 截面方程: \(x=h\)(为了不引起混淆设为h)................................................................................................................(5)
B类: 截面方程: \(\frac{x-h}{y}=\tan(\theta)\)...................................................................................................................(6)
C类: 截面方程: \(\frac{x-h}{z}=\tan(\theta)\)...................................................................................................................(6)
为了数学结果的简洁,我采纳sheng_jianguo 的建议,重新描述如下:
可以得到左边封头半椭球方程：
\(\frac{(x+a)^2}{a^2}+\frac{y^2}{d^2}+\frac{z^2}{c^2}=1\) ..................................(1)
椭圆柱方程：
\(\frac{x^2}{a^2}+\frac{z^2}{c^2}=1\) .................................................................................................(2)
\(-b\leq y \leqb\)......................................................................................................................................(3)
可以得到右边封头半椭球方程：
\(\frac{(x-a)^2}{a^2}+\frac{y^2}{d^2}+\frac{z^2}{c^2}=1\) .....................................(4)
A类: 截面方程: \(x=h\)...........................................................................................................................(5)
B类: 截面方程: \(\frac{x-h}{y}=\tan(\beta)\)...................................................................................................................(6)
C类: 截面方程: \(\frac{x-h}{z}=\tan(\theta)\)...................................................................................................................(6)
先求椭圆柱内液体的体积:
A.水平放置时
当 \(0 \leqslant h \lt a\)时
\(V_{10}=\frac{2bc}{a}((h-a)\sqrt{h(2a-h)}+a^2\arctan{(\frac{\sqrt{h(2a-h)}}{a-h})})\)
当 \(a\leqslant h \lt 2a\)时
\(V_{10}=\frac{2bc}{a}((a-h)\sqrt{h(2a-h)}+a^2(\pi-\arctan{(\frac{\sqrt{h(2a-h)}}{h-a})}))\)
B.当左右倾斜放置时:
记
\(k_1=\tan(\theta)\)
\(h_1=h+bk_1,h_2=h-bk_1\)
\(S_{10}=\frac{c}{a}((h-a)\sqrt{h(2a-h)}+a^2\arctan{(\frac{\sqrt{h(2a-h)}}{a-h})})\)
\(S_{11}=\frac{c}{a}((a-h)\sqrt{h(2a-h)}+a^2(\pi-\arctan{(\frac{\sqrt{h(2a-h)}}{h-a})}))\)
1.当 \(0 \leqslant k_1\lt \frac{h-a}{b}\)时
\(V(h)=b(S_{10}(h_1)+S_{10}(h_2))\)
2.当 \( \frac{h-a}{b} \leqslant k_1\leqslant \frac{h+a}{b}\)时
\(V(h)=b(S_{11}(h_1)+S_{10}(h_2))\)
3.当 \( \frac{h+a}{b} \leqslant k_1 \)时
\(V(h)=b(S_{11}(h_1)+S_{11}(h_2))\)
C.前后倾斜放置时
记:
\(k=\tan{\beta}\)
\(F_1(x)=ax-\frac{ax\sqrt{c^2-x^2}}{2c}-\frac{ac}{2}\arctan{(\frac{x}{\sqrt{c^2-x^2}})}\)
\(F_2(x)=2ab(\frac{x\sqrt{c^2-x^2}}{c}+c\arctan{(\frac{x}{\sqrt{c^2-x^2}})})\)
\(V_0(h)=\frac{(y_1+y_2)(x_1-x_2)}{2}=\frac{2a^2c(ha+c^2k^2)\sqrt{c^2k^2+2ah-h^2}}{(a^2+c^2k^2)^2}\)
\(y_1=\frac{a(ha+c^2k^2+ck\sqrt{2ha+c^2k^2-h^2})}{a^2+c^2k^2}\)
\(y_2=\frac{a(ha+c^2k^2-ck\sqrt{2ha+c^2k^2-h^2})}{a^2+c^2k^2}\)
\(x_1=\frac{c(kc(a-h)+a\sqrt{2ha+c^2k^2-h^2})}{a^2+c^2k^2}\)
\(x_2=\frac{c(kc(a-h)-a\sqrt{2ha+c^2k^2-h^2})}{a^2+c^2k^2}\)
1.当\(0 \leqslant h \lt a\),且\(0 \leqslant k \lt\frac{a-h}{c}\) 时
\(V_{11}(h)=2b(V_0(h)-(F_1(x_1)-F_2(x_2)))\)
2.当\(0 \leqslant h \lt a\),且\(\frac{a-h}{c} \leqslant k \)时
\(V_{12}(h)=2b(V_0(h)-(F_1(x_1)-F_1(x_2))+(F_2(c)-F_2(x_1)))\)
3.当\(a \leqslanth \lt 2a\),且 \( 0 \lt k\leqslant \frac{h-a}{c}\) 时
\(V_{13}(h)=2b(V_0(h)-(F_1(x_1)-F_1(x_2))+(F_2(c)-F_2(x_1)))\)
4.当\(a \leqslanth \lt 2a\),且 \(\frac{a-h}{c} \leqslant k\)时
\(V_{14}(h)=2b(V_0(h)-(F_1(x_1)-F_1(x_2))+(F_2(c)-F_2(x_1))+(F_2(x_2)-F_2(-c)))\)
下面是网络上截取的有关水平放置 时的容积计算公式
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http://pan.baidu.com/share/link?shareid=2359613377&uk=354741210
下载的计算软件,但没有找到注册码,谁有兴趣提供注册码?
有关椭球冠的体积与表面积公式很复杂: