# A Novel Breast-Volume Self-Measurement Method with Improved Convenient and Accuracy
## 3 Results
### Breast-Volume Mathematical Model
#### 3.1.2 Breast-Volume Calculation Formula  
To calculate the lower breast volume, we need to calculate the volume of the complete ellipsoid. The left region of the right breast is taken as an example. Let \(f(x,y,z)=\frac{x^{2}}{H_{Ai-Ai^{2}}}+\frac{y^{2}}{W_{Ai^{\prime}-Bi}}+\frac{z ^{2}}{L_{Di-Ai}}\). According to the features of triple integrals, the ellipsoid volume (\(V\)) can be expressed as:  
\[ \begin{array}{c}V=\underset{\Omega}{\iint}f(x,\ y,\ z)dxdydz\\ \Omega:\frac{x^{2}}{H_{Ai-Ai^{2}}}+\frac{y^{2}}{W_{Ai^{\prime}-Bi}}+\frac{z^{ 2}}{L_{Di-Ai}}\leq 1\end{array}  \tag{31}\]  
Let \(\frac{x}{H_{Ai-Ai^{2}}}=X,\frac{y}{W_{Ai^{\prime}-Bi}}=Y,\frac{z}{L_{Di-Ai}}=Z\), then Equation (31) can be rewritten as:  
\[ \begin{array}{ll}V&=\underset{\Omega}{\iint}f\left(H_{Ai-Ai^{\prime}}X,\ W_{Ai^{ \prime}-Bi}Y,\ L_{Di-Ai}Z\right)H_{Ai-Ai^{\prime}}dX\cdot W_{Ai^{\prime}-Bi} \cdot dY\cdot L_{Di-Ai}\cdot dZ\\ &=H_{Ai-Ai^{\prime}}W_{Ai^{\prime}-Bi}L_{Di-Ai}[\iint\ \left(X^{2}+Y^{2}+Z^{2} \right)dXdYdZ\end{array}  \tag{32}\]Then, the spherical coordinate system is used. Let  
\[\left\{ \begin{array}{l}X=rsin\theta sint\\ Y=rsin\theta cost\\ Z=rcos\theta\end{array} \right. \tag{33}\]  
where \(r\in[0,1],t\in[0,2\pi],\theta\in[0,\pi]\). According to Equation (33),  
\[dXdYdZ=r^{2}sin\theta drd\theta dt \tag{34}\]  
By substituting Equation (34) into Equation (32),  
\[ \begin{array}{ll}V&=H_{Ai-A^{\prime}}\cdot W_{[PERSON]^{\prime}-B1}\cdot L_{[PERSON]} \iint\limits_{\Omega}\big{(}X^{2}+Y^{2}+Z^{2}\big{)}dXdYdZ\\ &=\iint\limits_{\Omega}r^{2}sin\theta drd\theta dt=H_{[PERSON]^{\prime}}\cdot W_{[PERSON] ^{\prime}-Bi}\cdot L_{[PERSON]}\int_{0}^{2\pi}dt\int_{0}^{\pi}sin\theta d\theta \int_{0}^{1}r^{2}dt\\ &=\frac{4\pi}{3}H_{[PERSON]^{\prime}}\cdot W_{[PERSON]^{\prime}-Bi}\cdot L_{[PERSON]} \end{array}  \tag{35}\]  
Hence, the complete ellipsoid volume of the left region of the right breast can be determined by \(V=\frac{4\pi}{3}H_{[PERSON]^{\prime}}\cdot W_{[PERSON]^{\prime}-Bi}\cdot L_{Di-Ai}\). Likewise, the complete ellipsoid volume of the right region of the right breast can be determined by \(V=\frac{4\pi}{3}H_{[PERSON]^{\prime}}\cdot W_{[PERSON]^{\prime}-Ci}\cdot L_{Di-Ai}\); the complete ellipsoid volume of the right region of the left breast can be determined by \(V=\frac{4\pi}{3}H_{[PERSON]^{\prime}}\cdot W_{[PERSON]^{\prime}-Bi}\cdot L_{[PERSON]}\); and the complete ellipsoid volume of the left region of the left breast can be determined by \(V=\frac{4\pi}{3}H_{[PERSON]^{\prime}}\cdot W_{[PERSON]^{\prime}-Ci}\cdot L_{[PERSON]}\). Based on the above derived equations of the complete ellipsoid volume, we can obtain an equation for the practical volume of the lower breast: left /right breast volume  
\[ \begin{array}{ll}V=\frac{1}{8}\cdot\frac{4\pi}{3}H_{[PERSON]^{\prime}}\cdot W_ {[PERSON]^{\prime}-Bi}\cdot L_{[PERSON]}+\frac{1}{8}\cdot\frac{4\pi}{3}H_{[PERSON]^{ \prime}}\cdot W_{[PERSON]^{\prime}-Ci}\cdot L_{[PERSON]}\\ =\frac{\pi}{6}H_{[PERSON]^{\prime}}\cdot L_{[PERSON]}\big{(}W_{[PERSON]^{\prime}-Bi}+W_{ [PERSON]^{\prime}-Ci}\big{)}\end{array}  \tag{36}\]  
The volume of the upper breast is the partial elliptical-cone volume. Hence, to obtain the volume of the upper breast, we need to first derive the equation of the complete elliptical-cone volume. It is assumed that the semi-major axis of the elliptical-cone base is \(a\), the semi-minor axis of the elliptical-cone base is \(b\), and the height of the elliptical cone is \(h\). According to the triangle similarity theorems, we have  
\[\frac{A}{a}=\frac{h-z}{h} \tag{37}\]  
According to Equation (37),  
\[A=\frac{(h-z)}{h}a \tag{38}\]  
According to the triangle similarity theorems,  
\[\frac{B}{b}=\frac{h-z}{h} \tag{39}\]  
According to Equation (39),  
\[B=\frac{(h-z)}{h}b \tag{40}\]  
The complete elliptical-cone volume can thus be calculated as  
\[ \begin{array}{ll}V=\int_{0}^{h}\pi ABdz=\int_{0}^{h}\pi\cdot\frac{(h-z)}{h}a \cdot\frac{(h-z)}{h}bdz\\ =\frac{\pi ab}{b^{2}}\int_{0}^{h}{(h-z)}^{2}dz=\frac{1}{3}\pi abh\end{array}