问题

I am trying to analytically calculate the integral of a function using this code:

from sympy import sqrt, symbols, integrate, pi
x=symbols('x',real=True, positive=True)
rho, r=symbols('rho r', real=True, positive=True)
f=sqrt(rho**2-(x-r)**2)
integrate(f,(x,r,r+rho)).simplify()

SymPy tells me the result is the one in the picture:

however I want to evaluate only the second one. How can I do that?

from sympy import sqrt, symbols, integrate, pi
x=symbols('x',real=True, positive=True)
rho, r=symbols('rho r', real=True, positive=True)
f=sqrt(rho**2-(x-r)**2)
integrate(f,(x,r,r+rho)).simplify()

答案1

得分: 1

根据这个评论，你可以将不等式作为假设传递给refine：

print(refine(integrate(f, (x, r, r + rho)).simplify(), (r - x)**2/rho**2 > 1))
# Integral(I*sqrt(r**2 - 2*r*x - rho**2 + x**2), (x, r, r + rho))

(我不确定为什么这个没有文档记录)

print(refine(integrate(f,(x,r,r+rho)).simplify(), (r - x)**2/rho**2 > 1))
# Integral(I*sqrt(r**2 - 2*r*x - rho**2 + x**2), (x, r, r + rho))

答案2

得分: 1

以下是SymPy给出的内容：

In [10]: i = integrate(f, (x, r, r + rho)).simplify()
In [11]: i
Out[11]: 
r + ρ                                                                                                                                                              
  ⌠                                                                                                                                                                
  ⎮   ⎧        2        2                         2                             2                            4                          2                  2       
  ⎮   ⎪     ⅈ⋅ρ ⋅(r - x)                       ⅈ⋅ρ                           ⅈ⋅ρ                    ⅈ⋅(r - x)                3⋅ⅈ⋅(r - x)            (r - x)        
  ⎮   ⎪─────────────────────── - ─────────────────────────────── - ─────────────────────── - ─────────────────────── + ───────────────────────  for ──────── > 1   
  ⎮   ⎪                    3/2       ____________   ____________        __________________                       3/2        __________________          2          
  ⎮   ⎪  ⎛   2           2⎞      2⋅╲╱ -r - ρ + x ⋋⋓⋂⋈⋐⋏⋍⋚⋏⋈⋍⋌⋋⋚⋃⋔⋈⋋⋚⋌⋃⋏⋃⋍⋎⋂⋋⋚⋌⋍⋏⋃⋍⋌⋏⋈⋏⋐⋏⋋⋚⋍⋂⋏⋈⋋⋚⋏⋎⋏⋏⋎⋋⋚⋏⋍⋌⋏⋏⋃⋎⋋⋎⋋⋐⋍⋚⋏⋌⋏⋐⋍⋍⋃⋋⋎⋂⋚⋐⋍⋎⋋⋏⋎⋏⋏⋍⋋⋚⋌⋍⋏⋎⋏⋏⋎⋋⋚⋏⋍⋏⋍⋎⋋⋚⋏⋃⋍⋃⋏⋏⋐⋏⋍⋍⋃⋏⋎⋋⋚⋏⋃⋎⋋⋏⋎⋏⋏⋍⋋⋚⋌⋍⋏⋎⋏⋏⋎⋋⋚⋏⋍⋏⋍⋎⋋⋚⋏⋃⋍⋃⋏⋏⋐⋏⋍⋍⋃⋏⋎⋋⋚⋏⋃⋎⋋⋏⋎⋏⋏⋍⋋⋚⋌⋍⋏⋎⋏⋏⋎⋋⋚⋏⋍⋏⋍⋎⋋⋚⋏⋃⋍⋃⋏⋏⋐⋏⋍⋍⋃⋏⋎⋋⋚⋏⋃⋎⋋⋏⋎⋏⋏⋍⋋⋚⋌⋍⋏⋎⋏⋏⋎⋋⋚⋏⋍⋏⋍⋎⋋⋚⋏⋃⋍⋃⋏⋏⋐⋏⋍⋍⋃⋏⋎⋋⋚⋏⋃⋎⋋⋏⋎⋏⋏⋍⋋⋚⋌⋍⋏⋎⋏⋏⋎⋋⋚⋏⋍⋏⋍⋎⋋⋚⋏⋃⋍⋃⋏⋏⋐⋏⋍⋍⋃⋏⋎⋋⋚⋏⋃⋎⋋⋏⋎⋏⋏⋍⋋⋚⋌⋍⋏⋎⋏⋏⋎⋋⋚⋏⋍⋏⋍⋎⋋⋚⋏⋃⋍⋃⋏⋏⋐⋏⋍⋍⋃⋏⋎⋋⋚⋏⋃⋎⋋⋏⋎⋏⋏⋍⋋⋚⋌⋍⋏⋎⋏⋏⋎⋋⋚⋏⋍⋏⋍⋎⋋⋚⋏⋃⋍⋃⋏⋏⋐⋏⋍⋍⋃⋏⋎⋋⋚⋏⋃⋎⋋⋏⋎⋏⋏⋍⋋⋚⋌⋍⋏⋎⋏⋏⋎⋋⋚⋏⋍⋏⋍⋎⋋⋚⋏⋃⋍⋃⋏⋏⋐⋏⋍⋍⋃⋏⋎⋋⋚⋏⋃
<details>
<summary>英文:</summary>
Here is what SymPy gives:

In [10]: i = integrate(f,(x,r,r+rho)).simplify()

In [11]: i
Out[11]:
r + ρ
⌠
⎮ ⎧ 2 2 2 2 4 2 2
⎮ ⎪ ⅈ⋅ρ ⋅(r - x) ⅈ⋅ρ ⅈ⋅ρ ⅈ⋅(r - x) 3⋅ⅈ⋅(r - x) (r - x)
⎮ ⎪─────────────────────── - ─────────────────────────────── - ─────────────────────── - ─────────────────────── + ─────────────────────── for ──────── > 1
⎮ ⎪ 3/2 ____________ ____________ __________________ 3/2 __________________ 2
⎮ ⎪ ⎛ 2 2⎞ 2⋅╲╱ -r - ρ + x ⋅╲╱ -r + ρ + x ╱ 2 2 ⎛ 2 2⎞ ╱ 2 2 ρ
⎮ ⎨2⋅⎝- ρ + (-r + x) ⎠ 2⋅╲╱ - ρ + (-r + x) 2⋅⎝- ρ + (-r + x) ⎠ 2⋅╲╱ - ρ + (-r + x) dx
⎮ ⎪
⎮ ⎪ _______________
⎮ ⎪ ╱ 2 2
⎮ ⎪ ╲╱ ρ - (r - x) otherwise
⎮ ⎩
⌡
r

You say that you want the second case. In general we can extract cases from a Piecewise using .args but we need to bring the Piecewise out to top lvel using `piecewise_fold` first:

In [12]: piecewise_fold(i)
Out[12]:
⎧r + ρ
⎪ ⌠
⎪ ⎮ ⎛ 2 2 2 2 4 2 ⎞ 2
⎪ ⎮ ⎜ ⅈ⋅ρ ⋅(r - x) ⅈ⋅ρ ⅈ⋅ρ ⅈ⋅(r - x) 3⋅ⅈ⋅(r - x) ⎟ (r - x)
⎪ ⎮ ⎜─────────────────────── - ─────────────────────────────── - ─────────────────────── - ─────────────────────── + ───────────────────────⎟ dx for ──────── > 1
⎪ ⎮ ⎜ 3/2 ____________ ____________ __________________ 3/2 __________________⎟ 2
⎪ ⎮ ⎜ ⎛ 2 2⎞ 2⋅╲╱ -r - ρ + x ⋅╲╱ -r + ρ + x ╱ 2 2 ⎛ 2 2⎞ ╱ 2 2 ⎟ ρ
⎪ ⎮ ⎝2⋅⎝- ρ + (-r + x) ⎠ 2⋅╲╱ - ρ + (-r + x) 2⋅⎝- ρ + (-r + x) ⎠ 2⋅╲╱ - ρ + (-r + x) ⎠
⎪ ⌡
⎨ r
⎪
⎪ r + ρ
⎪ ⌠
⎪ ⎮ _______________
⎪ ⎮ ╱ 2 2
⎪ ⎮ ╲╱ ρ - (r - x) dx otherwise
⎪ ⌡
⎪ r
⎩

In [13]: piecewise_fold(i).args[0][0]
Out[13]:
r + ρ
⌠
⎮ ⎛ 2 2 2 2 4 2 ⎞
⎮ ⎜ ⅈ⋅ρ ⋅(r - x) ⅈ⋅ρ ⅈ⋅ρ ⅈ⋅(r - x) 3⋅ⅈ⋅(r - x) ⎟
⎮ ⎜─────────────────────── - ─────────────────────────────── - ─────────────────────── - ─────────────────────── + ───────────────────────⎟ dx
⎮ ⎜ 3/2 ____________ ____________ __________________ 3/2 __________________⎟
⎮ ⎜ ⎛ 2 2⎞ 2⋅╲╱ -r - ρ + x ⋅╲╱ -r + ρ + x ╱ 2 2 ⎛ 2 2⎞ ╱ 2 2 ⎟
⎮ ⎝2⋅⎝- ρ + (-r + x) ⎠ 2⋅╲╱ - ρ + (-r + x) 2⋅⎝- ρ + (-r + x) ⎠ 2⋅╲╱ - ρ + (-r + x) ⎠
⌡
r

In [14]: piecewise_fold(i).args[1][0]
Out[14]:
r + ρ
⌠
⎮ _______________
⎮ ╱ 2 2
⎮ ╲╱ ρ - (r - x) dx
⌡
r

That seems to be what you wanted. However it is just an unevaluated integral. If that is actually what you want then you should use `Integral` rather than `integrate` in the first place:

In [15]: Integral(f,(x,r,r+rho))
Out[15]:
r + ρ
⌠
⎮ ________________
⎮ ╱ 2 2
⎮ ╲╱ ρ - (-r + x) dx
⌡
r

The difference between `integrate` and `Integral` is that `Integral` just represents the integral symbolically whereas `integrate` will attempt to *compute* the integral ideally giving an expression that does not involve integrals:

In [16]: Integral(x, x)
Out[16]:
⌠
⎮ x dx
⌡

In [17]: integrate(x, x)
Out[17]:
2
x
──
2

In [18]: Integral(x, x).doit()
Out[18]:
2
x
──
2

</details>
# 答案3
**得分**: 1
好的，我明白你现在在问什么。在这里，我不确定为什么 `integrate` 返回它所返回的结果，但手动积分方法将计算这个积分：
```python
In [3]: f.integrate((x, r, r+rho), manual=True)
Out[3]: 
            ____________________________________
   2       ╱    2                  2          2 
π⋅ρ    ρ⋅╲╱  - r  + 2⋅r⋅(r + ρ) + ρ  - (r + ρ)  
──── + ─────────────────────────────────────────
 4                         2                    
In [4]: f.integrate((x, r, r+rho), manual=True).simplify()
Out[4]: 
   2
π⋅ρ 
────
 4  

这是使用 SymPy 1.11.1。

In [3]: f.integrate((x, r, r+rho), manual=True)
Out[3]: 
            ____________________________________
   2       ╱    2                  2          2 
π⋅ρ    ρ⋅╲╱  - r  + 2⋅r⋅(r + ρ) + ρ  - (r + ρ)  
──── + ─────────────────────────────────────────
 4                         2                    
In [4]: f.integrate((x, r, r+rho), manual=True).simplify()
Out[4]: 
   2
π⋅ρ 
────
 4