# integral of function with sgn (x,I) inside

HI

I encounter a problem on an integral calculation, and I do not know how to find a workaround. (I mean a mathematically correct workaround, not the questionable tinkering of replacing SGN (f (x, I)) by -1 !!! that I did to get a numeric value ;-) ). if I comment out lines (77,78) and (128,129) I get an error:

**ValueError: The function to be integrated depends on 2 variables (x, z),
and so cannot be integrated in one dimension.
Please fix additional variables with the 'params' argument**

Does somebody knows what 'params' I must set in function numerical_integral ? code on sagecell

```
show("###########################################################")
show("Compute Surface of the spherical triangle and XY plan")
show("###########################################################")
var('x,y')
rNum=1
var('xLow,xUp,yLow,yUp')
# rounding values
PtsL= [[0.13795, 0.37902], [0.33554, 0.92189], [0.84803, 0.2028], [0.80141, 0.37902]]
PtsNamesL = [ 'xy01', 'xy12', 'xy02', 'xyCut']
ellipse0 = 1/2*sqrt(-x^2 + 1)*sqrt(-sqrt(2) + 2)
ellipse1 = x*cos(1/9*pi)/sin(1/9*pi)
ellipse2 = sqrt(-4/3*x^2 + 1)
plt = list_plot(PtsL, color='blue', size=30)
pt_opt = {'color': 'blue', 'horizontal_alignment': 'left', 'vertical_alignment': 'bottom'}
plt += sum(text(name, vector(pt), **pt_opt) for name, pt in zip(PtsNamesL, PtsL))
plt += line([[0, PtsL[3][1]], PtsL[3]], color='brown', thickness=2, linestyle='dashed')
plt += line([[PtsL[1][0], 0], PtsL[1]], color='blue', thickness=2, linestyle='dashed')
plt+=implicit_plot(ellipse0-y,(x,0,1),(y,0,1),color='orange')
plt+=plot(ellipse1,(x,0.1,PtsL[1][0]),color='yellow')
plt+=implicit_plot(ellipse2-y,(x,0,1),(y,0,1),color='green')
rS(x,y,z)=(x,y,sqrt(rNum^2-x^2-y^2))
rSdx=diff(rS,x,1)
rSdy=diff(rS,y,1)
assume(0<x<1)
assume(0<y<1)
assume(rNum^2-x^2-y^2 >0)
show("r diff x : ",rSdx," \t r diff y : ",rSdy)
rSds=(vector(rSdx).cross_product(vector(rSdy)))
# I don't use rSds.norm(),because it is adding abs function
# because for us R^2-x^2-y^2 >=0
dsNorm=sqrt(rSds[0]^2+rSds[1]^2+rSds[2]^2)
show("ds : ",rSds," \t norm : ",dsNorm)
show("order of integration, Y,X : FIRST part, left side of the blue line ")
Pxy0 = plt
Pxy0+= plot([ellipse0, ellipse1], (PtsL[0][0], PtsL[1][0]), fill={0: [1]}, fillcolor='#ccc', fillalpha=.3)
Pxy0+=implicit_plot(ellipse0-y,(x,0,1),(y,0,1),color='orange')
Pxy0+=plot(ellipse1,(x,(PtsL[0][0]),(PtsL[1][0])),color='yellow')
Pxy0+=implicit_plot(ellipse2-y,(x,0,1),(y,0,1),color='green')
Pxy0.show(xmin=0, xmax=1, ymin=0, ymax=1, aspect_ratio=1, figsize=8)
show("first piece delimited by yellow curve ,orange curve and blue line in xy plan")
eqLimitL=[]
eqLimitL.append(yLow==ellipse0)
eqLimitL.append(yUp==ellipse1)
eqLimitL.append(xLow==PtsL[0][0])
eqLimitL.append(xUp==PtsL[1][0])
arrondi=10
# spherical triangle partaged in two pieces
# first piece delimited by yellow curve ,orange curve and blue line in xy plan
show(LatexExpr(r"\text{surface of first piece of spherical Triangle} \,=\,"
+r"\int_{"+latex(eqLimitL[0])+"}^{"+latex(eqLimitL[1])+"} " \
+r"\int_{xLow="+latex(round(eqLimitL[2].rhs(),arrondi))+"}^{xUp="+latex(round(eqLimitL[3].rhs(),arrondi))+"} " \
#+r", "+latex(rS[2])+latex(dsNorm)+r" \, \, dy \, dx"))
+r", "+latex(dsNorm)+r" \, \, dy \, dx"))
intY0=(integrate(dsNorm,y,eqLimitL[0].rhs(),eqLimitL[1].rhs(),algorithm="giac"))
print(intY0)
intY00=intY0
#show(" below we replace sgn function by fsgn(x)=-1 because numerical_integral can't integrate with sgn ")
fsgn(x)= -1
intY00=intY0.substitute_function(sgn,fsgn)
show("intY0 with sgn Function: ", intY0)
show("intY00 subs of sgn: ", intY00)
intYX00=numerical_integral(intY00,eqLimitL[2].rhs(),eqLimitL[3].rhs())
show("First spherical surface piece after numerical integration : ", intYX00)
var('xLow,xUp,yLow,yUp',domain='real')
show("order of integration, Y,X : Second part, rigth side of the blue line ")
Pxy1 = plt
Pxy1+= plot([ellipse0, ellipse2], (PtsL[1][0], PtsL[2][0]), fill={0: [1]}, fillcolor='#ccc', fillalpha=.3)
Pxy1+= plot([ellipse0, ellipse2], (PtsL[1][0], PtsL[2][0]), fill={0: [1]}, fillcolor='#ccc', fillalpha=.3)
Pxy1+=implicit_plot(ellipse0-y,(x,0,1),(y,0,1),color='orange')
Pxy1+=plot(ellipse1,(x,PtsL[0][0],PtsL[1][0]),color='yellow')
Pxy1+=implicit_plot(ellipse2-y,(x,0,1),(y,0,1),color='green')
Pxy1.show(xmin=0, xmax=1, ymin=0, ymax=1, aspect_ratio=1, figsize=8)
eqLimitL=[]
# delimited y area
#assume(PtsL[2][1]<=ellipse0 <=PtsL[1][1])
#assume(PtsL[2][1]<=ellipse2 <=PtsL[1][1])
# delimited x area
eqLimitL.append(yLow==ellipse0)
eqLimitL.append(yUp==ellipse2)
eqLimitL.append(xLow==PtsL[1][0])
eqLimitL.append(xUp==PtsL[2][0])
arrondi=10
# spherical triangle partaged in two pieces
# Second piece delimited by blue line ,orange curve and green curve in xy plan
show(LatexExpr(r"\text{surface of second piece of spherical Triangle} \,=\,"
+r"\int_{"+latex(eqLimitL[0])+"}^{"+latex(eqLimitL[1])+"} " \
+r"\int_{xLow="+latex(round(eqLimitL[2].rhs(),arrondi))+"}^{xUp="+latex(round(eqLimitL[3].rhs(),arrondi))+"} " \
#+r", "+latex(rS[2])+latex(dsNorm)+r" \, \, dy \, dx"))
+r", "+latex(dsNorm)+r" \, \, dy \, dx"))
#intY1=(integrate(dsNorm,y,eqLimitL[0].rhs(),eqLimitL[1].rhs(),algorithm="giac"))
intY1=(integrate(dsNorm,y,yLow,yUp,algorithm="giac"))
print(intY1)
intY11=intY1.subs(eqLimitL)
show("intY1 : ",intY11)
intY111=intY11
#show(" below we replace sgn function by fsgn(x)=-1 because numerical_integral can't integrate with sgn ")
fsgn(x)= -1
intY111=intY11.substitute_function(sgn,fsgn)
show("intY11 with sgn Function: ", intY11)
show("intY111 subs of sgn: ", intY111)
intYX111=numerical_integral(intY111,eqLimitL[2].rhs().n(),eqLimitL[3].rhs().n())
show("Second spherical surface piece after numerical integration : ", intYX111)
show("total Triangle area Computed by surface integration: ", intYX00[0]+intYX111[0])
#show('Total TriAngle Area Computed with Harriot Theorem : ',HarriotArea,' \t with R = ',rNum)
#show(" We can also verify the computation in multiplying the radius by 2(rNum at top of the file)")
#show(" and check that the surface is multiplied by 2^2=4 and the volume by 2^3=8")
#show(" you can also check the computation by changing the spherical surface with the condition,")
#show(" that this surface should be contained in first octant ")
```

I forgot to tell: do not reset the file in notebook, it will crash the notebook

I am wondering about this question.

No one answered, I know there is no obligation to answer a question, and I do not claim an answer, moreover maybe that my mathematical question is not relevant. and therefore I am wondering about my question, I would have liked an opinion . (One can tick several boxes)

1)I messed up in my integral calculation

2) my question is scatty kind of: https://ask.sagemath.org/question/553... (maybe a troll behind this question ? not sure )

3)my question is not really interesting

4)there is no simple solution to my question

I explain what I wanted to do in the entire code ,(probably naively and doomed to failure, but that's okay ;-),this is not the point ) the code above was an excerpt :

I wanted to try to obtain an algebraic length of an arc of ellipse using Harriot's theorem. step 1) verify that we find the surface area of any spherical triangle numerically located in the 1st octant, calculated by Harriot's theorem and by the surface integral dy dx ,1) dy algebraic integral, 2) dx numeric. step 2) use the algebraic surface value obtained by Harriot to attempt to find an algebraic value of the length of an arc of ellipse

I wanted to join the full code maybe it will help somebody to answer me but it is not accepted I think because the message seems too long.

(I'm a bit ashamed of this code because I don't know how to program correctly in Python code, and maybe somebody can propose me some improvement ?). in case of my question is relevant ! ;-)

I tried to put the code on Cocalc but it seems that now there is no more share button !