i need a python code that solves this question in general. and solves for bisection, secant and newtons method. because the code that was given by chegg last time was wrong so please edit the code that is below following the criteria of the question

You are designing a spherical tank as shown in Figure$($Q$1)$ below to hold water

for a small village in a country. The volume of liquid it can hold can be computed

as

$V=h^2\frac{[3R-h]}{3}$

where $V$ volume $($m$3),h$ depth of water in tank $($m$),$ and $R$ the tank radius $($m$).$

If $R=3m,$ to what depth must the tank be filled so that it holds $30m3?$ Use a

computer software program $($Python$)$ to determine your answer. You are

required to compare bisection method, Newton's method, and Secant method

solutions and recommend one of these software programs with justifications.

Employ initial guesses of $0$ and $R$

Program should show:

initial guess value input

second guess value input

indication of error if the interval inputs are not correct input values.

Tolerance$/$error input

All intermediate values of $h$ and its final value

import math

# Function to calculate the volume of water in the tank

def volume$($h$,$ R$)$:

return math.pi $*$ h $*2*(3*$ R $-$ h$)/3$

# Bisection method

def bisection$\_$method$($target$\_$volume, R$,$ tol$=1$e$-6,$ max$\_$iter$=100)$:

a $=0$

b $=$ R

iter$\_$count $=0$

if volume$($a$,$ R$)>$ target$\_$volume or volume$($b$,$ R$)$ target$\_$volume:

print$("$Error: Initial interval does not contain the solution."$)$

return None

while iter$\_$count $$ max$\_$iter:

c $=($a $+$ b$)/2$

vc $=$ volume$($c$,$ R$)$

if abs$($vc $-$ target$\_$volume$)$ tol:

print$("$Bisection method converged after", iter$\_$count, "iterations."$)$

return c

if vc $>$ target$\_$volume:

b $=$ c

else:

a $=$ c

iter$\_$count $+=1$

print$("$Bisection method did not converge within the maximum number of iterations."$)$

return None

# Newton's method

def newton$\_$method$($target$\_$volume, R$,$ tol$=1$e$-6,$ max$\_$iter$=100)$:

h $=$ R $/2$ # Initial guess

iter$\_$count $=0$

while iter$\_$count $$ max$\_$iter:

f $=$ volume$($h$,$ R$)-$ target$\_$volume

if abs$($f$)$ tol:

print$("$Newton$\text{'}$s method converged after", iter$\_$count, "iterations."$)$

return h

df $=2*$ math.pi $*($R $-$ h$)*$ h $+$ math.pi$*(3*$R $-2*$h$)$

# Derivative of the volume function

h $-=$ f $/$ df

iter$\_$count $+=1$

print$("$Newton$\text{'}$s method did not converge within the maximum number of iterations."$)$

return None

# Secant method

def secant$\_$method$($target$\_$volume, R$,$ tol$=1$e$-6,$ max$\_$iter$=100)$:

h$0=0$

h$1=$ R

iter$\_$count $=0$

while iter$\_$count $$ max$\_$iter:

f$0=$ volume$($h$0,$ R$)-$ target$\_$volume

f$1=$ volume$($h$1,$ R$)-$ target$\_$volume

h$\_$new $=$ h$1-$ f$1*($h$1-$ h$0)/($f$1-$ f$0)$

if abs$($h$\_$new $-$ h$1)$ tol:

print$("$Secant method converged after", iter$\_$count, "iterations."$)$

return h$\_$new

h$0,$ h$1=$ h$1,$ h$\_$new

iter$\_$count $+=1$

print$("$Secant method did not converge within the maximum number of iterations."$)$

return None

# Main function

def main$()$:

R $=3$ # Radius of the tank

target$\_$volume $=30$ # Target volume of water

print$("$Initial guess values: $0$ and", R$)$

# Bisection method

bisection$\_$result $=$ bisection$\_$method$($target$\_$volume, R$)$

print$("$Bisection method result:", bisection$\_$result, $"$m$")$

# Newton's method

newton$\_$result $=$ newton$\_$method$($target$\_$volume, R$)$

print$("$Newton$\text{'}$s method result:", newton$\_$result, $"$m$")$

# Secant method

secant$\_$result $=$ secant$\_$method$($target$\_$volume, R$)$

print$("$Secant method result:", secant$\_$result, $"$m$")$

if $\_\_$name$\_\_=="\_\_$main$\_\_"$:

main$()$