How do I solve this equation in MatLab?

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Find the area of the region bounded by the hyperbola 9x^2−4y^2=36 and the line x=3.
I cannot find out the proper way to input this into MatLab. It requires trig substitution so I believe the problem stems from there but I am new to MatLab and don't know what I can do to fix it.
Here is my code:
>> syms x
>> EQ = 3*sqrt(x^2-4)
EQ =
3*(x^2 - 4)^(1/2)
>> A = int(EQ,2,3)
A =
log(161 - 72*5^(1/2)) + (9*5^(1/2))/2
>>
Here is the actual answer to the problem:
(9/2)*sqrt(5) - 6*ln((3 + sqrt(5))/2)

Accepted Answer

KSSV
KSSV on 24 Jun 2021
Why worry the final answer is same right?
syms x
eq = 3*sqrt(x^2-4) ;
A1 = int(eq,2,3) ;
A2 = (9/2)*sqrt(5) - 6*log((3 + sqrt(5))/2) ;
[double(A1) A2]
ans = 1×2
4.2878 4.2878
  4 Comments
Isaac Hewitt
Isaac Hewitt on 24 Jun 2021
Edited: Isaac Hewitt on 24 Jun 2021
What does that mean? I just downloaded MatLab for a class a few days ago, I have no idea what that means.
I ask this because in all my calculations
log(161 - 72*5^(1/2)) + (9*5^(1/2))/2 = 7.55445
(9/2)*sqrt(5) - 6*ln((3 + sqrt(5))/2) = 4.28776
Walter Roberson
Walter Roberson on 24 Jun 2021
format long g
syms x
eq = 3*sqrt(x^2-4)
eq = 
A1 = int(eq,2,3)
A1 = 
A2 = (9/2)*sqrt(5) - 6*log((3 + sqrt(5))/2)
A2 =
4.28776399803381
simplify(A1 - A2)
ans = 
double(A1 - A2)
ans =
2.3429264568019e-16
log(161 - 72*5^(1/2)) + (9*5^(1/2))/2
ans =
4.28776399803129
.... not 7.55445

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