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)

 Akzeptierte Antwort

KSSV
KSSV am 24 Jun. 2021

0 Stimmen

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 Kommentare

Isaac Hewitt
Isaac Hewitt am 24 Jun. 2021
Bearbeitet: Isaac Hewitt am 24 Jun. 2021
Can you explain this more, how is this the same answer?
Yeah I just did this and I have no idea how you did this at all. Please explain what is happening.
KSSV
KSSV am 24 Jun. 2021
Answer is converted into double from symbolic class.
Isaac Hewitt
Isaac Hewitt am 24 Jun. 2021
Bearbeitet: Isaac Hewitt am 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
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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