Area bounded by hyperbola 9x² - 4y² = 36 and line x = 3

[tangur]

Find the area of the region bounded by the hyperbola $$9x^2-4y^2 = 36$$ and the line $$x = 3$$.

I'm thinking that I have to integrate for x, so I'll have the sum of twice the area from $$2$$ to $$3$$.
The function will be $$+ \sqrt {\frac {9x^2-36}{4}}$$

hence, the integral will be$$2\int_2^3 {\sqrt {\frac {9x^2-36}{4}}}dx$$
I just wanted to know if my reasoning is right


Thanks in advance

 

[mathman]

It looks right to me. Go ahead with it.

 

[tangur]

when I integrate $$\int \sqrt {\frac {9x^2-36}{4}}dx$$ I get as an answer $$\frac {3}{2} x{\sqrt {x^2-4}} - 6 ln({\frac {x}{2}}+\frac{\sqrt {x^2-4}}{2})$$ however maple gives me $$\frac {3}{2} x{\sqrt {x^2-4}} - 6 ln(x+\sqrt {x^2-4})$$

I used $$x=2sec(\theta)$$ hence $$\frac {x}{2}=sec(\theta)$$ so $$tan(\theta)=\frac{\sqrt{x^2-4}}{2}$$

which I substituted into $$6(sec(\theta)tan(\theta)-ln(sec(\theta)+tan(\theta)))$$

I know its Saturday night and any help will greatly be appreciated

Thanks

 

[tangur]

Weird, overall it gives the same answer, 4.2878 sq units, in both maple and on paper, however I don't understand how maple takes the 1/2 out of the ln.

 

[tangur]

Ok, I quadrupled checked my integral and it is right, it seems that even though maple does not display the 1/2 it still accounts for it.

 

[teclo]

i'm not sure but i don't think that the 1/2 over the guys inside the natural log have any impact on a definite integratl. combining the two you would get that sum of the numerators over 2. which would be the natural log of the top minus ln 2 -- which being a constant wouldn't be affected by variables.

why it's there, I'm clueless. the only thing i could think of is the values determined by the trig functions of theta, but that doesn't seem to be the case.
er edit: maybe maple simplified it out as it can be included in C indefinintly , and is extraneous in the case of a definant integral


i quickly worked out the same answer you had, i haven't ever used maple so i don't know.

 

[HallsofIvy]

The only difference between


$$ln({\frac {x}{2}}+\frac{\sqrt {x^2-4}}{2})$$
and
$$ln({x+ \srt{x^2- 4})$$
is -ln(2) which is a constant.
Anti-derivatives can have any constant added and, anyway, cancels when you evaluate at the limits of integration.