Find the area cut off by x=4 from the hyperbola x^2/9-y^2/4=1. Answer is 4.982 in the book. I have proceeded as under:

Y=2/3*sqrt(x^2-9) and rhe reqd. area is double of integral 2/3*sqrt(x^2-9) from 3 to 4.
Int= 2/3*[xsqrt(x^2-9)/2 – 9/2*log{x+sqrt(x^2-9)}] from 3 to 4
=x/3sqrt(x^2-9)-3log{x+sqrt(x^2-9) from 3 to 4
=4/3*sqrt(16-9)-3log{4+sqrt(16-9)} -0+3log(3+0)= 4/3*sqrt7-3log(4+2.646)+3log3
=4*2.646/3 - 3log6.646 + 3log3=3.528-5.67+3.297=1.155
Area cut off should be 2*1.155=2.31 which is widely away from 4.982 Where have I committed the mistake?

2 answers

As a first check, I went to

http://www.wolframalpha.com/input/?i=2%E2%88%AB[3%2C4]+%282%2F3+%E2%88%9A%28x^2-9%29%29+dx

and saw that they show the area as 2.28

So, I suspect there is an error in the problem or the answer.

Your calculation appears to be correct, within roundoff errors.
Thanks for the advice. I checked the problem statment and answer several times and got the same result. I also suspect it to be a print mistake in the book.
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