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If the circumference of the region bounded by the curve y=cosh(x) and the lines y=0 x=a
and x=-a is 2a+4, where a>0 find the area of the surface obtained by rotating the part of
the curve y=cosh(x) between x=a x=-a and around the x axis. This is my homework question.I tried to solve it.I get a result but ı'm not sure because
there is not number in my answer and this is area question.ı want to say my approach to
this question.
Firstly,ı found arc length interms of a. this is ea-e-a = 2a+4 ,i.e,
sinh(2a)=a+2 Secondly,ı found surface.∫2∏cosh(x)√1+sinh2(x) dx (from -a to a)
After calculations,S=sinh(2a)+a∏/2
Did ı do wrong something when solving the question,the answer is strange...
and x=-a is 2a+4, where a>0 find the area of the surface obtained by rotating the part of
the curve y=cosh(x) between x=a x=-a and around the x axis. This is my homework question.I tried to solve it.I get a result but ı'm not sure because
there is not number in my answer and this is area question.ı want to say my approach to
this question.
Firstly,ı found arc length interms of a. this is ea-e-a = 2a+4 ,i.e,
sinh(2a)=a+2 Secondly,ı found surface.∫2∏cosh(x)√1+sinh2(x) dx (from -a to a)
After calculations,S=sinh(2a)+a∏/2
Did ı do wrong something when solving the question,the answer is strange...