Question

surface area integration

Answer 1

\[y=\cosh2x\]\[y'=2\sinh2x\]
between 0 and 1\[A=2 \pi \int\limits \cosh2x \sqrt{ 1+4\sinh^{2}2x} dx\]

Answer 2

im not sure how to progress with the integral from there would it be u sub?

Answer 3

@iambatman

Answer 4

Yeah substitution is the way to go here. =) I'll go ahead and make a suggestion: \[\Large u = 1+4\sinh^2 2x\]

Answer 5

so \[du=8\sinh2xcosh2x=4\sinh4x\]?

Answer 6

where do I go from there or am I doing the identity or derivative wrong?

Answer 7

Whoops I made a mistake in making that suggestion I think haha.

Answer 8

I think instead we need to exploit this relationship: \[\Large \cosh^2x-\sinh^2x=1\] and figure out a substitution to make it work this way instead I think. I'm a little busy at the moment sorry for the delayed responses.

Answer 9

I don't see how to fit that specific instance into the identity though

Answer 10

I believe I did it right, you might want to just do it on your own to make sure I didn't mess up. \[\Large 2 \sinh 2x = \sinh 2u \\ \Large 4 \cosh 2x dx = 2 \cosh 2u du\] \[\Large \pi \int\limits_0^{1}\cosh 2u \sqrt{1+\sinh^2 2u} du = \pi \int\limits_0^{2\pi} \cosh^22u du\]

Answer 11

what was u?

Answer 12

@iambatman

Answer 13

@jim_thompson5910

Answer 14

@zepdrix @myininaya