**Calc2 Help Needed** Find the area of the surface generated by revolving the curve x = (e^y + e^-y) / 2 from x = 0 to x = ln(2).

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**Calc2 Help Needed** Find the area of the surface generated by revolving the curve x = (e^y + e^-y) / 2 from x = 0 to x = ln(2).

Mathematics
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This is around the y-axis..
Yeah it is surface area not volume, though.
oh dang...

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hmm.. Still confused.. do i need u-substitution?
Hi ... Ummm its nice equ :D i have somethings to say .. i dont know whither its true or not .. but i like to share it x = cosh y while x cannot be zero ... but we can say when y = 0 , x = 1 you need to find area .. so we need to enclose the draw |dw:1328090291826:dw| so what is range ? cause x cannot be zero ???? in other word what is the area that we want to cal. ^_^!
First off, I agree with waleed that it is strange to have one of the bounds be x=0, but this shouldn't be a major problem because we want surface area, not area under the curve. So, from the top... We should notice that our function is special: it is a hyperbolic cosine function\[x=\frac{e^y-e^{-y}}2=\cosh y\]and we are going to the following derivation\[\cosh^2x=(\frac{e^x+e^{-x}}2)^2=\frac{e^{2x}+2+e^{-2x}}4=\frac12[1+\cosh(2x)]\]which implies that\[\sqrt{\frac12(1+\cosh x)}=\cos\frac x2\]Using the formula for the arc length differential ds we have\[ds=\sqrt{1+(\frac{dx}{dy})^2}dy=\sqrt{1+(\frac{e^y-e^{-y}}2)^2}dy\]\[=\sqrt{1+(\frac{e^{2y}-2+e^{-2y}}4)}dy=\sqrt{\frac12(1+\frac{e^{2y}+e^{-2y}}2})dy\]\[=\sqrt{\frac12[1+\cosh(2y)]}dy=\cosh ydy\]actually, this proves another theorem which we could have just used straight away, but I couldn't find until after I had derived it!\[\sqrt{1+(\frac d{dx}\cosh x)^2}=\cosh x\]I wish I had found that sooner...
Now look at the picture to find our bounds|dw:1328219640977:dw|So along the y-axis it looks like our bounds are\[-5/4\le y\le5/4\]Now we need to do our revolution, which is done about the y-axis by the formula\[S=\int_{a}^{b}2\pi xds==\int_{-5/4}^{5/4}2\pi\cosh y\cdot\cosh ydy=\int_{-5/4}^{5/4}2\pi\cosh^2ydy\]Since our graph is symmetric about the x-axis we can put the bounds from 0 to 5/4, then double our answer. Doing that and remembering the 'half-angle' formula for cosh^2x we derived earlier we get:\[S=2\int_{0}^{5/4}2\pi\cdot\frac12[1+\cosh(2y)]dy=2\pi\int_{0}^{5/4}1+\cosh(2y)dy\]\[=2\pi[y+\frac12\sinh(2y)]|_{0}^{5/4}=\pi[2y+\sinh(2y)]|_{0}^{5/4}=\pi(\frac52+\sinh\frac52)\]Please let me know if this is right, thanks :D
I can't get the final answer right.. You are plugging in 0 and 5/4 right?
yes
I got 2.5436pi. But it says incorrect.. :/
well I make no guarantees I didn't make a mistake, let me look it over... :/
I get about 8.55pi
I see what you did, you just put \[\frac52+\sin(\frac{5}2^{\circ})\]But that is a very different thing. What we have in this problem are HYPERBOLIC trig functions. Notice the h!\[\cosh x=\frac{e^x+e^{-x}}2\]\[\sinh x=\frac{e^x-e^{-x}}2\]http://en.wikipedia.org/wiki/Hyperbolic_function Definitely have to note the difference between the functions sin and sinh. Hopefully my original answer is correct. http://www.wolframalpha.com/input/?i=%285%2F2%2Bsinh%285%2F2%29%29 ...times pi
It isn't... ahhhh :/
1 Attachment
shucks...
Haha.. it is a big pain in the retrice
I have my first calc2 exam tomorrow, and hopefully nothing like that is on there.. I'm a computer science major, so I don't need to know this much math!
Well it's going to drive me crazy now, so I may have to appeal to someone else on here to find my mistake. I'll keep looking in the meantime though.
Thanks! I posted another problem if you want to give it a shot. It's much easier.. I just cant find the mistake in my work.
your link says the interval is \[0\le y\le\ln2\]but you wrote x in the post. so everthing may work if we just evaluate the integral with those bounds
Oh wow.. that is my bad!!!
\[S=\int_{a}^{b}2\pi xds=\int_{0}^{\ln2}2\pi\cosh y\cdot\cosh ydy=\int_{0}^{\ln2}2\pi\cosh^2ydy\]\[=\int_{0}^{\ln2}2\pi\cdot\frac12[1+\cosh(2y)]dy=\pi\int_{0}^{\ln2}1+\cosh(2y)dy\]\[=\pi[y+\frac12\sinh(2y)]|_{0}^{\ln2}=\pi[\ln2+\frac12\sinh(\ln2)]\]\[=\pi\ln2+\frac{\pi}2(e^2-e^{-2})=\]http://www.wolframalpha.com/input/?i=pi%28ln2%2B%28e%5E2-e%5E%28-2%29%29%2F2%29 (fingers crossed)
nope :(
maybe I made an arithmetic mistake? try this to see if my setup is right: http://www.wolframalpha.com/input/?i=integral+2*pi*%28coshy%29%5E2dy+from+0+to+ln2
YAYY!!!!! 5.12283 was correct!!
oh great correct setup, messed up doing the integral somehow well I'll investigate that
I've got one last one.. If you wanna help?? I'll have a 100/100..
worth a try :D
Everything I have is right up until the last line.\[S=\int_{a}^{b}2\pi xds=\int_{0}^{\ln2}2\pi\cosh y\cdot\cosh ydy=\int_{0}^{\ln2}2\pi\cosh^2ydy\]\[=\int_{0}^{\ln2}2\pi\cdot\frac12[1+\cosh(2y)]dy=\pi\int_{0}^{\ln2}1+\cosh(2y)dy\]\[=\pi[y+\frac12\sinh(2y)]|_{0}^{\ln2}=\pi[\ln2+\frac12\sinh(2\ln2)]\]and then I simplified wrong, but this is the right answer
\[\pi[\ln2+\frac12\sinh(2\ln2)]=\pi[\ln2+\frac14(2^2-2^{-2})]\]\[=\pi[\ln2+\frac14(4-\frac14)]=\pi(\ln2+\frac{15}{16})\approx5.122\]hooray! You probably don't care but it makes me happy to figure it out :D
Turninng Test ... I spent much of time looking for this ... Good 2 see u Trying all of this ... Thnx :)

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