## anonymous 4 years ago **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).

1. anonymous

This is around the y-axis..

2. anonymous

Yeah it is surface area not volume, though.

3. TuringTest

oh dang...

4. anonymous

hmm.. Still confused.. do i need u-substitution?

5. anonymous

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. ^_^!

6. TuringTest

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...

7. TuringTest

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

8. anonymous

I can't get the final answer right.. You are plugging in 0 and 5/4 right?

9. TuringTest

yes

10. anonymous

I got 2.5436pi. But it says incorrect.. :/

11. TuringTest

well I make no guarantees I didn't make a mistake, let me look it over... :/

12. TuringTest

13. TuringTest

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

14. anonymous

It isn't... ahhhh :/

15. TuringTest

shucks...

16. anonymous

Haha.. it is a big pain in the retrice

17. anonymous

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!

18. TuringTest

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.

19. anonymous

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.

20. anonymous
21. TuringTest

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

22. anonymous

Oh wow.. that is my bad!!!

23. TuringTest

$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)

24. anonymous

nope :(

25. TuringTest

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

26. anonymous

YAYY!!!!! 5.12283 was correct!!

27. TuringTest

oh great correct setup, messed up doing the integral somehow well I'll investigate that

28. anonymous

I've got one last one.. If you wanna help?? I'll have a 100/100..

29. TuringTest

worth a try :D

30. anonymous
31. TuringTest

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

32. TuringTest

$\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

33. anonymous

Turninng Test ... I spent much of time looking for this ... Good 2 see u Trying all of this ... Thnx :)