## anonymous 3 years ago Use the method of cylindrical shells to find the volume V generated by rotating the region bounded by the given curves about the specified axis. y = 7e^x, y = 7e−x, x = 1; about the y-axis

1. anonymous

I don't get what I did wrong in my work. Let me post it.

2. anonymous

|dw:1358999196019:dw|

3. anonymous

Did I set up the integral properly?

4. anonymous

@Mimi_x3 @satellite73

5. tkhunny

What did you get for your points of intersection? y = 7e^x vs x = 1 is obviously (1,7e) y = 7e−x vs x = 1 is obviously (1,7e-1) But what about y = 7e^x vs y = 7e−x ? Since this cannot be solved by direct methods, we get (0.9488478,18.0791250) Maybe the problem statement is not written correctly?

6. anonymous

Let me send a pic.

7. anonymous

8. anonymous

I used the e points of intersection so I used 0 to 1.

9. anonymous

x not e.

10. tkhunny

$$y = 7e^{x}$$ -- You wrote this one okay. $$y = 7e^{-x}$$ -- This is quite a different story. Please learn to code LaTeX or to use the Equation Editor. It will be a great benefit to you now and in the future. You're f(x) is not correct. You just picked one of the curves. You need the distance between them.

11. anonymous

Ohh I can write in latex. It just takes too long lol.

12. tkhunny

It would have avoided much time and error. 7e-x is VERY different from 7e^(-x) or $$7e^{-x}$$

13. anonymous

So I wrote my integral as: $14\pi \int\limits_{}^{}xe^x dx$

14. anonymous

How do I incorporate the e^-x then?

15. tkhunny

$$e^{x}$$ is no good. You need $$e^{x} - e^{-x}$$.

16. anonymous

Huh? Why?

17. anonymous

Nevermind. I see why.

18. anonymous

The limits of intergation should still be the same right?

19. tkhunny

Did you draw a picture? The height of your cyllinders is the difference of the two curves.

20. anonymous

Yeah I did, One moment.

21. anonymous

|dw:1359002030111:dw|

22. anonymous

Something like that.

23. tkhunny

That's it. And the final answer is?

24. anonymous

I see why it's the difference though.

25. anonymous

Sec.

26. anonymous

$\frac{ 28\pi }{ e}$ is what I got.

27. tkhunny

I like it. Now, your punishment for writing one of the equations incorrectly in the beginning is to find the point of intersection for what you wrote. Find the value of x such that $$7e^{x} = 7e - x$$. It's a wonderful exploration.

28. anonymous

Eww... Natural logs...

29. tkhunny

No, no. It's far worse than that. :-) When you are ready...

30. anonymous

Ahh I see... Hehe...

31. anonymous

Yeah... I have to use a calculator lol.