## Dido525 Group Title 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 one year ago one year ago

1. Dido525 Group Title

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

2. Dido525 Group Title

|dw:1358999196019:dw|

3. Dido525 Group Title

Did I set up the integral properly?

4. Dido525 Group Title

@Mimi_x3 @satellite73

5. tkhunny Group Title

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. Dido525 Group Title

Let me send a pic.

7. Dido525 Group Title

8. Dido525 Group Title

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

9. Dido525 Group Title

x not e.

10. tkhunny Group Title

$$y = 7e^{x}$$ -- 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. Dido525 Group Title

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

12. tkhunny Group Title

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

13. Dido525 Group Title

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

14. Dido525 Group Title

How do I incorporate the e^-x then?

15. tkhunny Group Title

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

16. Dido525 Group Title

Huh? Why?

17. Dido525 Group Title

Nevermind. I see why.

18. Dido525 Group Title

The limits of intergation should still be the same right?

19. tkhunny Group Title

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

20. Dido525 Group Title

Yeah I did, One moment.

21. Dido525 Group Title

|dw:1359002030111:dw|

22. Dido525 Group Title

Something like that.

23. tkhunny Group Title

That's it. And the final answer is?

24. Dido525 Group Title

I see why it's the difference though.

25. Dido525 Group Title

Sec.

26. Dido525 Group Title

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

27. tkhunny Group Title

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. Dido525 Group Title

Eww... Natural logs...

29. tkhunny Group Title

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

30. Dido525 Group Title

Ahh I see... Hehe...

31. Dido525 Group Title

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