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 one year 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 yaxis
 one year 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 yaxis

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Dido525
 one year ago
Best ResponseYou've already chosen the best response.0I don't get what I did wrong in my work. Let me post it.

Dido525
 one year ago
Best ResponseYou've already chosen the best response.0Did I set up the integral properly?

tkhunny
 one year ago
Best ResponseYou've already chosen the best response.0What 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 (1,7e1) 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?

Dido525
 one year ago
Best ResponseYou've already chosen the best response.0I used the e points of intersection so I used 0 to 1.

tkhunny
 one year ago
Best ResponseYou've already chosen the best response.0\(y = 7e^{x}\)  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.

Dido525
 one year ago
Best ResponseYou've already chosen the best response.0Ohh I can write in latex. It just takes too long lol.

tkhunny
 one year ago
Best ResponseYou've already chosen the best response.0It would have avoided much time and error. 7ex is VERY different from 7e^(x) or \(7e^{x}\)

Dido525
 one year ago
Best ResponseYou've already chosen the best response.0So I wrote my integral as: \[14\pi \int\limits_{}^{}xe^x dx\]

Dido525
 one year ago
Best ResponseYou've already chosen the best response.0How do I incorporate the e^x then?

tkhunny
 one year ago
Best ResponseYou've already chosen the best response.0\(e^{x}\) is no good. You need \(e^{x}  e^{x}\).

Dido525
 one year ago
Best ResponseYou've already chosen the best response.0The limits of intergation should still be the same right?

tkhunny
 one year ago
Best ResponseYou've already chosen the best response.0Did you draw a picture? The height of your cyllinders is the difference of the two curves.

Dido525
 one year ago
Best ResponseYou've already chosen the best response.0Yeah I did, One moment.

tkhunny
 one year ago
Best ResponseYou've already chosen the best response.0That's it. And the final answer is?

Dido525
 one year ago
Best ResponseYou've already chosen the best response.0I see why it's the difference though.

Dido525
 one year ago
Best ResponseYou've already chosen the best response.0\[\frac{ 28\pi }{ e}\] is what I got.

tkhunny
 one year ago
Best ResponseYou've already chosen the best response.0I 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.

Dido525
 one year ago
Best ResponseYou've already chosen the best response.0Eww... Natural logs...

tkhunny
 one year ago
Best ResponseYou've already chosen the best response.0No, no. It's far worse than that. :) When you are ready...

Dido525
 one year ago
Best ResponseYou've already chosen the best response.0Yeah... I have to use a calculator lol.
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