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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
 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
 one year ago

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Dido525Best ResponseYou've already chosen the best response.0
I don't get what I did wrong in my work. Let me post it.
 one year ago

Dido525Best ResponseYou've already chosen the best response.0
Did I set up the integral properly?
 one year ago

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

Dido525Best ResponseYou've already chosen the best response.0
I used the e points of intersection so I used 0 to 1.
 one year ago

tkhunnyBest 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.
 one year ago

Dido525Best ResponseYou've already chosen the best response.0
Ohh I can write in latex. It just takes too long lol.
 one year ago

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

Dido525Best ResponseYou've already chosen the best response.0
So I wrote my integral as: \[14\pi \int\limits_{}^{}xe^x dx\]
 one year ago

Dido525Best ResponseYou've already chosen the best response.0
How do I incorporate the e^x then?
 one year ago

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

Dido525Best ResponseYou've already chosen the best response.0
The limits of intergation should still be the same right?
 one year ago

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

Dido525Best ResponseYou've already chosen the best response.0
Yeah I did, One moment.
 one year ago

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

Dido525Best ResponseYou've already chosen the best response.0
I see why it's the difference though.
 one year ago

Dido525Best ResponseYou've already chosen the best response.0
\[\frac{ 28\pi }{ e}\] is what I got.
 one year ago

tkhunnyBest ResponseYou've already chosen the best response.0
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.
 one year ago

Dido525Best ResponseYou've already chosen the best response.0
Eww... Natural logs...
 one year ago

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

Dido525Best ResponseYou've already chosen the best response.0
Yeah... I have to use a calculator lol.
 one year ago
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