exp volume q (1 Viewer)

[Rhanoct]

The region under the graph y = (1/2)[e^x + e^-x] and between the lines x=-2 and x=2 is rotated aroudn the x-axis. Find the volume of the solid of revolution formed.

ZZzzz... trying to learn exp stuff from the text book because I missed class... I tried a number of methods and can't get the answer (49.150u^3)

halp plx

 

[veloc1ty]

Hmm I'm interested in this too, I tried just then and got 22.79 units^3.

 

[lyounamu]

The region under the graph y = (1/2)[e^x + e^-x] and between the lines x=-2 and x=2 is rotated aroudn the x-axis. Find the volume of the solid of revolution formed.

ZZzzz... trying to learn exp stuff from the text book because I missed class... I tried a number of methods and can't get the answer (49.150u^3)

halp plx

I got it. I will post up solution here.

y=1/2(e^x + e^-x)
= 1/2e^x + 1/2e^-x
y^2 = 1/4 e^(2x) + 1/4e^(-2x) +1/2
Integral of that is

I = 1/8e^(2x) - 1/8e^(-2x) + 1/2x

You substitute x=2 and x=-2 separately.
A = pi ((1/8e^4 - 1/8e^(-4) + 1) - (1/8e^(-4) -1/8e^4 -1))
= 49.15008...
= 49.150 units^3

 

[Rhanoct]

thanks <3

 

[Forbidden.]

Oh my god I just realised the function y = (1/2)[e^x + e^-x] is actually the hyperbolic function y = sinh x

 

[veloc1ty]

Oh wow I forgot to square it... I am thinking perhaps I should revise for my trials.

 

[lyounamu]

Oh wow I forgot to square it... I am thinking perhaps I should revise for my trials.

Don't worry. I forgot to times it by pi when I first tried it. LOL