Question

Let R be the region in the xy-plane between the graphs of y = ex and y = e-x from x = 0 to x = 2.
a) Find the volume of the solid generated when R is revolved about the x-axis.
(This assignment must be done on graph paper)

Answer 1

simple enough, you know the formula for the area of a circle?

Answer 2

yes do i do top - bottom as in \[\int\limits_{0}^{2}\](7.3891-e^x) ^2

Answer 3

\[\int \pi(e^x)^2dx-\int \pi (e^{-x})^2dx \]
\[\int \pi e^{2x}dx-\int \pi e^{-2x}dx \]

Answer 4

its hallow though. wouldnt you do R^2-r^2

Answer 5

I have see this problem three times in two days
it's like a plague apparently

Answer 6

seen*

Answer 7

\[\pi (\frac{e^{2(2)}}{2}-\frac{e^{2(0)}}{2})+ \pi (\frac{e^{-2(2)}}{2}-\frac{e^{-2(0)}}{2}) \]

\[\pi (\frac{e^{4}-1}{2})+ \pi (\frac{e^{-4}-1}{2}) \]

Answer 8

by subtracting the top from the bottom function you create the "hole"

Answer 9

ok thats what i thought i had to do thanks . so whats the answer

Answer 10

im sure you are capable enough to finish it up from where I left off ....

Answer 11

why are you dividing by 2 ?

Answer 12

integrate e^2x

what do you get?

Answer 13

or better yet

derive e^2x and we get 2 e^2x

so we need to catch that "2" that pops out by the /2