Question

can anyone help me answer this question???
Let R be the region in the xy-plane between the graphs of y = e^x 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.

Answer 1

Do you know how to find the volume of the function rotated around the x-axis if we only had one function, like e^x only?

Answer 2

not really...

Answer 3

http://i41.tinypic.com/154yoic.png
If we found the sum of the volumes of those thin sections made into cylinders over the interval [0,2], we could find the volume of the function rotated around the x-axis.

\[ V_{cylinder} = \pi r^{2} h \]
r would be the function value at x, itself if we rotated it around the x-axis, and the height would be the 'dx' (meaning, it's really, really small but not 0). The integral over the interval [0,2] would be the sum of these cylinders' volumes.

\[ \int_{0}^{2} \pi y^{2} dx = \pi \int_{0}^{2} e^{x} dx\]

This is for one function around the x-axis. For the region between two functions, we'd find the volume of the outer one and then subtract the volume of the inner.

Answer 4

http://openstudy.com/study#/updates/4f496108e4b00c3c5d330507

might save some writers cramp

Answer 5

thx lol i kno that person

Answer 6

Oh, it seems somebody had the exact same question. :P

Also, my last integral should be of e^(2x), not e^x, my bad, :P

Answer 7

they apparently have a teacher that tells them to come here to check it out; but gives them all the same problem to ask :)

Answer 8

r teacher is pretty nice XD ill just go n look at his work but thx guys i got the first half of it

Answer 9

if you have any questions about it, you can post them in this thread ... for Access to figure out :)

Answer 10

I knew I remembered the exact question from somewhere, and I was thinking about trying to make an intuitive explanation for how to do these kinds of problems (for both others and myself), except then I ran into difficulties making Google Sketchup separate the shape I made into a few disks. lol

Answer 11

π(e2(2)/2+e−2(2)/2)−π(e2(0)/2+e−2(0)/2) im a lil confused bout here why did u divide by 2??

Answer 12

becasue thats the result of integrating e^2x

Answer 13

try this:

take the derivative of e^2x, what do you get?

Answer 14

2e^2x

Answer 15

right so we need that /2 to catch the 2 that pops out

derive 1/2 e^2x

what do we get?

Answer 16

e^2x

Answer 17

the longer method is to say:

e^2x

u = 2x

du = 2 dx

du/2 = dx

e^u
---- du integrates to: e^u/2
2


substitute in u again for: e^2x/2

Answer 18

i just cut the useless redefining of stuff and go straight for the juggular lol

Answer 19

i got it thx again :D

Answer 20

can u guys help me with this question too?
2: let f be the function given by : \[f(x)=\sqrt{x ^4-16x ^{2}}\]
a) Find the domain of f.
b) Find f '(x)
c) Find the slope of the line normal to the graph of f at x = 5

Answer 21

part b i dont need help with only a and c

Answer 22

ohh, you don't need help with part b... nvm on that

a) you want the values that you can input into that function to have an existing output (in the reals, not complex)

So, the square root has the restriction that everything under it must be positive or zero.

x^4 - 16x^2 >= 0
x^2 (x^2 - 16) >= 0
x^2 (x + 4)(x - 4) >= 0 x^2 is always positive or zero
(x + 4)(x - 4) >= 0 x + 4 is always larger than x - 4, so let's only deal with the number that is closer to being negative.

x - 4 >= 0
x >= 4

This would be the domain.

Answer 23

c) You already have the derivative (assuming you found it in part b)
The slope of the line normal is perpendicular to the tangent. That is, if you find the derivative at x=5, you can then take the opposite reciprocal for the perpendicular slope.

Answer 24

thx