Question

consider the region bounded by y=1/x, x=1 and x=M(M is a real number greater then 1)0
Now, calculate the volume of the solid obtained by rotating the region about the x-axis. what happens the the volume of the solid when M gets larger?

Answer 1

is this a correct formula to calculate th volume?....\[\int\limits_{1}^{m}\pi(1/x)^2dx\]

Answer 2

now just test values of m?

Answer 3

as m gets larger the volume oviousy get larger?

Answer 4

ok why shells? thanks

Answer 5

v=2pi\[ 2 \Pi r h \Delta r=v\]

Answer 6

i see shells give better estimate.....i must rotate around x-axis

Answer 7

oops. I am rotating around y-axis.

Answer 8

shells are just are easier that's all?

Answer 9

so discs! pi r^2 where r= y = 1/x
and volume is pi x^-2 dx

Answer 10

which is what you did.

Answer 11

cool...\[\pi \int\limits_{1}^{m}(1/x)^2 dx?....\]

Answer 12

yes. (sorry for that detour). integrate.

for what happens the the volume of the solid when M gets larger?
let M-> inf and see what the limit is for the volume

Answer 13

dude ...its fine im thankfull for the help iv got more if your havin fun?

Answer 14

ok how do i do that??

Answer 15

integral x^-2 ? use the power rule

Answer 16

m approaches zero?

Answer 17

x^-2?

Answer 18

i see
(1/x)^2=x^-2

Answer 19

1/x is x^-1 and (x^-1)^2 = x^-2
integrating that is as simple as integration gets

Answer 20

power rule=f prime of g(x) times g prime of x...

Answer 21

so indefinite integral give limit?

Answer 22

\[\int\limits_{1}^{m}x^{-2} dx= -x^{-1}|_{1}^{m}\]

Answer 23

ok i use m!!!?]

Answer 24

i plugged in m values...it increases pi/2,2pi/3....

Answer 25

use the m values with out pi?

Answer 26

you plug in the limits
pi(- m^-1 - -1^-1 )= pi*(-1/m +1) or pi*(1- 1/m)

Answer 27

shoot i dont get it?

Answer 28

this is very basic calculus
maybe this will help
http://www.khanacademy.org/math/calculus/integral-calculus/v/introduction-to-definite-integrals

Answer 29

\[ \pi \int\limits_{1}^{m}(1/x)^2 dx = \pi (-\frac{1}{x})|_{1}^{m} = \pi(-\frac{1}{m}+1)\]

Answer 30

i did that and got....when m is 2 i get pi/2...when m is 3 i get2pi/3

Answer 31

yes. that is right. when m is 1000 you get 999 pi/1000
you get (m-1)/m * pi as m gets big, this approaches 1*pi or pi

Answer 32

or if you like
pi*(1-1/m) = pi when m -> inf (1/very big number --> 0)

Answer 33

do you see that the volume of this shape does not grow, but rather approaches pi as m gets bigger

Answer 34

can you explain more.......i typed in 100000000000000 and got 4.4?

Answer 35

you typed that into what?

Answer 36

ti-89

Answer 37

the volume really doesn't grow? its width is growing?

Answer 38

so you are doing the integration numerically?

symbolically, you get this answer
http://www.wolframalpha.com/input/?i=pi+*+int+x%5E%28-2%29+dx+where+x%3D+1+to+m

Answer 39

if you replace m with infinity in the limit, you get

http://www.wolframalpha.com/input/?i=pi+*+int+x%5E%28-2%29+dx+where+x%3D+1+to+inf

Answer 40

what is going on is the y value (1/x) is getting closer and closer to the x-axis
so though the shape is getting taller,the radius y is getting smaller. the two "cancel" each other.