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i think i got it , integral 2pi ( x - x^2) (6-x) dx from 0 to 1 ?

the part i dont get is the 6 - x

isnt there an inner radius and an outer radius ?

like with the washer method , integral pi f(x)^2 - pi (g(x)) ^2 , where f(x) > g(x)

where exactly does 6-x come from

volume will be \[2\pi \int\limits_{0}^{1} (5+x)(x-x^2).dx -\pi 5^2 .1\]

i believe thats wrong

integral 2pi ( x - x^2) (6-x) dx from 0 to 1

i'm giving you the graph...

thats not correct, what do you get as an answer

I agree with cantor set. The 6-x coming from the distance from a slice to the axis of symmetry.

thats wrong too

the answer is 96 pi/5

the question is here http://tutorial.math.lamar.edu/Classes/CalcI/VolumeWithCylinder.aspx

oh nevermind

ok small error

but i should have said, use shell method :)

i make it 48pi

shell method stumps me on this problem

Shell method
\(\int_0^2 2\pi y (8-y^3)\, dy\)

\[2\pi \int\limits_{x=0}^{8}y (8-x)dy\]

yeah, how come its 8-y^3

use x=y^3

watchman has already given it.....

sorry - i made the same mistake as dipank

you need one more boundary...

Because the axis of rotation now is the x-axis (before the rotation axis was parallel to the y-axis)

is there a similiar problem to this revolving about y axis

shell method for region y= x^2 bounded by y=0, x=2, revovling about y=-2

\(\int_0^4 2\pi (y+2)(2-\sqrt{y})\,dy\)

thx

whats your trick for shell methods, i have trouble setting them up

\[2\pi \int\limits_{0}^{2}x e^{-x^2/2}dx\]

i think this will be the ans...

Give it a try Cantorset for you very last problem.

Oh sorry, the height should be 2- sqrt(y)

dont you mean, width of the cylinder?

oh height, if you look at it sideways

ok one sec, before i do that one. what if i have a region bounded y = x^2 , y=4, x = 2, about x axis

then it would be integral 2pi * y *sqrt y dy ?

from 0 to 4

please tell me is the integral correct for the problem y=e^(-x^2/2)

no

for which one? first solve for x

Need to go ... I'll be back in 20 minute(ish)

ok

graph the curve y = e^(-x^2)

and we are revolving about x axis

for the volume bounded by the curve y=e^(-x^2/2), x=2,y=0, around x-axis..

oh wait, i meant
y = e^(-x^2)

i have done a rough graph can you check it...

not the right graph, should look like a bell shape

ok i will try it.......

here http://www.wolframalpha.com/input/?i=graph+e^%28-x^2%29

thanx

its easier to check with disc method, so lets try that

integral pi *e^(-x^2) ^2 from 0 to 2

918800

sorry I mean to the right by \(x=\sqrt{-2\ln y}\)

to the left ... :)

for the sake of graphing and checking i changed the problem

y = 4 e^(-x^2/4) (which stretches it horizontally and vertically)

bounded by x = 2 , x = 0 , y =0

sorry i keep changing problem , im checking this with calculator

using the disc method, just to make sure we are doing the shell method right (to check later)

and we rotate about which axis?

disc method is int pi ( 4 e^(-x^2/4) ) ^2 , from 0 to 2 , about x axis

i get 60.13 as my volume

ok shell method ... x= sqrt (- 4 ln (y/4) )

It is not nice using the shell method here. You need to split into two integrals.

so there are two pieces to the shell method
int 2pi* y*sqrt (- 4 ln (y/4) ) from 2 to 4 +

+ int 2pi * y* 2 from 0 to 2

0 to 4e^(-1) and from 4e^(-1) to 4

oh

int 2pi * y* 2 from 0 to 4e^-1 + int 2pi* y*sqrt (- 4 ln (y/4) ) from 4e^-1 to 4

answer does not match disc method

ok it does,

wow, what a problem !!!

i guess i have problems with upper and lower limits, i confuse them , which ones we are doing

im just doing these problems to be thorough,