Find volume of revolution for region bounded by y = x^2, y = x , in first quadrant, revolve it about the line x = 6 using shell method

- anonymous

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- anonymous

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

- anonymous

the part i dont get is the 6 - x

- anonymous

isnt there an inner radius and an outer radius ?

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## More answers

- anonymous

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

- anonymous

where exactly does 6-x come from

- anonymous

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

- anonymous

i believe thats wrong

- anonymous

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

- anonymous

i'm giving you the graph...

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- anonymous

thats not correct, what do you get as an answer

- watchmath

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

- anonymous

first find the volume made by revolving only y=x, then subtract the volume made by y=x^2. and that also due to the cylinder of radius 5.

- anonymous

, watchmath, what about this question
Determine the volume of the solid obtained by rotating the region bounded by x^(1/3), x=8, y=0, about the x-axis.

- anonymous

\[\pi \int\limits_{x=0}^{8}y^2 dx \] where y=x^1/3
\[\pi \int\limits_{0}^{8}x ^{2/3}dx =3\pi 8^{5/3}/2=48\pi\]

- anonymous

thats wrong too

- anonymous

the answer is 96 pi/5

- anonymous

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

- anonymous

sorry i did a mistake in doing the integration...\[\pi \int\limits_{0}^{8}x ^{2/3}dx=3\pi (x^{5/3}|_{0}^{8})/5=3\pi 8^{5/3}/5\]\[=3\pi 2^5 /5 =96\pi /5\]

- anonymous

oh nevermind

- anonymous

ok small error

- anonymous

but i should have said, use shell method :)

- anonymous

i make it 48pi

- anonymous

shell method stumps me on this problem

- watchmath

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

- anonymous

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

- anonymous

yeah, how come its 8-y^3

- anonymous

use x=y^3

- anonymous

watchman has already given it.....

- anonymous

sorry - i made the same mistake as dipank

- anonymous

how come normally we dont do 8 - y^3 , like shell method for region bounded by y = x^2 , x axis, y axis

- anonymous

you need one more boundary...

- watchmath

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

- anonymous

is there a similiar problem to this revolving about y axis

- watchmath

Assuming all the information are the same but now we rotate about the y-axis then the volume is given by
\(2\pi\int_0^8 x\cdot x^{1/3}\, dx\)

- anonymous

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

- watchmath

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

- anonymous

thx

- anonymous

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

- anonymous

whats the shell method for region bounded by y = e^(-x^2) , x = 2, y = 0. revolved about x axis . only need to set up

- anonymous

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

- watchmath

Well there is not trick. First make a slice that is parallel to the axis of symmetry. In your last problem we slice along the y-axis. Then imagine that you rotate that slice. You will have a cyllinder. Try to find the radius of the cyllinder. If the slice is of distance y from the x-axis, then it is of distance y+2 from the axis of rotation (y=-2). So the circumference of the base of the cyllinder is 2pi(y+2). Now the height of the cyllinder is bounded to the right by x=8 and to the left by x= sqrt(y). So the height is (8-sqrt(y)).

- anonymous

i think this will be the ans...

- watchmath

Give it a try Cantorset for you very last problem.

- anonymous

wait, which problem are you explaining , you said the height of the cylinder is bounded by x = 8, for which problem

- anonymous

So the circumference of the base of the cyllinder is 2pi(y+2). Now the height of the cyllinder is bounded to the right by x=8 and to the left by x= sqrt(y). So the height is (8-sqrt(y)). for which problem was this?

- watchmath

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

- anonymous

dont you mean, width of the cylinder?

- anonymous

oh height, if you look at it sideways

- watchmath

Try this one. Find the volume of solid resulted from rotating the region bounded by the circle (x-2)^2+y^2=1 around the y-axis (the solid is a torus).

- anonymous

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

- anonymous

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

- anonymous

from 0 to 4

- anonymous

for the second problem the height is same for all the shells and it is 2 and the radius is equal to y so volume will \[2\pi \int\limits_{y=0}^{1} y *2dy\]be

- anonymous

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

- anonymous

no

- anonymous

for which one? first solve for x

- watchmath

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

- anonymous

ok

- anonymous

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

- anonymous

and we are revolving about x axis

- anonymous

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

- anonymous

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- anonymous

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

- anonymous

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

- anonymous

not the right graph, should look like a bell shape

- anonymous

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

- anonymous

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

- anonymous

thanx

- anonymous

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

- anonymous

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

- anonymous

918800

- watchmath

If you want to use the cyllindrical shell to the e^(-x^2/2) problem, then remember we need to make a slice parallel to the x-axis. This slice is bounded to the right by x=2 and to the left by \(x=\sqrt{-\ln y}\).

- watchmath

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

- watchmath

to the left ... :)

- anonymous

for the sake of graphing and checking i changed the problem

- anonymous

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

- anonymous

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

- anonymous

sorry i keep changing problem , im checking this with calculator

- anonymous

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

- watchmath

and we rotate about which axis?

- anonymous

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

- anonymous

i get 60.13 as my volume

- anonymous

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

- watchmath

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

- anonymous

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

- anonymous

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

- watchmath

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

- anonymous

oh

- anonymous

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

- anonymous

answer does not match disc method

- anonymous

ok it does,

- anonymous

wow, what a problem !!!

- anonymous

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

- anonymous

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

- anonymous

im just doing these problems to be thorough,

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