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

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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

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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 ?

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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...
1 Attachment
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.
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.
, 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.
\[\pi \int\limits_{x=0}^{8}y^2 dx \] where y=x^1/3 \[\pi \int\limits_{0}^{8}x ^{2/3}dx =3\pi 8^{5/3}/2=48\pi\]
thats wrong too
the answer is 96 pi/5
the question is here http://tutorial.math.lamar.edu/Classes/CalcI/VolumeWithCylinder.aspx
sorry i did a mistake in doing the integration...\[\pi \int\limits_{0}^{8}x ^{2/3}dx=3\pi (x^{5/3}|_{0}^{8})/5=3\pi 8^{5/3}/5\]\[=3\pi 2^5 /5 =96\pi /5\]
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
how come normally we dont do 8 - y^3 , like shell method for region bounded by y = x^2 , x axis, y axis
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
Assuming all the information are the same but now we rotate about the y-axis then the volume is given by \(2\pi\int_0^8 x\cdot x^{1/3}\, dx\)
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
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
\[2\pi \int\limits_{0}^{2}x e^{-x^2/2}dx\]
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)).
i think this will be the ans...
Give it a try Cantorset for you very last problem.
wait, which problem are you explaining , you said the height of the cylinder is bounded by x = 8, for which problem
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?
Oh sorry, the height should be 2- sqrt(y)
dont you mean, width of the cylinder?
oh height, if you look at it sideways
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).
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
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
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..
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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
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 \(x=\sqrt{-\ln y}\).
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 !!!
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
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,

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