anonymous
  • anonymous
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
Mathematics
  • Stacey Warren - Expert brainly.com
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jamiebookeater
  • jamiebookeater
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anonymous
  • anonymous
i think i got it , integral 2pi ( x - x^2) (6-x) dx from 0 to 1 ?
anonymous
  • anonymous
the part i dont get is the 6 - x
anonymous
  • anonymous
isnt there an inner radius and an outer radius ?

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anonymous
  • anonymous
like with the washer method , integral pi f(x)^2 - pi (g(x)) ^2 , where f(x) > g(x)
anonymous
  • anonymous
where exactly does 6-x come from
anonymous
  • anonymous
volume will be \[2\pi \int\limits_{0}^{1} (5+x)(x-x^2).dx -\pi 5^2 .1\]
anonymous
  • anonymous
i believe thats wrong
anonymous
  • anonymous
integral 2pi ( x - x^2) (6-x) dx from 0 to 1
anonymous
  • anonymous
i'm giving you the graph...
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anonymous
  • anonymous
thats not correct, what do you get as an answer
watchmath
  • watchmath
I agree with cantor set. The 6-x coming from the distance from a slice to the axis of symmetry.
anonymous
  • 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
  • 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
  • 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
  • anonymous
thats wrong too
anonymous
  • anonymous
the answer is 96 pi/5
anonymous
  • anonymous
the question is here http://tutorial.math.lamar.edu/Classes/CalcI/VolumeWithCylinder.aspx
anonymous
  • 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
  • anonymous
oh nevermind
anonymous
  • anonymous
ok small error
anonymous
  • anonymous
but i should have said, use shell method :)
anonymous
  • anonymous
i make it 48pi
anonymous
  • anonymous
shell method stumps me on this problem
watchmath
  • watchmath
Shell method \(\int_0^2 2\pi y (8-y^3)\, dy\)
anonymous
  • anonymous
\[2\pi \int\limits_{x=0}^{8}y (8-x)dy\]
anonymous
  • anonymous
yeah, how come its 8-y^3
anonymous
  • anonymous
use x=y^3
anonymous
  • anonymous
watchman has already given it.....
anonymous
  • anonymous
sorry - i made the same mistake as dipank
anonymous
  • 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
  • anonymous
you need one more boundary...
watchmath
  • watchmath
Because the axis of rotation now is the x-axis (before the rotation axis was parallel to the y-axis)
anonymous
  • anonymous
is there a similiar problem to this revolving about y axis
watchmath
  • 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
  • anonymous
shell method for region y= x^2 bounded by y=0, x=2, revovling about y=-2
watchmath
  • watchmath
\(\int_0^4 2\pi (y+2)(2-\sqrt{y})\,dy\)
anonymous
  • anonymous
thx
anonymous
  • anonymous
whats your trick for shell methods, i have trouble setting them up
anonymous
  • 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
  • anonymous
\[2\pi \int\limits_{0}^{2}x e^{-x^2/2}dx\]
watchmath
  • 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
  • anonymous
i think this will be the ans...
watchmath
  • watchmath
Give it a try Cantorset for you very last problem.
anonymous
  • anonymous
wait, which problem are you explaining , you said the height of the cylinder is bounded by x = 8, for which problem
anonymous
  • 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
  • watchmath
Oh sorry, the height should be 2- sqrt(y)
anonymous
  • anonymous
dont you mean, width of the cylinder?
anonymous
  • anonymous
oh height, if you look at it sideways
watchmath
  • 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
  • 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
  • anonymous
then it would be integral 2pi * y *sqrt y dy ?
anonymous
  • anonymous
from 0 to 4
anonymous
  • 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
  • anonymous
please tell me is the integral correct for the problem y=e^(-x^2/2)
anonymous
  • anonymous
no
anonymous
  • anonymous
for which one? first solve for x
watchmath
  • watchmath
Need to go ... I'll be back in 20 minute(ish)
anonymous
  • anonymous
ok
anonymous
  • anonymous
graph the curve y = e^(-x^2)
anonymous
  • anonymous
and we are revolving about x axis
anonymous
  • anonymous
for the volume bounded by the curve y=e^(-x^2/2), x=2,y=0, around x-axis..
anonymous
  • anonymous
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anonymous
  • anonymous
oh wait, i meant y = e^(-x^2)
anonymous
  • anonymous
i have done a rough graph can you check it...
anonymous
  • anonymous
not the right graph, should look like a bell shape
anonymous
  • anonymous
ok i will try it.......
anonymous
  • anonymous
here http://www.wolframalpha.com/input/?i=graph+e^%28-x^2%29
anonymous
  • anonymous
thanx
anonymous
  • anonymous
its easier to check with disc method, so lets try that
anonymous
  • anonymous
integral pi *e^(-x^2) ^2 from 0 to 2
anonymous
  • anonymous
918800
watchmath
  • 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
  • watchmath
sorry I mean to the right by \(x=\sqrt{-2\ln y}\)
watchmath
  • watchmath
to the left ... :)
anonymous
  • anonymous
for the sake of graphing and checking i changed the problem
anonymous
  • anonymous
y = 4 e^(-x^2/4) (which stretches it horizontally and vertically)
anonymous
  • anonymous
bounded by x = 2 , x = 0 , y =0
anonymous
  • anonymous
sorry i keep changing problem , im checking this with calculator
anonymous
  • anonymous
using the disc method, just to make sure we are doing the shell method right (to check later)
watchmath
  • watchmath
and we rotate about which axis?
anonymous
  • anonymous
disc method is int pi ( 4 e^(-x^2/4) ) ^2 , from 0 to 2 , about x axis
anonymous
  • anonymous
i get 60.13 as my volume
anonymous
  • anonymous
ok shell method ... x= sqrt (- 4 ln (y/4) )
watchmath
  • watchmath
It is not nice using the shell method here. You need to split into two integrals.
anonymous
  • anonymous
so there are two pieces to the shell method int 2pi* y*sqrt (- 4 ln (y/4) ) from 2 to 4 +
anonymous
  • anonymous
+ int 2pi * y* 2 from 0 to 2
watchmath
  • watchmath
0 to 4e^(-1) and from 4e^(-1) to 4
anonymous
  • anonymous
oh
anonymous
  • anonymous
int 2pi * y* 2 from 0 to 4e^-1 + int 2pi* y*sqrt (- 4 ln (y/4) ) from 4e^-1 to 4
anonymous
  • anonymous
answer does not match disc method
anonymous
  • anonymous
ok it does,
anonymous
  • anonymous
wow, what a problem !!!
anonymous
  • 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
  • anonymous
i guess i have problems with upper and lower limits, i confuse them , which ones we are doing
anonymous
  • anonymous
im just doing these problems to be thorough,

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