Question

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

Answer 1

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

Answer 2

the part i dont get is the 6 - x

Answer 3

isnt there an inner radius and an outer radius ?

Answer 4

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

Answer 5

where exactly does 6-x come from

Answer 6

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

Answer 7

i believe thats wrong

Answer 8

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

Answer 9

i'm giving you the graph...

Answer 10

thats not correct, what do you get as an answer

Answer 11

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

Answer 12

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.

Answer 13

, 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.

Answer 14

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

Answer 15

thats wrong too

Answer 16

the answer is 96 pi/5

Answer 17

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

Answer 18

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

Answer 19

oh nevermind

Answer 20

ok small error

Answer 21

but i should have said, use shell method :)

Answer 22

i make it 48pi

Answer 23

shell method stumps me on this problem

Answer 24

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

Answer 25

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

Answer 26

yeah, how come its 8-y^3

Answer 27

use x=y^3

Answer 28

watchman has already given it.....

Answer 29

sorry - i made the same mistake as dipank

Answer 30

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

Answer 31

you need one more boundary...

Answer 32

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

Answer 33

is there a similiar problem to this revolving about y axis

Answer 34

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

Answer 35

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

Answer 36

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

Answer 37

thx

Answer 38

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

Answer 39

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

Answer 40

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

Answer 41

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)).

Answer 42

i think this will be the ans...

Answer 43

Give it a try Cantorset for you very last problem.

Answer 44

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

Answer 45

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?

Answer 46

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

Answer 47

dont you mean, width of the cylinder?

Answer 48

oh height, if you look at it sideways

Answer 49

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).

Answer 50

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

Answer 51

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

Answer 52

from 0 to 4

Answer 53

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

Answer 54

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

Answer 55

no

Answer 56

for which one? first solve for x

Answer 57

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

Answer 58

ok

Answer 59

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

Answer 60

and we are revolving about x axis

Answer 61

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

Answer 62

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

Answer 63

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

Answer 64

not the right graph, should look like a bell shape

Answer 65

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

Answer 66

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

Answer 67

thanx

Answer 68

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

Answer 69

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

Answer 70

918800

Answer 71

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 \(x=\sqrt{-\ln y}\).

Answer 72

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

Answer 73

to the left ... :)

Answer 74

for the sake of graphing and checking i changed the problem

Answer 75

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

Answer 76

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

Answer 77

sorry i keep changing problem , im checking this with calculator

Answer 78

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

Answer 79

and we rotate about which axis?

Answer 80

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

Answer 81

i get 60.13 as my volume

Answer 82

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

Answer 83

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

Answer 84

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

Answer 85

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

Answer 86

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

Answer 87

oh

Answer 88

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

Answer 89

answer does not match disc method

Answer 90

ok it does,

Answer 91

wow, what a problem !!!

Answer 92

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

Answer 93

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

Answer 94

im just doing these problems to be thorough,