Question

find the volume of the region bounded by the graphs y = sqrt(x), y = 0, and x = 6, revolved about the line x = 6

Answer 1

:) again :)

Answer 2

hello! This problem is similar to the one we did together yesterday, but for some reason I am still doing this wrong

Answer 3

I know that I am supposed to do the disk method, and that I need to solve for each equation in terms of y

Answer 4

rewrite y= sqrt(x) in terms of x = f(y)

y = sqrt(x) ; ^2 both side

y^2 = x this we can solve for easily right?

Answer 5

so I have v = pi (the integral from 0 to 6) (y+9)^2 - (9)^2 dx

Answer 6

graph this and see that it is the same

Answer 7

yes

Answer 8

x = 6 is a pain, we want to move it to the x=0 line to examine it; how do we move x = 6 to x=0?

x= 6 ; -6 from both sides

x-6 = 0

do the same for the x=y^2 to keep it in the same position

Answer 9

minus 6 from both sides in the equation I wrote?

Answer 10

x = y^2 ; -6 both sides

x-6 = y^2 -6

Answer 11

whoops---I am revolving around x = 9, sorry about that

Answer 12

brb..

Answer 13

okay

Answer 14

back...

Answer 15

hi

Answer 16

since it 9 we just -9 fro itall

Answer 17

so v = pi (intergral from 0 to 6) (y + 9)^2 - 9^2 ?

Answer 18

move it all to the left by 9 to examine it at x=0; does that make sense?

Its over on a shelf and we want to examine it at the table where we can see it clearly; the table is the origin of the graph where x=0 and y=0, so we move it

Answer 19

So now I am looking at a graph that is shifted backwards by 9 on my calculator

Answer 20

y = sqrt(x) ; change to an equal form to accomodate for the y axis

y^2 = x ; now move it over to the x=0 by subtracting 9

y^2 -9 = x-9 ; the x-9 can be rewritten as just x

Answer 21

sqrt(x +9)

Answer 22

okay so it is y^2 -9 not +9

Answer 23

sqrt(x+9) is good; now reform it

y = sqrt(x+9)

y^2 = x+9

y^2 -9 = x

Answer 24

whats our bounds now: for y=0 to how far?

Answer 25

x = 6....?

Answer 26

I've never seen anyone do transformations of a graph this way..

Answer 27

y = sqrt(9)

y = 3

Answer 28

we have to redo them when we solve for a different variable...keep forgetting

Answer 29

integrate the discs from 0 to 3

Answer 30

okay

Answer 31

pi {S} y^2-9 dy ; [0,3] = volume of revolution

Answer 32

forgot to square the function lol

Answer 33

pi {S} [y^2-9]^2 dy ; [0,3]

Answer 34

so you don't have to subtract another 9?

Answer 35

why would you subtract two 9s?

Answer 36

we only move the graph over to x=0 from x=9 right?

Answer 37

one from the first eqn and one from eqn y = 0?

Answer 38

if i tellyou to move 9 steps to the left; what do you do? move 18?

Answer 39

ohhhh

Answer 40

y=0 is already bounding the volume; no need to adjust that part lol

Answer 41

answer is 648/5?

Answer 42

dunno; lets check..

Answer 43

no, its' not

Answer 44

pi {S} (y^2-9)^2 dy

pi {S} (y^4 -18y^2 +81) dy

pi(y^5/5 -18y^3/3 +81y) at y=3

Answer 45

had that part

Answer 46

and got 648/5, but the homework grader marked it wrong

Answer 47

i get 129.6pi

Answer 48

same answer, and the wrong one. What did we do wrong?

Answer 49

Hang on. You're rotating the graph of y= sqrt(x) from x=6 to x=9 about the line x=9 ? Is that right?

Answer 50

243 27 243
pi( ---- - --- + ---) right?
5 3 1

Answer 51

x = 0 to 9

Answer 52

right, and we are rotating the region about the line x = 9

Answer 53

ilove; retype the problem in its entirety so that we have a clear picture

Answer 54

Find the volumes of the solids generated by revolving the regions bounded by the graphs of the equations about the given lines.
y = sqrt(x)
y = 0
x = 6

about the line x = 9

Answer 55

ohhh... so its a donut shape; i assumed you meant y=0; x=9 about the x=9

Answer 56

oh, nope sorry about that

Answer 57

I don't think it's a doughnut

Answer 58

its a torus alright :)

Answer 59

More like a mountain.

Answer 60

we have to subtract the middle of what we found to get the volume then

Answer 61

oh wait. I'm looking at wrong graph

Answer 62

Small hill. Not a doughnut. There is no hole in the middle.

Answer 63

x=6 to x=9 is empty tho

Answer 64

there is no hole it the middle? isn't the shape a bunt cake?

Answer 65

no

Answer 66

graph y = sqrt(x) from x = 6 to x=9 y = 0 to y = 3

Answer 67

but the region is cut off at x = 6, and we have to get it to revolve about line x = 9

Answer 68

that's where it starts

Answer 69

not where it ends. it doesn't say it's bounded by x=0

Answer 70

hmmm...then how to we caluclate where it begins and ends?

Answer 71

http://www.wolframalpha.com/input/?i=graph+y+%3D+sqrt%28x%29+from+x+%3D+6+to+x%3D9+y+%3D+0+to+y+%3D+3

Rotate that around x=9

Answer 72

If the problem said bounded by x=0, x=6 and y= 0 and y = sqrt(x) I would agree with that picture. It doesn't say x=0.

Answer 73

so using what you just said, the boundries would be?

Answer 74

x = 6, x = 9, y = 0, y = sqrt(x) and rotate around x = 9

Answer 75

then we would double our results right?

Answer 76

no, you have the wrong part of the graph I think.

Answer 77

with this boundary; y = [0,sqrt(6)]

Answer 78

You want to integrate over the white part in your picture rather than the grey part.

Answer 79

what wouuld the equation look like?

Answer 80

\[v = \pi \int\limits_{0}^{3}...\]

Answer 81

You're using disks or shells?

Answer 82

We can only use the disk method, right?

Answer 83

where did you get the boundaries 0 to 3 from , again?

Answer 84

do you have the value for the answer so we can be sure we're doing the right region?

Answer 85

nope----online h/w problem

Answer 86

dumb.

Answer 87

I'm pretty sure I'm reading it right though. For this I would use cylindrical shells

Answer 88

y = sqrt(x); curve of sqrt(x)
y = 0 ; the x axis is y=0
x = 6 ; the left boundary

x=9; axis to spin around

Answer 89

x=6 the right bound

Answer 90

x = 6 is the left bound in my interpretation of the problem, and the axis of rotation would be the right bound.

Answer 91

???? what do you get as your answer?

Answer 92

I thought the entire region was left bound, and the axis is right bound.

Answer 93

like left-right or top-bottom

Answer 94

the region to cut from my original; if im right is:

is the cylindar of r = 3, and height = sqrt(6)

216 sqrt(6) pi
------------
5

Answer 95

okay, this is the right answer. Can you go over it one more time?

Answer 96

http://www.wolframalpha.com/input/?i=%28int%28pi%28y^2-9%29^2%29+dy+from+0+to+sqrt%286%29%29++-+%289sqrt%286%29pi%29+