Question

consider the region bounded by the graphs of f(x)=x^1/2, y=0, and x=2. find the volume of the soild formed by rotating the region about the x-axis

Answer 1

ok.... y = sqrt(x) good

Answer 2

between [0,2]? right

Answer 3

yes

Answer 4

we find an area of rotation by taking the radius and adding it up as we go thru the interval

(S) 2pi(f(x)) dx

or:

2pi (S) f(x) dx ... right?

Answer 5

i have no idea....

Answer 6

lol .... ok... I can work with that :)

Answer 7

is the graph bounded by x \[\in\] [0, 2]?

Answer 8

if we take one section of this graph, we have a radius = to the value of f(x) at the point x right?

Answer 9

\[x \in [0, 2]\] ?

Answer 10

yea

Answer 11

rdoshi, yes

Answer 12

alrighty...

Answer 13

if we take this radius and integrate 2pi(r) we will get the area of this "slice"

lets use an example we are familiar with ok?

Answer 14

lets say the radius = 5; the area of a circle would be equal to pi(r^2) right?

Answer 15

yess

Answer 16

I understand that, i'm just confused on how to do it with this problem...because I just AM DUMB

Answer 17

if we take the Circumference of a circle and "integrate" it we get this:

(S) 2pi(r) dr

2pi (S) r dr

2pi (r^2/2) = pi(r^2) which gives us the area of our circle.... 25pi

Answer 18

\[\int\limits_{0}^{2}\pi \sqrt{x}\]
That should do it?

Answer 19

what we do by integrating along an interval is taking each "slice" finding its area and then adding up all the slices to get the whole volume

Answer 20

sorry sqrt(x) is squared for area

Answer 21

so its just integral (pi * x) from 0 to 2?

Answer 22

our radius is equal to y = x^(1/2)

does that make sense to you?

Answer 23

you find the equation for the cross sectional area of the solenoid that is formed by rotating the graph
which is A(x) = pi (sqrt x)^2
then you just find the integral of that area along x for the volume

Answer 24

but to find the integral do you find the anti derviative?

Answer 25

yes... an integral and an antiderivative are the same thing...

Answer 26

beth: whats the antiderivative of x^(1/2)

Answer 27

nd then i just plug in 2 and 0 and solve right

Answer 28

actually, 0 will equal 0 so you just have to solve for x=2

Answer 29

you dont take integral of sqrt(x)

Answer 30

because to find area you have to square the radius

Answer 31

sqrt(x) ^ 2 is just x... so you find the integral of pi(x) from 0 to 2

Answer 32

but i am trying to find volume not area...or does that even matter?

Answer 33

volume is just 2d area.... add up all the volumes of the slices to get the 3d area.... same thing

Answer 34

alright here's how it works.... integrating area will give you volume.
in order to find area A = pi radius^2...
radius in this case is sqrt(x)

Answer 35

2pi (S) x^(1/2) dx [0,2]

2pi(x^3/2)(2/3) when x = 2 is all you need to find

Answer 36

(4/3)pi 2^(3/2) = ??

Answer 37

pi (S) (x^1/2)^2 dx from 0 to 2

Answer 38

What rdoshi said. Your should only need integrals for this question. by using the formula below it will find you the volume.

when rotating around x:
\[V = \pi \int\limits_{b}^{a} y^{2} dx\]
[therefore this means f(x) is squared]

when rotating around the y:
\[V = \pi \int\limits_{b}^{a} x^{2} dy\]
[therefore this means you make x the subject of the function and then square it]

Answer 39

so the final answer should be 2pi?

Answer 40

we already went thru what we need to do to get the "area"... so follow along :)

Answer 41

4 pi 2^(3/2)
-------------
3

Answer 42

you guys are telling me two different things...i dont know which one is right haaha

Answer 43

its 2pi

Answer 44

for area you just exclude pi from the formula while taking into account that if the function goes into a negative area you use absolute values (seeing you can't have negative an area).

Answer 45

yea but i am trying to find volume, but wait that doesnt matter right?

Answer 46

the area of each slice is the antiderivative of 2pi(x^(1/2))

Answer 47

i thought the area of each slice is the antiderivative of pi((x^1/2)^2)

Answer 48

the antiderivative of 2pi(r) = 2pi(r^2)/2

Answer 49

2piR is circumfrence not area

Answer 50

pi R^2 is area

Answer 51

why would it be the antiderivative of pi(x^(1/2)^2)?? that makes no sense

Answer 52

R in this case is y = sqrt(x)

Answer 53

rdoshi... you antiderive the circumference of a circle to get the formula for the area of a circle.... its just that simple

Answer 54

anti deravative of 2pi R is pi R^2

Answer 55

yes... and the R value is y = x^(1/2) BEFORE you get the antiderivative...

Answer 56

that doesn't change anything

Answer 57

it does in my mind :)

Answer 58

you can substitute before or after... method of substitution

Answer 59

I perfer before... makes my life easier :)

Answer 60

nonetheless you end up to solve for pi * integral of x from [0,2]

Answer 61

amistre64 does the equation look like this \[ \int\limits_{2}^{0} \pi x^3/2 divided by 3/\]

Answer 62

that should give you (x^2)/2 evaluated from 0 to 2
thus (2^2)/2 - (0^2)/2

Answer 63

can someone just type out the work from start to finish so i undertand it bplease....:)

Answer 64

the volume of a cone is 1/3 basearea times h.... because the "3" in 1/3 comes from the antiderivative of the equation for the line.... never tried it the other way around tho

Answer 65

I know that the area of each slice is (S) 2pi(x^(1/2)) dx ... so ill use that:

Answer 66

2pi (2/3) (x^(3/2)) is what I will use to find the volume of rotation...

Answer 67

area of each slice A = pi * (Radius = sqrt x)^2

Answer 68

thus A = pi*x

Answer 69

4pi (2)^(2/3) 0
----------- - --- = area under the curve
3 3

Answer 70

4pi sqrt(8)
--------- = area
3

fliiped my exponent around there by accident...

Answer 71

8pi sqrt(2)
-------- should be the area
3

Answer 72

about 11.85 if I did it right, but maybe im wrong... ;)

Answer 73

are you normally right...god i hope so

Answer 74

for a solenoid with maximum radius 2?

Answer 75

lol ..... me too :)

Answer 76

ok well can you now teach me how to find the volume of the solid form by rotating the region sbout the y axis

Answer 77

dude its pretty much the same

Answer 78

put your equation in terms of "y" and solve it the same way

Answer 79

when you rotate a graph, and want to find volume, you take the cross section of the 3D solid formed and find the equation for that section. Then you integrate that equation for cross section.

Answer 80

In your given case, your y = sqrt (x)
You are rotating it around x-axis

Answer 81

it will give you a solenoid... kinda like a sideways bowl

Answer 82

but now it need to be rotated about the y....

Answer 83

i'm saying what amistre64 said is not quite right

Answer 84

when you have a cross sectional area of a solenoid, it is always a circle

Answer 85

and Area of that circle is A = pi * y(x)

Answer 86

when you do integral of A from given bounds for x, you get the volume

Answer 87

y(x)^2 sorry... i forgot the power in my A equation

Answer 88

if you are rotating about Y axis, all you need to do is to put the given equation in x(y) form

Answer 89

and evaluate for your range... i.e y belongs to [a, b]

Answer 90

ill admit I could be wrong... so I will go "re-look" at my understanding of this :)

Answer 91

ok that makes sense but what if i fine the volme of the solid ford b rotating the region about the line y= -2

Answer 92

yeah, I was mistaken.... apparently I was confusing a couple a methods...

for instance, take the line y=5 on the interval of [0,10]. this should produce a volume of 250pi. But if I do it the way I had in my head I get:

2pi (S) 5 dx

2pi(5x) -> 2pi5(10) = 100pi....

So I was wrong.....