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

- anonymous

- schrodinger

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- amistre64

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

- amistre64

between [0,2]? right

- anonymous

yes

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## More answers

- amistre64

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?

- anonymous

i have no idea....

- amistre64

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

- anonymous

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

- amistre64

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

- anonymous

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

- anonymous

yea

- amistre64

rdoshi, yes

- anonymous

alrighty...

- amistre64

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?

- amistre64

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

- anonymous

yess

- anonymous

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

- amistre64

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

- anonymous

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

- amistre64

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

- anonymous

sorry sqrt(x) is squared for area

- anonymous

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

- amistre64

our radius is equal to y = x^(1/2)
does that make sense to you?

- anonymous

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

- anonymous

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

- amistre64

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

- amistre64

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

- anonymous

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

- amistre64

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

- anonymous

you dont take integral of sqrt(x)

- anonymous

because to find area you have to square the radius

- anonymous

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

- anonymous

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

- amistre64

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

- anonymous

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)

- amistre64

2pi (S) x^(1/2) dx [0,2]
2pi(x^3/2)(2/3) when x = 2 is all you need to find

- amistre64

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

- anonymous

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

- anonymous

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]

- anonymous

so the final answer should be 2pi?

- amistre64

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

- amistre64

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

- anonymous

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

- anonymous

its 2pi

- anonymous

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).

- anonymous

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

- amistre64

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

- anonymous

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

- amistre64

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

- anonymous

2piR is circumfrence not area

- anonymous

pi R^2 is area

- amistre64

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

- anonymous

R in this case is y = sqrt(x)

- amistre64

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

- anonymous

anti deravative of 2pi R is pi R^2

- amistre64

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

- anonymous

that doesn't change anything

- amistre64

it does in my mind :)

- anonymous

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

- amistre64

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

- anonymous

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

- anonymous

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

- anonymous

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

- anonymous

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

- amistre64

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

- amistre64

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

- amistre64

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

- anonymous

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

- anonymous

thus A = pi*x

- amistre64

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

- amistre64

4pi sqrt(8)
--------- = area
3
fliiped my exponent around there by accident...

- amistre64

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

- amistre64

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

- anonymous

are you normally right...god i hope so

- anonymous

for a solenoid with maximum radius 2?

- amistre64

lol ..... me too :)

- anonymous

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

- anonymous

dude its pretty much the same

- amistre64

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

- anonymous

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.

- anonymous

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

- anonymous

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

- anonymous

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

- anonymous

i'm saying what amistre64 said is not quite right

- anonymous

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

- anonymous

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

- anonymous

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

- anonymous

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

- anonymous

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

- anonymous

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

- amistre64

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

- anonymous

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

- amistre64

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.....

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