anonymous
  • anonymous
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
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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chestercat
  • chestercat
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amistre64
  • amistre64
ok.... y = sqrt(x) good
amistre64
  • amistre64
between [0,2]? right
anonymous
  • anonymous
yes

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amistre64
  • 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
  • anonymous
i have no idea....
amistre64
  • amistre64
lol .... ok... I can work with that :)
anonymous
  • anonymous
is the graph bounded by x \[\in\] [0, 2]?
amistre64
  • 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
  • anonymous
\[x \in [0, 2]\] ?
anonymous
  • anonymous
yea
amistre64
  • amistre64
rdoshi, yes
anonymous
  • anonymous
alrighty...
amistre64
  • 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
  • amistre64
lets say the radius = 5; the area of a circle would be equal to pi(r^2) right?
anonymous
  • anonymous
yess
anonymous
  • anonymous
I understand that, i'm just confused on how to do it with this problem...because I just AM DUMB
amistre64
  • 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
  • anonymous
\[\int\limits_{0}^{2}\pi \sqrt{x}\] That should do it?
amistre64
  • 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
  • anonymous
sorry sqrt(x) is squared for area
anonymous
  • anonymous
so its just integral (pi * x) from 0 to 2?
amistre64
  • amistre64
our radius is equal to y = x^(1/2) does that make sense to you?
anonymous
  • 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
  • anonymous
but to find the integral do you find the anti derviative?
amistre64
  • amistre64
yes... an integral and an antiderivative are the same thing...
amistre64
  • amistre64
beth: whats the antiderivative of x^(1/2)
anonymous
  • anonymous
nd then i just plug in 2 and 0 and solve right
amistre64
  • amistre64
actually, 0 will equal 0 so you just have to solve for x=2
anonymous
  • anonymous
you dont take integral of sqrt(x)
anonymous
  • anonymous
because to find area you have to square the radius
anonymous
  • anonymous
sqrt(x) ^ 2 is just x... so you find the integral of pi(x) from 0 to 2
anonymous
  • anonymous
but i am trying to find volume not area...or does that even matter?
amistre64
  • amistre64
volume is just 2d area.... add up all the volumes of the slices to get the 3d area.... same thing
anonymous
  • 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
  • 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
  • amistre64
(4/3)pi 2^(3/2) = ??
anonymous
  • anonymous
pi (S) (x^1/2)^2 dx from 0 to 2
anonymous
  • 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
  • anonymous
so the final answer should be 2pi?
amistre64
  • amistre64
we already went thru what we need to do to get the "area"... so follow along :)
amistre64
  • amistre64
4 pi 2^(3/2) ------------- 3
anonymous
  • anonymous
you guys are telling me two different things...i dont know which one is right haaha
anonymous
  • anonymous
its 2pi
anonymous
  • 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
  • anonymous
yea but i am trying to find volume, but wait that doesnt matter right?
amistre64
  • amistre64
the area of each slice is the antiderivative of 2pi(x^(1/2))
anonymous
  • anonymous
i thought the area of each slice is the antiderivative of pi((x^1/2)^2)
amistre64
  • amistre64
the antiderivative of 2pi(r) = 2pi(r^2)/2
anonymous
  • anonymous
2piR is circumfrence not area
anonymous
  • anonymous
pi R^2 is area
amistre64
  • amistre64
why would it be the antiderivative of pi(x^(1/2)^2)?? that makes no sense
anonymous
  • anonymous
R in this case is y = sqrt(x)
amistre64
  • amistre64
rdoshi... you antiderive the circumference of a circle to get the formula for the area of a circle.... its just that simple
anonymous
  • anonymous
anti deravative of 2pi R is pi R^2
amistre64
  • amistre64
yes... and the R value is y = x^(1/2) BEFORE you get the antiderivative...
anonymous
  • anonymous
that doesn't change anything
amistre64
  • amistre64
it does in my mind :)
anonymous
  • anonymous
you can substitute before or after... method of substitution
amistre64
  • amistre64
I perfer before... makes my life easier :)
anonymous
  • anonymous
nonetheless you end up to solve for pi * integral of x from [0,2]
anonymous
  • anonymous
amistre64 does the equation look like this \[ \int\limits_{2}^{0} \pi x^3/2 divided by 3/\]
anonymous
  • anonymous
that should give you (x^2)/2 evaluated from 0 to 2 thus (2^2)/2 - (0^2)/2
anonymous
  • anonymous
can someone just type out the work from start to finish so i undertand it bplease....:)
amistre64
  • 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
  • amistre64
I know that the area of each slice is (S) 2pi(x^(1/2)) dx ... so ill use that:
amistre64
  • amistre64
2pi (2/3) (x^(3/2)) is what I will use to find the volume of rotation...
anonymous
  • anonymous
area of each slice A = pi * (Radius = sqrt x)^2
anonymous
  • anonymous
thus A = pi*x
amistre64
  • amistre64
4pi (2)^(2/3) 0 ----------- - --- = area under the curve 3 3
amistre64
  • amistre64
4pi sqrt(8) --------- = area 3 fliiped my exponent around there by accident...
amistre64
  • amistre64
8pi sqrt(2) -------- should be the area 3
amistre64
  • amistre64
about 11.85 if I did it right, but maybe im wrong... ;)
anonymous
  • anonymous
are you normally right...god i hope so
anonymous
  • anonymous
for a solenoid with maximum radius 2?
amistre64
  • amistre64
lol ..... me too :)
anonymous
  • 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
  • anonymous
dude its pretty much the same
amistre64
  • amistre64
put your equation in terms of "y" and solve it the same way
anonymous
  • 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
  • anonymous
In your given case, your y = sqrt (x) You are rotating it around x-axis
anonymous
  • anonymous
it will give you a solenoid... kinda like a sideways bowl
anonymous
  • anonymous
but now it need to be rotated about the y....
anonymous
  • anonymous
i'm saying what amistre64 said is not quite right
anonymous
  • anonymous
when you have a cross sectional area of a solenoid, it is always a circle
anonymous
  • anonymous
and Area of that circle is A = pi * y(x)
anonymous
  • anonymous
when you do integral of A from given bounds for x, you get the volume
anonymous
  • anonymous
y(x)^2 sorry... i forgot the power in my A equation
anonymous
  • anonymous
if you are rotating about Y axis, all you need to do is to put the given equation in x(y) form
anonymous
  • anonymous
and evaluate for your range... i.e y belongs to [a, b]
amistre64
  • amistre64
ill admit I could be wrong... so I will go "re-look" at my understanding of this :)
anonymous
  • 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
  • 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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