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anonymous
 5 years ago
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 xaxis
anonymous
 5 years ago
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 xaxis

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amistre64
 5 years ago
Best ResponseYou've already chosen the best response.0ok.... y = sqrt(x) good

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.0we 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?

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.0lol .... ok... I can work with that :)

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0is the graph bounded by x \[\in\] [0, 2]?

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.0if we take one section of this graph, we have a radius = to the value of f(x) at the point x right?

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.0if 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
 5 years ago
Best ResponseYou've already chosen the best response.0lets say the radius = 5; the area of a circle would be equal to pi(r^2) right?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0I understand that, i'm just confused on how to do it with this problem...because I just AM DUMB

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.0if 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
 5 years ago
Best ResponseYou've already chosen the best response.0\[\int\limits_{0}^{2}\pi \sqrt{x}\] That should do it?

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.0what 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
 5 years ago
Best ResponseYou've already chosen the best response.0sorry sqrt(x) is squared for area

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0so its just integral (pi * x) from 0 to 2?

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.0our radius is equal to y = x^(1/2) does that make sense to you?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0you 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
 5 years ago
Best ResponseYou've already chosen the best response.0but to find the integral do you find the anti derviative?

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.0yes... an integral and an antiderivative are the same thing...

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.0beth: whats the antiderivative of x^(1/2)

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0nd then i just plug in 2 and 0 and solve right

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.0actually, 0 will equal 0 so you just have to solve for x=2

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0you dont take integral of sqrt(x)

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0because to find area you have to square the radius

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0sqrt(x) ^ 2 is just x... so you find the integral of pi(x) from 0 to 2

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0but i am trying to find volume not area...or does that even matter?

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.0volume is just 2d area.... add up all the volumes of the slices to get the 3d area.... same thing

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0alright 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
 5 years ago
Best ResponseYou've already chosen the best response.02pi (S) x^(1/2) dx [0,2] 2pi(x^3/2)(2/3) when x = 2 is all you need to find

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0pi (S) (x^1/2)^2 dx from 0 to 2

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0What 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
 5 years ago
Best ResponseYou've already chosen the best response.0so the final answer should be 2pi?

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.0we already went thru what we need to do to get the "area"... so follow along :)

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.04 pi 2^(3/2)  3

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0you guys are telling me two different things...i dont know which one is right haaha

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0for 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
 5 years ago
Best ResponseYou've already chosen the best response.0yea but i am trying to find volume, but wait that doesnt matter right?

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.0the area of each slice is the antiderivative of 2pi(x^(1/2))

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0i thought the area of each slice is the antiderivative of pi((x^1/2)^2)

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.0the antiderivative of 2pi(r) = 2pi(r^2)/2

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.02piR is circumfrence not area

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.0why would it be the antiderivative of pi(x^(1/2)^2)?? that makes no sense

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0R in this case is y = sqrt(x)

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.0rdoshi... you antiderive the circumference of a circle to get the formula for the area of a circle.... its just that simple

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0anti deravative of 2pi R is pi R^2

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.0yes... and the R value is y = x^(1/2) BEFORE you get the antiderivative...

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0that doesn't change anything

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.0it does in my mind :)

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0you can substitute before or after... method of substitution

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.0I perfer before... makes my life easier :)

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0nonetheless you end up to solve for pi * integral of x from [0,2]

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0amistre64 does the equation look like this \[ \int\limits_{2}^{0} \pi x^3/2 divided by 3/\]

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0that should give you (x^2)/2 evaluated from 0 to 2 thus (2^2)/2  (0^2)/2

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0can someone just type out the work from start to finish so i undertand it bplease....:)

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.0the 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
 5 years ago
Best ResponseYou've already chosen the best response.0I know that the area of each slice is (S) 2pi(x^(1/2)) dx ... so ill use that:

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.02pi (2/3) (x^(3/2)) is what I will use to find the volume of rotation...

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0area of each slice A = pi * (Radius = sqrt x)^2

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.04pi (2)^(2/3) 0    = area under the curve 3 3

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.04pi sqrt(8)  = area 3 fliiped my exponent around there by accident...

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.08pi sqrt(2)  should be the area 3

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.0about 11.85 if I did it right, but maybe im wrong... ;)

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0are you normally right...god i hope so

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0for a solenoid with maximum radius 2?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0ok well can you now teach me how to find the volume of the solid form by rotating the region sbout the y axis

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0dude its pretty much the same

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.0put your equation in terms of "y" and solve it the same way

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0when 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
 5 years ago
Best ResponseYou've already chosen the best response.0In your given case, your y = sqrt (x) You are rotating it around xaxis

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0it will give you a solenoid... kinda like a sideways bowl

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0but now it need to be rotated about the y....

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0i'm saying what amistre64 said is not quite right

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0when you have a cross sectional area of a solenoid, it is always a circle

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0and Area of that circle is A = pi * y(x)

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0when you do integral of A from given bounds for x, you get the volume

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0y(x)^2 sorry... i forgot the power in my A equation

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0if you are rotating about Y axis, all you need to do is to put the given equation in x(y) form

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0and evaluate for your range... i.e y belongs to [a, b]

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.0ill admit I could be wrong... so I will go "relook" at my understanding of this :)

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0ok that makes sense but what if i fine the volme of the solid ford b rotating the region about the line y= 2

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.0yeah, 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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