Consider the solid obtained by rotating the region bounded by the given curves about the line y = 5.
y = 5x , y = 5sqrt(x)
find the volume

- anonymous

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- amistre64

noooooooo!!!!!!!!!!......well.......maybe ;)

- anonymous

;) helloo life saver xD

- amistre64

the interval is x=[0,1] again

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## More answers

- amistre64

pi {S} [5x-5]^2 - [5sqrt(x)-5]^2 dx ; [0,1]

- amistre64

\[\pi \int\limits_{0}^{1} (25x^2 +25 -50x)-(25x +25 -50\sqrt(x)).dx\]

- anonymous

the boundary wouldnt be from 0, 5 ? cuz it intersects at (1,5) i think

- amistre64

5*5 = 25; 5*sqrt(5) != 25 :)

- amistre64

5*1 = 5 ; 5*sqrt(1) = 5

- amistre64

the x interval is from 0 to 1; th ey interval would be from 0 to 5

- amistre64

but we only need use one interval

- anonymous

ohh 5 is the y interval. oops lol ><

- amistre64

http://www.wolframalpha.com/input/?i=pi+*+int%28+[5x-5]^2+-+[5sqrt%28x%29-5]^2%29+dx++from+0+to+1

- amistre64

well that cut short lol

- amistre64

copy paste that whole thing itno your address bar to see the results

- amistre64

Ï€ âˆ« (25x2âˆ’75x+50âˆšx) dx is what the integral simplifies to before integrating

- anonymous

got it :) i have one quick question. when youre given a line to rotate around, like lets say y = 1, and youre given two equations

- anonymous

you would plug in 1 for y, and move everything to one side right?

- amistre64

1 for y?

- anonymous

like you know how for this prob, we were told to rotate the region about the line y = 5 right

- anonymous

so you simply plugged in 5 for y into both the equations

- amistre64

yes; so subtract 5 from the functions to get it to the 0 line

- anonymous

and moved everything onto one side ?

- amistre64

y= 5x -5
y = 5sqrt(x) -5

- anonymous

gotcha. and you would do the same if we were told to rotate around the line lets say for example, x = 4 right?

- amistre64

if it were x = 4; we would most likely modify the equations to satiisfy a y variable and subtrat 4 from them

- anonymous

ooh okay, i understand now :) sorry i have another question xD when it says to rotate around lets say the y or x axis, how would you go about solving for that type of question?

- amistre64

weve been focusing on disk methods; but the shell method is just a good for volume of rotation

- amistre64

when we rotate around y; have your equations and interval ready for 'x'
when we rotate around x; just have the equations and interal in terms of 'y'

- amistre64

the shell method is pretty cool to imagine; we take areas of flattened tin cans and add them up :)

- amistre64

imagine a tin can; it has a height of" f(x)
when we cut it up the side and spread it out we have the area of a flat rectangular sheet right?

- anonymous

correct

- amistre64

the width of this sheet is the circumference of the bottom circle; 2pi [x]

- amistre64

the area of the sheet is widht * height: 2pi x[f(x)]

- amistre64

add up all the sheets to get the area of the solid...volume of the solid that is

- amistre64

it cleans up the integral alot when youcan do it

- anonymous

ooh. i should practice the shell method xD im so used to squaring the big R and little r for these volume type of problems lol
i have one last exam tomorrow and a final next week for calculus. wish me luck ! hehe xD

- amistre64

Gluck :)

- anonymous

thank you :D and thank you for all your help, i truly appreciate it :')

- anonymous

*off to study i go* ;)

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