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

;) helloo life saver xD

the interval is x=[0,1] again

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

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

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

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

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

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

but we only need use one interval

ohh 5 is the y interval. oops lol ><

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

well that cut short lol

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

π ∫ (25x2−75x+50√x) dx is what the integral simplifies to before integrating

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

1 for y?

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

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

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

and moved everything onto one side ?

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

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

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

correct

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

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

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

it cleans up the integral alot when youcan do it

Gluck :)

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

*off to study i go* ;)