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

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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

Mathematics
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noooooooo!!!!!!!!!!......well.......maybe ;)
;) helloo life saver xD
the interval is x=[0,1] again

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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
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
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
gotcha. and you would do the same if we were told to rotate around the line lets say for example, x = 4 right?
if it were x = 4; we would most likely modify the equations to satiisfy a y variable and subtrat 4 from them
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?
weve been focusing on disk methods; but the shell method is just a good for volume of rotation
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'
the shell method is pretty cool to imagine; we take areas of flattened tin cans and add them up :)
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?
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
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
Gluck :)
thank you :D and thank you for all your help, i truly appreciate it :')
*off to study i go* ;)

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