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anonymous
 5 years ago
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
 5 years ago
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

This Question is Closed

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.0noooooooo!!!!!!!!!!......well.......maybe ;)

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0;) helloo life saver xD

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.0the interval is x=[0,1] again

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.0pi {S} [5x5]^2  [5sqrt(x)5]^2 dx ; [0,1]

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.0\[\pi \int\limits_{0}^{1} (25x^2 +25 50x)(25x +25 50\sqrt(x)).dx\]

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0the boundary wouldnt be from 0, 5 ? cuz it intersects at (1,5) i think

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.05*5 = 25; 5*sqrt(5) != 25 :)

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.05*1 = 5 ; 5*sqrt(1) = 5

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.0the x interval is from 0 to 1; th ey interval would be from 0 to 5

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.0but we only need use one interval

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0ohh 5 is the y interval. oops lol ><

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.0http://www.wolframalpha.com/input/?i=pi+*+int%28+ [5x5]^2++[5sqrt%28x%295]^2%29+dx++from+0+to+1

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.0well that cut short lol

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.0copy paste that whole thing itno your address bar to see the results

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.0π ∫ (25x2−75x+50√x) dx is what the integral simplifies to before integrating

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0got 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
 5 years ago
Best ResponseYou've already chosen the best response.0you would plug in 1 for y, and move everything to one side right?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0like you know how for this prob, we were told to rotate the region about the line y = 5 right

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0so you simply plugged in 5 for y into both the equations

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.0yes; so subtract 5 from the functions to get it to the 0 line

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0and moved everything onto one side ?

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

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0gotcha. and you would do the same if we were told to rotate around the line lets say for example, x = 4 right?

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.0if it were x = 4; we would most likely modify the equations to satiisfy a y variable and subtrat 4 from them

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0ooh 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
 5 years ago
Best ResponseYou've already chosen the best response.0weve been focusing on disk methods; but the shell method is just a good for volume of rotation

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.0when 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
 5 years ago
Best ResponseYou've already chosen the best response.0the shell method is pretty cool to imagine; we take areas of flattened tin cans and add them up :)

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

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.0the width of this sheet is the circumference of the bottom circle; 2pi [x]

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.0the area of the sheet is widht * height: 2pi x[f(x)]

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.0add up all the sheets to get the area of the solid...volume of the solid that is

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.0it cleans up the integral alot when youcan do it

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0ooh. 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

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0thank you :D and thank you for all your help, i truly appreciate it :')

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0*off to study i go* ;)
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