Question

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

Answer 1

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

Answer 2

;) helloo life saver xD

Answer 3

the interval is x=[0,1] again

Answer 4

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

Answer 5

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

Answer 6

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

Answer 7

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

Answer 8

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

Answer 9

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

Answer 10

but we only need use one interval

Answer 11

ohh 5 is the y interval. oops lol ><

Answer 12

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

Answer 13

well that cut short lol

Answer 14

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

Answer 15

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

Answer 16

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

Answer 17

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

Answer 18

1 for y?

Answer 19

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

Answer 20

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

Answer 21

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

Answer 22

and moved everything onto one side ?

Answer 23

y= 5x -5

y = 5sqrt(x) -5

Answer 24

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

Answer 25

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

Answer 26

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?

Answer 27

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

Answer 28

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'

Answer 29

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

Answer 30

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?

Answer 31

correct

Answer 32

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

Answer 33

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

Answer 34

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

Answer 35

it cleans up the integral alot when youcan do it

Answer 36

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

Answer 37

Gluck :)

Answer 38

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

Answer 39

*off to study i go* ;)