## 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

1. amistre64

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

2. anonymous

;) helloo life saver xD

3. amistre64

the interval is x=[0,1] again

4. amistre64

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

5. amistre64

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

6. anonymous

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

7. amistre64

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

8. amistre64

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

9. amistre64

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

10. amistre64

but we only need use one interval

11. anonymous

ohh 5 is the y interval. oops lol ><

12. amistre64

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

13. amistre64

well that cut short lol

14. amistre64

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

15. amistre64

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

16. 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

17. anonymous

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

18. amistre64

1 for y?

19. anonymous

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

20. anonymous

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

21. amistre64

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

22. anonymous

and moved everything onto one side ?

23. amistre64

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

24. anonymous

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

25. amistre64

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

26. 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?

27. amistre64

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

28. 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'

29. amistre64

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

30. 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?

31. anonymous

correct

32. amistre64

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

33. amistre64

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

34. amistre64

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

35. amistre64

it cleans up the integral alot when youcan do it

36. 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

37. amistre64

Gluck :)

38. anonymous

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

39. anonymous

*off to study i go* ;)