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hmmm....we just did one like like this earlier :)
hello :)) ya its the exact same one. im a little stuck
y = 5; means that we take the functions and -5 to get them to y=0
5x-5 and sqrt(x)-5 rotated about the x axis
yeah, both of those are squared right. then i take the anti derivative of it correct ?
5x = sqrt(x) 25x^2 = x 25x^2 -x = 0 x(25x - 1) = 0 x = 0 and x = 1/25 right?
then our bounds of integration are 0 to 1/25
im a little confused now because i thought our bounds were from 0 to 5 from before
before you had 5sqrt(x). Did you typo it here?
no thats correct but when we first graphed the two equations, werent the bounds from 0 to 5 ?
cant recall, been doing alot of them lately :) but the steps are the same... i these equations are good; then the results will be x = 0 and x = 1/25
then i do the same thing for 5x - 5 5x = 5 x = 1 so the bounds are from 0 to 1 for that part of the integral?
when does 5x = sqrt(x) wehn x = 0 and x = 1/25 5/25 = 1/5 and sqrt(1/25) = 1/5 right?
we arent double integrating this are we?
ohh nooo we arent. sorry sorry. i was mixing it up with something else
now we want to add up all the standard areas thru integration; and the area of each circle we generate by spinning is: pi [f(x)-5]^2 right?
we can do each function seperately and subtract them in the end; or do them together. pi [f(x)-5]^2 - pi[g(x)-5]^2 will give us the total volume then
pi [S] [5x-5]^2 - [sqrt(x)-5]^2 dx ; [0,1/25]
25x^2 -50x +25 - [x-10sqrt(x) +25] 25x^2 -50x +25 - x+10sqrt(x) -25 pi [S] [25x^2 -51x +10sqrt(x)] dx ; [0,1/25] then right?
25x^3 51x^2 20sqrt(x^3) ------ - ------- + ---------- 3 2 3
and then we plug in the boundaries, and subtract them right?
yes; but the easy thing here is that 0 = 0 so we only determine 1/25 for our answer
ah. gotcha. thank you so much for your help. made it a lot clearer for me. thank you :)
youre welcome :)
dot worry if the volume is negative; just take the absolute value of it then :)