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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 = sqrtx
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 = sqrtx find the volume

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amistre64
 5 years ago
Best ResponseYou've already chosen the best response.0hmmm....we just did one like like this earlier :)

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0hello :)) ya its the exact same one. im a little stuck

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.0y = 5; means that we take the functions and 5 to get them to y=0

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.05x5 and sqrt(x)5 rotated about the x axis

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0yeah, both of those are squared right. then i take the anti derivative of it correct ?

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.05x = sqrt(x) 25x^2 = x 25x^2 x = 0 x(25x  1) = 0 x = 0 and x = 1/25 right?

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.0then our bounds of integration are 0 to 1/25

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0im a little confused now because i thought our bounds were from 0 to 5 from before

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.0before you had 5sqrt(x). Did you typo it here?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0no thats correct but when we first graphed the two equations, werent the bounds from 0 to 5 ?

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

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

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.0when does 5x = sqrt(x) wehn x = 0 and x = 1/25 5/25 = 1/5 and sqrt(1/25) = 1/5 right?

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.0we arent double integrating this are we?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0ohh nooo we arent. sorry sorry. i was mixing it up with something else

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

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

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.0pi [S] [5x5]^2  [sqrt(x)5]^2 dx ; [0,1/25]

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.025x^2 50x +25  [x10sqrt(x) +25] 25x^2 50x +25  x+10sqrt(x) 25 pi [S] [25x^2 51x +10sqrt(x)] dx ; [0,1/25] then right?

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.025x^3 51x^2 20sqrt(x^3)    +  3 2 3

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0and then we plug in the boundaries, and subtract them right?

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.0yes; but the easy thing here is that 0 = 0 so we only determine 1/25 for our answer

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0ah. gotcha. thank you so much for your help. made it a lot clearer for me. thank you :)

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.0dot worry if the volume is negative; just take the absolute value of it then :)
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