anonymous
  • anonymous
Consider the solid obtained by rotating the region bounded by the given curves about the line y = 5. y = 5x , y = 5 * squareroot of x Find volume.
Mathematics
  • Stacey Warren - Expert brainly.com
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chestercat
  • chestercat
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amistre64
  • amistre64
considered....now what?
anonymous
  • anonymous
find volume
anonymous
  • anonymous
is it solid of revolution problem in calc

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amistre64
  • amistre64
you dont say..... ;)
anonymous
  • anonymous
yup
anonymous
  • anonymous
Ok, we first have to find intersections
anonymous
  • anonymous
have you guys ever heard of the disk or washer method to finding volume?
anonymous
  • anonymous
Yes
amistre64
  • amistre64
and a shell
amistre64
  • amistre64
first things first; whats our boundaries for integration
anonymous
  • anonymous
http://www4d.wolframalpha.com/Calculate/MSP/MSP46219fbd79927442221000038ec20h29g0g1b81?MSPStoreType=image/gif&s=25&w=399&h=188&cdf=Coordinates&cdf=Tooltips
anonymous
  • anonymous
thank you for that link :) how do you find the boundaries though?
anonymous
  • anonymous
We look the the intersection point which are (1,5)(0,0)
anonymous
  • anonymous
so would the bounaries be from 0 to 5?
anonymous
  • anonymous
yes
anonymous
  • anonymous
So what method would you use?
amistre64
  • amistre64
the y=5 line seems spurious
anonymous
  • anonymous
hmm, im not sure which method to choose
anonymous
  • anonymous
because im looking through my notes right now, it looks like we could use the disk method
anonymous
  • anonymous
Since you are going to be rotating around an horizontal point(y=5) you are going to use disk method( or washer)
anonymous
  • anonymous
in Some cases you can use either but it is matter convienience
amistre64
  • amistre64
the disk method is adding up all the areas of the circles; [S] 2pi [f(x)]^2 dx - [S] 2pi [g(x)]^2 dx
amistre64
  • amistre64
ohhh....rotating AROUND y=5; haha :) helps to read the question lol
anonymous
  • anonymous
I would advise against blindly memorizing the formula
anonymous
  • anonymous
i have in my notes that the disk method formula is pi (R(x))^2 dx
amistre64
  • amistre64
the formula is jsut the qnqlytic representation of what you are actually accomplishing;
amistre64
  • amistre64
R(x) - 5 in this case
amistre64
  • amistre64
analytic...
anonymous
  • anonymous
So the bigger circle have radius of 5-5x the smaller circle have radius of 5-5Sqrt[x]
amistre64
  • amistre64
i was combining the shell and the disk ....my mistake
anonymous
  • anonymous
ohh i think i get it. so would it be (5-5sqrt)^2-(5-5x)^2 dx
amistre64
  • amistre64
you simply find the volume of the outer function and cut out the volume of the inner function
anonymous
  • anonymous
I think you have it reversed
amistre64
  • amistre64
[5x-5] ^2 - [5sqrt(x)-5]^2
anonymous
  • anonymous
how come its that way rather than the way i put it? i remember my professor i think tellung us that its the bottom function subtracted from the top one, and in this case, the top function is the 5sqrt of x
anonymous
  • anonymous
i might be confusing it with something else x_X
amistre64
  • amistre64
sine y = 5; you want to bring the function down to y = 0 therefore: f(x) - 5
amistre64
  • amistre64
since y = 5...yadayada
amistre64
  • amistre64
at y=5; the function is out of reach up on a shelf; you want to get it down to where you can work with it on the table.... bring it down to y=0 by subtracting 5 from it
anonymous
  • anonymous
ohhh, i see what you mean now
anonymous
  • anonymous
thank you guys so much for all the help but im running late for class haha so ill try to work this with what you guys helped me with so far. thank you guys times infinityy !
amistre64
  • amistre64
youre welcome :)

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