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is it solid of revolution problem in calc
you dont say..... ;)
Ok, we first have to find intersections
have you guys ever heard of the disk or washer method to finding volume?
and a shell
first things first; whats our boundaries for integration
thank you for that link :) how do you find the boundaries though?
We look the the intersection point which are (1,5)(0,0)
so would the bounaries be from 0 to 5?
So what method would you use?
the y=5 line seems spurious
hmm, im not sure which method to choose
because im looking through my notes right now, it looks like we could use the disk method
Since you are going to be rotating around an horizontal point(y=5) you are going to use disk method( or washer)
in Some cases you can use either but it is matter convienience
the disk method is adding up all the areas of the circles; [S] 2pi [f(x)]^2 dx - [S] 2pi [g(x)]^2 dx
ohhh....rotating AROUND y=5; haha :) helps to read the question lol
I would advise against blindly memorizing the formula
i have in my notes that the disk method formula is pi (R(x))^2 dx
the formula is jsut the qnqlytic representation of what you are actually accomplishing;
R(x) - 5 in this case
So the bigger circle have radius of 5-5x the smaller circle have radius of 5-5Sqrt[x]
i was combining the shell and the disk ....my mistake
ohh i think i get it. so would it be (5-5sqrt)^2-(5-5x)^2 dx
you simply find the volume of the outer function and cut out the volume of the inner function
I think you have it reversed
[5x-5] ^2 - [5sqrt(x)-5]^2
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
i might be confusing it with something else x_X
sine y = 5; you want to bring the function down to y = 0 therefore: f(x) - 5
since y = 5...yadayada
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
ohhh, i see what you mean now
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 !
youre welcome :)