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.

- anonymous

- katieb

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- amistre64

considered....now what?

- anonymous

find volume

- anonymous

is it solid of revolution problem in calc

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## More answers

- amistre64

you dont say..... ;)

- anonymous

yup

- anonymous

Ok, we first have to find intersections

- anonymous

have you guys ever heard of the disk or washer method to finding volume?

- anonymous

Yes

- amistre64

and a shell

- amistre64

first things first; whats our boundaries for integration

- anonymous

http://www4d.wolframalpha.com/Calculate/MSP/MSP46219fbd79927442221000038ec20h29g0g1b81?MSPStoreType=image/gif&s=25&w=399&h=188&cdf=Coordinates&cdf=Tooltips

- anonymous

thank you for that link :)
how do you find the boundaries though?

- anonymous

We look the the intersection point which are (1,5)(0,0)

- anonymous

so would the bounaries be from 0 to 5?

- anonymous

yes

- anonymous

So what method would you use?

- amistre64

the y=5 line seems spurious

- anonymous

hmm, im not sure which method to choose

- anonymous

because im looking through my notes right now, it looks like we could use the disk method

- anonymous

Since you are going to be rotating around an horizontal point(y=5) you are going to use disk method( or washer)

- anonymous

in Some cases you can use either but it is matter convienience

- 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

ohhh....rotating AROUND y=5; haha :) helps to read the question lol

- anonymous

I would advise against blindly memorizing the formula

- anonymous

i have in my notes that the disk method formula is pi (R(x))^2 dx

- amistre64

the formula is jsut the qnqlytic representation of what you are actually accomplishing;

- amistre64

R(x) - 5 in this case

- amistre64

analytic...

- anonymous

So the bigger circle have radius of 5-5x
the smaller circle have radius of 5-5Sqrt[x]

- amistre64

i was combining the shell and the disk ....my mistake

- anonymous

ohh i think i get it. so would it be (5-5sqrt)^2-(5-5x)^2 dx

- amistre64

you simply find the volume of the outer function and cut out the volume of the inner function

- anonymous

I think you have it reversed

- amistre64

[5x-5] ^2 - [5sqrt(x)-5]^2

- 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

i might be confusing it with something else x_X

- amistre64

sine y = 5; you want to bring the function down to y = 0 therefore:
f(x) - 5

- amistre64

since y = 5...yadayada

- 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

ohhh, i see what you mean now

- 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

youre welcome :)

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