Question

For some reason I am getting 64pi/5 instead of 64pi/15, please help!
Graph:http://imgur.com/epnj7
Problem:http://imgur.com/tgGbq

Answer 1

i have to go now - ill come back to you in 10 minures

Answer 2

Thanks, Ill be here

Answer 3

your biggest radius is 2; and your inner radius is 2 - 1/8x^2

Answer 4

we are integrating from 0 to 4; pi (outer^2 - inner^2)

Answer 5

Ahh, ok

Answer 6

I am seeing what you are saying

Answer 7

I have been making this mistake on all of my washer problems

Answer 8

the cylindar created pi 2^2/2 * 4 = 8 pi for the rotation right? now we gouge out the inner

Answer 9

I am not subtracting outer radius - inner radius

Answer 10

I am subtracting inner radius - volume under the lower function

Answer 11

(2 - 1/8x^2)^2:

2 - 1/8x^2
2 - 1/8x^2
----------
4 -1/4x^2
-1/4x^2 + 1/64x^4
--------------------
4 -1/2x^2 +1/64x^4

int(4 -1/2x^2 +1/64x^4)dx from 0 to 4 .... well zero is poitless so do 4 :)

Answer 12

Thanks

Answer 13

pi (4x -x^3/6 +x^5/320) = 128pi/15 if i did it right

Answer 14

did i miss it ? lol

Answer 15

64pi/15

Answer 16

in any case; did you get it right now :)

Answer 17

Try\[\pi \int\limits_{0}^{4}[(2-(1/8)x ^{2})^{2}-(2-\sqrt{x})^{2}]\]

Answer 18

I think so! I def. have to work some problems, i will post another question if i am having trouble :)

Answer 19

R3 is just bound by y=0; y=1/8x^2 and x=0 x=4 right?

Answer 20

i always had a tendency to redraw it at the o axis... but found out that it was just extra work :)