Can someone have a look at the attached problem . I get the answer 112pi/15 but the document states its 64pi/15, and I can't see where I've gone wrong.
Graph:http://imgur.com/epnj7
Problem:http://imgur.com/tgGbq

Question

anonymous · Answer

I dont know the washer method, but there is another way of doing it , but getting the whole cylinder and subtracting the volume of the curve above the y axis.

anonymous · Answer

i got the following integral

pi INT[( 2^2 - (2-(1/8)x^2)] between 4 an 0

anonymous · Answer

ok'

anonymous · Answer

wait I have had a little look at these before

anonymous · Answer

I think you might have the integral wrong the wrong way

anonymous · Answer

i 'm not familiar with washer method so i looked it up in a textbook!

anonymous · Answer

oh - well i figured that the R referred to the radius of the cylinder or outer washer that is 2

anonymous · Answer

inner radius is (2 -(1/8)x^2 )  , outer radius is 4

anonymous · Answer

and the second part was the nonshaded area above the curve

anonymous · Answer

exactly

anonymous · Answer

I think you missed a square somewhere

anonymous · Answer

$$= \pi \int\limits_{0}^{4} ( 2^2- ( 2- \frac{1}{8} x^2)^2 dx$$

anonymous · Answer

yes - thats the integral i used

anonymous · Answer

the previous one was a typo

anonymous · Answer

now i calculated that as 112pi / 15