Question

find the area of the graph bounded by y=2x^2 , y=0 , and x=2 , revolved around the line x=2

Answer 1

Area? You mean surface area of the solid of revolution? Or do you want the volume?

Answer 2

I got \[(\frac{ 256\sqrt{2}-192 }{ 12 })\]

Answer 3

the volume I am sorry your right

Answer 4

always draw it first

Answer 5

Did you use disks or rings method?

Answer 6

I would recommend shell methid

Answer 7

method

Answer 8

I set up the problem like this\[V=\int\limits_{0}^{8}2^2 -[2-(\sqrt{\frac{ y }{ 2 }})^2 dy\] and I am suppose to use the shell method

Answer 9

Can you check my integral, Turing? I think that's right for rings.
\[Vol.= \pi \int\limits (2-x)^2 \cdot 2x^2 dx\]