Question

Find volume of solid generated by revolving the region between the graph:

y=4-x^2 and 0 13 years ago

Answer 1

I'm pretty sure you can do this @Hoa

Answer 2

:-) !

Answer 3

come on. do it, please

Answer 4

If you revolve this figure around the x-axis you take a look at a small differential dx, it's height will be f(x), so the resulting figure of this differential is a disc. The Area of a disc is pi r^2. Sum all this rings up, that's the part where the integral comes in:

\[\pi \int\limits_0^2 (4-x^2)^2dx \]

Answer 5

oh ok so u just foil it out and then integrate it right?

Answer 6

exactly, that will be the easiest approach in my opinion.

Answer 7

i got (x^5/5)-(8x^3/3)+(16x). Is that right so far?

Answer 8

yes that seems right to me

Answer 9

ok sweet thanx for the help! oh and how do u tell the difference between the disk and shell method?

Answer 10

Hmm I think this can best explained by looking at the graph, unfortunately I don't have any good graphic applications here, but let me try by making up one out of my mind:



You can try to see for your self that if you use the ring method to rotate this graph around the y-axis, then the resulting volume will turn out to be the region between the y-axis and the graph itself revolved around the y-axis.

If you use the shell method however, you can compute the region enclosed by the zeroes of the graph revolved around the y-axis.

I just made this example up, so it's not completely flawless. But mainly you will see which method suits you best for your problem. I prefer using the shell method when the resulting volume turns out to be the Area underneath a graph revolved around the y-axis.