Question

Find the volume of the solid generated from revolving y = sqrt(4-x^2) around the x axis.

Answer 1

amistre, please help me asap.

Answer 2

:/

Answer 3

i think this would be infinite without bounds

Answer 4

Ben he will be back dont worry.
i have 3mins. ;(

Answer 5

well it has zeros at -2 and 2 i think. i ended up getting an answer i just dont know how to check it. i got 32pi/3

Answer 6

to check, wolframalpha does a wonderful job

Answer 7

2pi is what the wolf gives us

http://www.wolframalpha.com/input/?i=integrate+sqrt%284-x%5E2%29+from+-2+to+2

Answer 8

Amistre, another question please?

Answer 9

if we do what is commonly refered to as a disc method we add up the areas of a bunch of circles; A = pi r^2, where r = our given function.\[pi\int _{0}^{2}(\sqrt{4-x^2})^2dx\]
\[pi\int _{0}^{2}(4-x^2)dx=pi(4x-\frac{1}{3}x^3)\]
\[pi(4(2)-\frac{1}{3}2^3)-0\]
\[pi(8-\frac{8}{3})\]i see the conundrum alright :/

Answer 10

what is wrong with that?

Answer 11

and souldent it be 2pi on the outside of the integral? you would either have to do it from -2 to 2 or you would need to do 2pi integral from 0 to 2

Answer 12

nah, the area of each circle is: pi [f(x)]^2 and we add all those up from -2 to 2, or just from 0 to 2 and double it

Answer 13

(sqrt(a))^2 = a so we can forgo the sqrt in the integration

int 4 - x^2

4 ups to 4x

-x^2 ups to -x^3/3

4x - x^3/3 at 0 = 0

at 2 = 8- 8/3 = 16/3 and not to forget the pi

http://www.wolframalpha.com/input/?i=integrate+pi%28sqrt%284-x%5E2%29%29%5E2+from+0+to+2

now the wolf agrees

Answer 14

so yeah, 32/3 pi should be it

Answer 15

i see, i entered it wrong at the start of all this .... it sucks getting old lol

Answer 16

haha ok cool thank you