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

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

Mathematics
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amistre, please help me asap.
:/
i think this would be infinite without bounds

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Ben he will be back dont worry. i have 3mins. ;(
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
to check, wolframalpha does a wonderful job
2pi is what the wolf gives us http://www.wolframalpha.com/input/?i=integrate+sqrt%284-x%5E2%29+from+-2+to+2
Amistre, another question please?
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 :/
what is wrong with that?
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
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
(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
so yeah, 32/3 pi should be it
i see, i entered it wrong at the start of all this .... it sucks getting old lol
haha ok cool thank you

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