anonymous 5 years ago Find the volume of the solid generated from revolving y = sqrt(4-x^2) around the x axis.

1. saifoo.khan

2. anonymous

:/

3. amistre64

i think this would be infinite without bounds

4. saifoo.khan

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

5. anonymous

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

6. amistre64

to check, wolframalpha does a wonderful job

7. amistre64

2pi is what the wolf gives us http://www.wolframalpha.com/input/?i=integrate+sqrt%284-x%5E2%29+from+-2+to+2

8. saifoo.khan

9. amistre64

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 :/

10. anonymous

what is wrong with that?

11. anonymous

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

12. amistre64

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

13. amistre64

(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

14. amistre64

so yeah, 32/3 pi should be it

15. amistre64

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

16. anonymous

haha ok cool thank you

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