## anonymous 5 years ago revolve the region bounded by x axis , y = sqrt ( x+ 4) , and x=0 about y axis using shell method. compare it to the disc method

1. dumbcow

$=2 \pi \int\limits_{0}^{4}x \sqrt{x+4} dx$

2. anonymous

no thats not the region

3. anonymous

the region is bounded by y = sqrt ( x + 4) , the y axis, its in quadrant 2

4. anonymous

so using that integral , its from -4 to 0

5. dumbcow

yes but we are revolving around y-axis so it covers both quadrants symmetrically and this way you get a positive volume

6. anonymous

it isnt symmertric though

7. dumbcow

x represents radius of 3-d figure, as you move away from y-axis the radius goes from 0 to 4

8. anonymous

the area from 0 to 4 is not equal to the area from -4 to 0

9. anonymous

10. dumbcow

oh i see, ok think of you reflected the function over y-axis replace x with (-x) f(x) = sqrt(4-x) i didn't do that in the intergral

11. anonymous

yes, that will do it,

12. dumbcow

sorry

13. anonymous

im trying to avoid these cases. are there other situations i can make up,

14. anonymous

no you did fine :)

15. anonymous

i dont seem to have a problem with disc method, since we have pi*r^2, but the shell method does produce negative

16. dumbcow

i guess always try to reflect onto positive quadrant when possible with volumes you want to integrate over positive values

17. dumbcow

18. anonymous

but i think the disc method in that case works

19. anonymous

using horizontal slices, we have x = y^2 - 4

20. anonymous

for y = sqrt ( x + 4

21. dumbcow

yeah discmethod $=\pi \int\limits_{0}^{2}(y^{2}-4)^{2}dy$

22. anonymous

so we get integral pi (

23. anonymous

pi integral ( 0 - (y^2-4))^2

24. anonymous

from -4 to 0

25. dumbcow

i prefer disc method

26. anonymous