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## 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

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1. anonymous

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

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. anonymous

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. anonymous

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

how about this, revolve region bounded by y = sqrt x, x = 4, axis, about the line y = -4

10. anonymous

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. anonymous

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. anonymous

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

17. anonymous

make radius positive in shell method then adjust f(x) accordingly

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. anonymous

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. anonymous

i prefer disc method

26. anonymous

oh my bad

27. anonymous

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