monroe17 Group Title Find the volume of the solid obtained by rotating the region bounded by the given curves about the specified axis. y=1/x^4, y=0, x=2, x=6, about y=-4 one year ago one year ago

1. math>philosophy Group Title

I would use the method of cylindrical shells. At any f(x) value between 2 and 6, what is the radius from that y point to y = -4?

2. ByteMe Group Title

Which method u prefer? shell or washer?

3. monroe17 Group Title

I just need help setting it up to integrate then i can take it from there...

4. ByteMe Group Title

The setup of the integral depends on the method you use....

5. monroe17 Group Title

^ I didn't see your comment.. the washer method :)

6. math>philosophy Group Title

7. monroe17 Group Title

isn't easier? or not? usually for me it is easier..

8. ByteMe Group Title

i agree that washer is appropriate here.... hang on....

9. math>philosophy Group Title

You guys have fun with that >_>

10. ByteMe Group Title

|dw:1361159214153:dw|

11. ByteMe Group Title

Can you give me the expression for R and r?

12. monroe17 Group Title

errr.. ;/ crap.

13. ByteMe Group Title

???

14. ByteMe Group Title

no... i don't think it's that.. :)

15. monroe17 Group Title

haha give me a sec

16. ByteMe Group Title

consider this...|dw:1361159501396:dw|

17. monroe17 Group Title

would it be 1/x^4-4

18. monroe17 Group Title

squared

19. ByteMe Group Title

for R, shouldn't it be 1/x^4 + 4 ???

20. monroe17 Group Title

- - + .. err yeah my bad

21. ByteMe Group Title

ok... you also agree that r = 4 (inner radius) ???

22. monroe17 Group Title

yupp it is

23. ByteMe Group Title

so... $$\large dV=\pi(R^2-r^2)dx$$ so $$\large V=\pi \int_2^6(R^2-r^2)dx$$ you replace the expressions for R and r ....

24. ByteMe Group Title

sorry... i gotta leave... :(

25. ByteMe Group Title

i'm pretty sure what we have is correct....