## anonymous 5 years ago if you are given the terms y = sqrt(x), y = 0, and x = 6, how do you find the volume of the solid generated by revolving it about the y-axis?

1. amistre64

you can shell it

2. amistre64

2pi {S} x[sqrt(x)] dx ; [0,6]

3. anonymous

yeah, that's how i did it

4. anonymous

but the answer is a decimal: 35.27...

5. amistre64

x[sqrt(x)] can be re written as: sqrt(x^3) = x^(3/2) and integrate

6. amistre64

it gonna be a decimal regardless; pi is irrational/transcendental

7. anonymous

I am using a math homework software that won't take decimals for this answer. Should I just do it by hand?

8. anonymous

like have the fractional, un-evaluated part as and answer?

9. amistre64

if it dont take decimals, then use the exact version of it...

10. anonymous

I will try right now

11. amistre64

2x^(5/2) -------- * 2pi at x = 6 5

12. anonymous

that's the answer, but I need to evaluate at 6. and the computer won't take my decimal answer

13. amistre64

4.pi. sqrt(6.6.6.6.6) = 4.pi.36 sqrt(6) ---------------- ------------ 5 5

14. amistre64

144pi sqrt(6) ----------- should be an exact answer then 5

15. anonymous

you can evalutate it that way???

16. amistre64

yeah; thats what it equate to right?

17. anonymous

yes, and it takes that answer! Thanks everyone!

18. amistre64

:) just glad I was right lol :)

19. watchmath

Let me just use the washer method for comparison $$\int_0^{\sqrt{6}}\pi(36-y^4)\,dy=\pi(36y-\frac{y^5}{5})\mid_0^\sqrt{6}=\pi(36\sqrt{6}-\frac{36\sqrt{6}}{5})=\frac{144\pi\sqrt{6}}{5}$$

20. anonymous

okay, so I can see how you can do that a little more mathematically

21. anonymous

:)

22. anonymous

Thanks!