## anonymous 3 years ago The region bounded by the given curves is rotated about the specified axis. Find the volume V of the resulting solid by any method. x = (y − 5)^2, x = 4; about y = 3 I am trying to use the disc method but I am not sure how I would.

1. anonymous

Anyone want to help me set up the integral?

2. anonymous

Anyone? :( .

3. anonymous

Well I found the intersection points for y which are y=3 and y=7.

4. tkhunny

Good. This is a nice problem. Have you heard of Pappus's Theorem?

5. anonymous

No.

6. tkhunny

Too bad. It is much simpler with the Centroid. Another lesson on another day, perhaps. The Whole thing: $$\pi\int\limits_{0}^{4}(y_{1}-3)^{2}\;dx$$ The Extra thing: $$\pi\int\limits_{0}^{4}(y_{2}-3)^{2}\;dx$$ Do you see the nature of the problem that requires two definitions for 'y'? It's NOT a function!

7. anonymous

Yeah, I realized that.

8. tkhunny

Perfect. What are the two definitions?

9. anonymous

Wait you lost me. Never mind.

10. anonymous

@tkhunny

11. tkhunny

$$x = (y-5)^{2}$$ Solve for $$y$$. You should get TWO results.

12. anonymous

yeah y=y-5 and y=5-y

13. anonymous

Wait. WHat am I doing...

14. anonymous

15. tkhunny

?? Where did the $$x$$ go? $$\sqrt{x} = y-5\;for\;y>5$$ $$\sqrt{x} = 5-y\;for\;y<5$$ Keep solving!

16. anonymous

|dw:1358491765588:dw|

17. anonymous

Sorry. I had a stupid moment.

18. tkhunny

Super. You almost beat me to it. Now, which one is on top and which one is on the bottom? These are my little y-functions.

19. anonymous

Top function is $\sqrt{x}+5$ Bottom function is $5-\sqrt{x}$

20. tkhunny

The Whole thing: $$\pi\int\limits_{0}^{5}((\sqrt{x}+5) −3)^{2}\;dx$$ The Extra part: Oh, you do that one!

21. anonymous

Woudn't it be from 0 to 4?

22. tkhunny

Yes. I just spotted that typo. Sorry about that. Good eye!

23. anonymous

Haha thanks :) .

24. tkhunny

25. anonymous

136/3 which is wrong :( .

26. anonymous

I forgot pi whoops.

27. tkhunny

That's the whole thing. You have to subtract the small piece. That's why we had two integrals.

28. anonymous

So it would be:$\pi \int\limits_{0}^{4}((\sqrt{x}+5)-3)^2-((5-\sqrt{x})-3)^2dx$

29. tkhunny

That should do it. Or, you can do them as separate integrals.

30. anonymous

THanks so much! Lets see if I get the right answer...

31. anonymous

Yeah it's right :) . Thanks so much!

32. anonymous

I know. I did it out on maple.

33. tkhunny

Interestingly, if you do a little algebra, you can do a whole lot less calculus. That wonderful integrand you created simplifies to $$8\sqrt{x}$$.

34. tkhunny

Well, for future reference, if the ENTIRE area is on one side of the axis of rotation, you can pull Pappus' Theorem out of your hat! Good luck when the time comes!

35. anonymous

When I learn it XD .