## Dido525 Group Title 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. one year ago one year ago

1. Dido525

Anyone want to help me set up the integral?

2. Dido525

Anyone? :( .

3. Dido525

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. Dido525

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. Dido525

Yeah, I realized that.

8. tkhunny

Perfect. What are the two definitions?

9. Dido525

Wait you lost me. Never mind.

10. Dido525

@tkhunny

11. tkhunny

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

12. Dido525

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

13. Dido525

Wait. WHat am I doing...

14. Dido525

15. tkhunny

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

16. Dido525

|dw:1358491765588:dw|

17. Dido525

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. Dido525

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. Dido525

Woudn't it be from 0 to 4?

22. tkhunny

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

23. Dido525

Haha thanks :) .

24. tkhunny

25. Dido525

136/3 which is wrong :( .

26. Dido525

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. Dido525

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. Dido525

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

31. Dido525

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

32. Dido525

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. Dido525

When I learn it XD .