## 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 Group Title

Anyone want to help me set up the integral?

2. Dido525 Group Title

Anyone? :( .

3. Dido525 Group Title

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

4. tkhunny Group Title

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

5. Dido525 Group Title

No.

6. tkhunny Group Title

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 Group Title

Yeah, I realized that.

8. tkhunny Group Title

Perfect. What are the two definitions?

9. Dido525 Group Title

Wait you lost me. Never mind.

10. Dido525 Group Title

@tkhunny

11. tkhunny Group Title

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

12. Dido525 Group Title

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

13. Dido525 Group Title

Wait. WHat am I doing...

14. Dido525 Group Title

15. tkhunny Group Title

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

16. Dido525 Group Title

|dw:1358491765588:dw|

17. Dido525 Group Title

Sorry. I had a stupid moment.

18. tkhunny Group Title

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 Group Title

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

20. tkhunny Group Title

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

21. Dido525 Group Title

Woudn't it be from 0 to 4?

22. tkhunny Group Title

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

23. Dido525 Group Title

Haha thanks :) .

24. tkhunny Group Title

25. Dido525 Group Title

136/3 which is wrong :( .

26. Dido525 Group Title

I forgot pi whoops.

27. tkhunny Group Title

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

28. Dido525 Group Title

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

29. tkhunny Group Title

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

30. Dido525 Group Title

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

31. Dido525 Group Title

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

32. Dido525 Group Title

I know. I did it out on maple.

33. tkhunny Group Title

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 Group Title

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 Group Title

When I learn it XD .