## anonymous 3 years ago Set up an intergal for the colume of the solid obtained by ratating the region bounded by the parabolas x=8y-2y^2, x= 4y-y^2 about the x-axis.

1. anonymous

I have no idea how to do this :( .

2. tkhunny

The maximum x-value for each is at y = 2. The intersections of the two are at (0,0) and (0,4). Would you like to chop it up into 3 pieces or simply apply Pappus' Theorem?

3. anonymous

I can't use Pappus' theorem.

4. tkhunny

Why not? Specifically proscribed? Not yet presented? Or just not able?

5. anonymous

We won't learn it.

6. anonymous

@saifoo.khan

7. anonymous

First of all, I need to sketch this thing. I can't figure out how the thing looks like.

8. tkhunny

Think it up yourself and surprise everyone! 3 pieces then. 1) Biggest Piece x from 4 to 8 and y from $$2 - \sqrt{\dfrac{8-x}{2}}$$ to $$2 + \sqrt{\dfrac{8-x}{2}}$$ 2) Top Piece x from 0 to 4 and y from $$2 + \sqrt{4-x}$$ to $$2 + \sqrt{\dfrac{8-x}{2}}$$ 3) Bottom Piece x from 0 to 4 and y from $$2 - \sqrt{\dfrac{8-x}{2}}$$ to $$2 + \sqrt{4-x}$$ Go!

9. tkhunny

Why not flip it around to where you are more familiar and solve that problem. y = 8x-2x^2 and y= 4x-x^2 and wrap it around the y-axis!

10. anonymous

But that changes the entire question.

11. tkhunny

No. The question is to calculate the volume. Unique answers do not care how you find them. It MUST be considered a valid methodology.

12. tkhunny

Anyway, you can flip it around just to get a look at it, if you can't relate to the present condition.

13. anonymous

But if you switch the x and y's that should change the entire shape of the graph.

14. tkhunny

That's silly. Why would you think that? Absolutely not the case. It is the EXACT SAME problem.

15. anonymous

|dw:1359609779642:dw| Something like that? REALLY bad sketch though.

16. tkhunny

|dw:1359609887178:dw| A little better. You crossed the x-axis. That's no good.

17. anonymous

yeah I know. I felt lazy.

18. tkhunny

Well, go ahead and do it, then. I gave you all the limits up above. Good luck. gtg

19. anonymous

Thanks!