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## 3psilon one year ago rotate the region bounded by y = 6 - 2x - x^2 and y = x+6 about the line y=3

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1. 3psilon

My answer differs from that of my teacher's. Can someone just show me the integral

2. zepdrix

Can you provide the teacher's answer a sec? I wanna see if I made some terrible mistake before I post this integral :P

3. 3psilon

ok ok

4. 3psilon

5. 3psilon

Problem 21 part D

6. zepdrix

Hmm this is setup really strangly... Clearly the parabola is our upper function. So why is the parabola being subtracted from the line... This is what I came up with, $\large \pi \int\limits_{-3}^0 \left[(6-2x-x^2)-3\right]^2-\left[(x+6)-3\right]^2\;dx$ I should look over my work again though.. bit of a tricky problem. I may have made a mistake somewhere.

7. 3psilon

But I thought the outer radius was at a height 3 plus the parabola function so why wouldn't it be 3+(6-2x-x^2)?

8. zepdrix

|dw:1362721421835:dw|So the outer radius $$\large R$$ is going to be $$\large y_1-3$$. Where $$\large y_1=6-2x-x^2$$.

9. zepdrix

I hope that drawing makes sense... :\ I took a slice and spun it around y=3. And drew some lines to figure out the radius.

10. 3psilon

It makes sense but can you explain why we subtract 3?

11. zepdrix

|dw:1362721856820:dw|That new length is the outer radius.

12. zepdrix

If we were to draw a disk (not hollowed out) with that radius, it would look like this.|dw:1362722006047:dw| This area of this disk, minus the area of, |dw:1362722103066:dw| Gives us the area of our disk hollowed out,\|dw:1362722226356:dw|[\large A=\pi\left[R^2-r^2\right]\]

13. zepdrix

$\large A=\pi\left[R^2-r^2\right]$

14. zepdrix

Bah I wish the drawing tool had more functionality :) lol

15. 3psilon

ohhhh okay okay ! I always had trouble seeing that part! Thank you ! hhha I really appreciate the drawings they helped a lot! Thanks for helping me get ready for my test!

16. zepdrix

Makes a bit of sense? :D Yay!

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