## monroe17 2 years ago Find the volume formed by rotating the region enclosed by: y=x^2, x=y^2 about the line x=-5

1. tkhunny

Find the points of intersection. This will also be your integral limits with this lovely region.

2. monroe17

how do i do that?

3. tkhunny

$$x^{2} = \sqrt{x}$$ -- Solve!

4. monroe17

0 and 1 :)

5. tkhunny

Perfect. Now, decide disks or shells. With this well-behaved region, it really doesn't much matter. Just pick one.

6. monroe17

disks

7. tkhunny

Disks it is, then. $$\pi\int\limits_{0}^{1} OuterRadius^{2} - InnerRadius^{2} d(Height)$$ The "Height" is the x-direction, so we have: $$\pi\int\limits_{0}^{1} OuterRadius^{2} - InnerRadius^{2} dx$$ Now indentify that inner and outer radii. What say you?

8. monroe17

x^2+5

9. tkhunny

Well, I would prefer the more literal (-5-x^2), but it will do since we are just going to square it. Is that the inner or the outer radius?

10. monroe17

11. tkhunny

No worries. x^2 is the inner radius. Are we calling that 'r'? What is the outer radius, "R"?

12. monroe17

isn't (-5-x^2) the outer radius? and yes

13. tkhunny

No, $$x^{2}$$ and $$\sqrt{x}$$ are a little funny on [0,1] $$\sqrt{x} \ge x^2\;for\;x\in[0,1]$$

14. monroe17

uhm.. im not sure ;/

15. tkhunny

r is $$(-5 - x^{2})$$ R is $$(-5 - \sqrt{x})$$ That's all. Did you draw a picture? You will see it at a glance.

16. monroe17

I'm still confused toward the sqrt(x) where'd that come from?

17. tkhunny

x = y^2 ==> y = sqrt(x) We have integration "dx", so we need the representation in terms of x.

18. monroe17

got it ;p lol just had to see the steps..

19. tkhunny

Well, there you have it. Let's see what you get.

20. monroe17

I got 109pi/30?

21. tkhunny

THAT is the right answer. Good work! Wading through it is FUN, isn't it?!

22. monroe17

haha.. totally ;p

23. tkhunny

Once we realized what the region was, and where the square root came from, you were all over it. This is encouraging! Move on to the next one!

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