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## anonymous one year ago Could someone tell me what I did wrong in this volume problem? (Washer)

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1. anonymous

2. phi

it looks like you get 32+ 32/5 = 38.4 what makes you think that is wrong?

3. anonymous

Well did I set up the integral according to the graph correctly?

4. phi

I would have to see the original question to be sure of that.

5. anonymous

It's at the top of the image

6. anonymous

x=y^2, x=4, about x=6

7. phi

is the lower bound known to be the x-axis ?

8. anonymous

Yes

9. phi

yes, you set it up correctly. as a double check, if we find the volume of the whole region with center at x=6, and radius (6-y^2) $\pi \int_0^2 (6-y^2)^2 dy = \pi\int_0^2 36+y^4 -12y^2 \ dy \\ = \pi(36y + \frac{y^5}{5} - 4y^3) \bigg|_0^2 = \pi(72 + \frac{32}{5}- 32)= \pi(40+ 6.4)$ now subtract off the volume of the "center cylinder" that has r=2 and h=2. Its volume is $$\pi r^2 h = \pi\cdot 4 \cdot 2= 8 \pi$$ we get $32\pi+6.4\pi = 38.4 \pi$ that is consistent with what you did.

10. anonymous

Hmm the answer in the book is 384π/5

11. phi

that is twice as large as what we found. it implies they were finding the volume of |dw:1441656377364:dw|

12. phi

the lower limit appears to be $$-\sqrt{x}$$ rather than 0

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