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

1. MASTERCHIEF

2. phi

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

3. MASTERCHIEF

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. MASTERCHIEF

It's at the top of the image

6. MASTERCHIEF

7. phi

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

8. MASTERCHIEF

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. MASTERCHIEF

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