Could someone tell me what I did wrong in this volume problem? (Washer)

Question

masterchief · Answer

attachment

phi · Answer

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

masterchief · Answer

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

phi · Answer

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

masterchief · Answer

It's at the top of the image

masterchief · Answer

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

phi · Answer

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

masterchief · Answer

Yes

phi · Answer

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.

masterchief · Answer

Hmm the answer in the book is 384π/5

phi · Answer

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