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

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phi
 one year ago
Best ResponseYou've already chosen the best response.3it looks like you get 32+ 32/5 = 38.4 what makes you think that is wrong?

MASTERCHIEF
 one year ago
Best ResponseYou've already chosen the best response.0Well did I set up the integral according to the graph correctly?

phi
 one year ago
Best ResponseYou've already chosen the best response.3I would have to see the original question to be sure of that.

MASTERCHIEF
 one year ago
Best ResponseYou've already chosen the best response.0It's at the top of the image

MASTERCHIEF
 one year ago
Best ResponseYou've already chosen the best response.0x=y^2, x=4, about x=6

phi
 one year ago
Best ResponseYou've already chosen the best response.3is the lower bound known to be the xaxis ?

phi
 one year ago
Best ResponseYou've already chosen the best response.3yes, 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 (6y^2) \[\pi \int_0^2 (6y^2)^2 dy = \pi\int_0^2 36+y^4 12y^2 \ dy \\ = \pi(36y + \frac{y^5}{5}  4y^3) \bigg_0^2 = \pi(72 + \frac{32}{5} 32)= \pi(40+ 6.4) \] 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
 one year ago
Best ResponseYou've already chosen the best response.0Hmm the answer in the book is 384π/5

phi
 one year ago
Best ResponseYou've already chosen the best response.3that is twice as large as what we found. it implies they were finding the volume of dw:1441656377364:dw

phi
 one year ago
Best ResponseYou've already chosen the best response.3the lower limit appears to be \( \sqrt{x} \) rather than 0
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