MasterChief
  • MasterChief
Could someone tell me what I did wrong in this volume problem? (Washer)
Mathematics
  • Stacey Warren - Expert brainly.com
Hey! We 've verified this expert answer for you, click below to unlock the details :)
SOLVED
At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga. Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus. Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.
chestercat
  • chestercat
I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!
MasterChief
  • MasterChief
1 Attachment
phi
  • phi
it looks like you get 32+ 32/5 = 38.4 what makes you think that is wrong?
MasterChief
  • MasterChief
Well did I set up the integral according to the graph correctly?

Looking for something else?

Not the answer you are looking for? Search for more explanations.

More answers

phi
  • phi
I would have to see the original question to be sure of that.
MasterChief
  • MasterChief
It's at the top of the image
MasterChief
  • MasterChief
x=y^2, x=4, about x=6
phi
  • phi
is the lower bound known to be the x-axis ?
MasterChief
  • MasterChief
Yes
phi
  • 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.
MasterChief
  • MasterChief
Hmm the answer in the book is 384π/5
phi
  • phi
that is twice as large as what we found. it implies they were finding the volume of |dw:1441656377364:dw|
phi
  • phi
the lower limit appears to be \( -\sqrt{x} \) rather than 0

Looking for something else?

Not the answer you are looking for? Search for more explanations.