anonymous
  • anonymous
The region bounded by the given curves is rotated about the specified axis. Find the volume V of the resulting solid by any method. x = (y − 5)^2, x = 4; about y = 3 I am trying to use the disc method but I am not sure how I would.
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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katieb
  • katieb
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anonymous
  • anonymous
Anyone want to help me set up the integral?
anonymous
  • anonymous
Anyone? :( .
anonymous
  • anonymous
Well I found the intersection points for y which are y=3 and y=7.

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More answers

tkhunny
  • tkhunny
Good. This is a nice problem. Have you heard of Pappus's Theorem?
anonymous
  • anonymous
No.
tkhunny
  • tkhunny
Too bad. It is much simpler with the Centroid. Another lesson on another day, perhaps. The Whole thing: \(\pi\int\limits_{0}^{4}(y_{1}-3)^{2}\;dx\) The Extra thing: \(\pi\int\limits_{0}^{4}(y_{2}-3)^{2}\;dx\) Do you see the nature of the problem that requires two definitions for 'y'? It's NOT a function!
anonymous
  • anonymous
Yeah, I realized that.
tkhunny
  • tkhunny
Perfect. What are the two definitions?
anonymous
  • anonymous
Wait you lost me. Never mind.
anonymous
  • anonymous
@tkhunny
tkhunny
  • tkhunny
\(x = (y-5)^{2}\) Solve for \(y\). You should get TWO results.
anonymous
  • anonymous
yeah y=y-5 and y=5-y
anonymous
  • anonymous
Wait. WHat am I doing...
anonymous
  • anonymous
Braid dead lol.
tkhunny
  • tkhunny
?? Where did the \(x\) go? \(\sqrt{x} = y-5\;for\;y>5\) \(\sqrt{x} = 5-y\;for\;y<5\) Keep solving!
anonymous
  • anonymous
|dw:1358491765588:dw|
anonymous
  • anonymous
Sorry. I had a stupid moment.
tkhunny
  • tkhunny
Super. You almost beat me to it. Now, which one is on top and which one is on the bottom? These are my little y-functions.
anonymous
  • anonymous
Top function is \[\sqrt{x}+5\] Bottom function is \[5-\sqrt{x}\]
tkhunny
  • tkhunny
The Whole thing: \(\pi\int\limits_{0}^{5}((\sqrt{x}+5) −3)^{2}\;dx\) The Extra part: Oh, you do that one!
anonymous
  • anonymous
Woudn't it be from 0 to 4?
tkhunny
  • tkhunny
Yes. I just spotted that typo. Sorry about that. Good eye!
anonymous
  • anonymous
Haha thanks :) .
tkhunny
  • tkhunny
And the answer is????????
anonymous
  • anonymous
136/3 which is wrong :( .
anonymous
  • anonymous
I forgot pi whoops.
tkhunny
  • tkhunny
That's the whole thing. You have to subtract the small piece. That's why we had two integrals.
anonymous
  • anonymous
So it would be:\[\pi \int\limits_{0}^{4}((\sqrt{x}+5)-3)^2-((5-\sqrt{x})-3)^2dx\]
tkhunny
  • tkhunny
That should do it. Or, you can do them as separate integrals.
anonymous
  • anonymous
THanks so much! Lets see if I get the right answer...
anonymous
  • anonymous
Yeah it's right :) . Thanks so much!
anonymous
  • anonymous
I know. I did it out on maple.
tkhunny
  • tkhunny
Interestingly, if you do a little algebra, you can do a whole lot less calculus. That wonderful integrand you created simplifies to \(8\sqrt{x}\).
tkhunny
  • tkhunny
Well, for future reference, if the ENTIRE area is on one side of the axis of rotation, you can pull Pappus' Theorem out of your hat! Good luck when the time comes!
anonymous
  • anonymous
When I learn it XD .

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