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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
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Anyone want to help me set up the integral?
Anyone? :( .
Well I found the intersection points for y which are y=3 and y=7.

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Other answers:

Good. This is a nice problem. Have you heard of Pappus's Theorem?
No.
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!
Yeah, I realized that.
Perfect. What are the two definitions?
Wait you lost me. Never mind.
\(x = (y-5)^{2}\) Solve for \(y\). You should get TWO results.
yeah y=y-5 and y=5-y
Wait. WHat am I doing...
Braid dead lol.
?? Where did the \(x\) go? \(\sqrt{x} = y-5\;for\;y>5\) \(\sqrt{x} = 5-y\;for\;y<5\) Keep solving!
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Sorry. I had a stupid moment.
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.
Top function is \[\sqrt{x}+5\] Bottom function is \[5-\sqrt{x}\]
The Whole thing: \(\pi\int\limits_{0}^{5}((\sqrt{x}+5) −3)^{2}\;dx\) The Extra part: Oh, you do that one!
Woudn't it be from 0 to 4?
Yes. I just spotted that typo. Sorry about that. Good eye!
Haha thanks :) .
And the answer is????????
136/3 which is wrong :( .
I forgot pi whoops.
That's the whole thing. You have to subtract the small piece. That's why we had two integrals.
So it would be:\[\pi \int\limits_{0}^{4}((\sqrt{x}+5)-3)^2-((5-\sqrt{x})-3)^2dx\]
That should do it. Or, you can do them as separate integrals.
THanks so much! Lets see if I get the right answer...
Yeah it's right :) . Thanks so much!
I know. I did it out on maple.
Interestingly, if you do a little algebra, you can do a whole lot less calculus. That wonderful integrand you created simplifies to \(8\sqrt{x}\).
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!
When I learn it XD .

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