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Dido525
 3 years ago
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.
Dido525
 3 years ago
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.

This Question is Closed

Dido525
 3 years ago
Best ResponseYou've already chosen the best response.0Anyone want to help me set up the integral?

Dido525
 3 years ago
Best ResponseYou've already chosen the best response.0Well I found the intersection points for y which are y=3 and y=7.

tkhunny
 3 years ago
Best ResponseYou've already chosen the best response.1Good. This is a nice problem. Have you heard of Pappus's Theorem?

tkhunny
 3 years ago
Best ResponseYou've already chosen the best response.1Too 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!

tkhunny
 3 years ago
Best ResponseYou've already chosen the best response.1Perfect. What are the two definitions?

Dido525
 3 years ago
Best ResponseYou've already chosen the best response.0Wait you lost me. Never mind.

tkhunny
 3 years ago
Best ResponseYou've already chosen the best response.1\(x = (y5)^{2}\) Solve for \(y\). You should get TWO results.

Dido525
 3 years ago
Best ResponseYou've already chosen the best response.0Wait. WHat am I doing...

tkhunny
 3 years ago
Best ResponseYou've already chosen the best response.1?? Where did the \(x\) go? \(\sqrt{x} = y5\;for\;y>5\) \(\sqrt{x} = 5y\;for\;y<5\) Keep solving!

Dido525
 3 years ago
Best ResponseYou've already chosen the best response.0Sorry. I had a stupid moment.

tkhunny
 3 years ago
Best ResponseYou've already chosen the best response.1Super. You almost beat me to it. Now, which one is on top and which one is on the bottom? These are my little yfunctions.

Dido525
 3 years ago
Best ResponseYou've already chosen the best response.0Top function is \[\sqrt{x}+5\] Bottom function is \[5\sqrt{x}\]

tkhunny
 3 years ago
Best ResponseYou've already chosen the best response.1The Whole thing: \(\pi\int\limits_{0}^{5}((\sqrt{x}+5) −3)^{2}\;dx\) The Extra part: Oh, you do that one!

Dido525
 3 years ago
Best ResponseYou've already chosen the best response.0Woudn't it be from 0 to 4?

tkhunny
 3 years ago
Best ResponseYou've already chosen the best response.1Yes. I just spotted that typo. Sorry about that. Good eye!

tkhunny
 3 years ago
Best ResponseYou've already chosen the best response.1And the answer is????????

Dido525
 3 years ago
Best ResponseYou've already chosen the best response.0136/3 which is wrong :( .

tkhunny
 3 years ago
Best ResponseYou've already chosen the best response.1That's the whole thing. You have to subtract the small piece. That's why we had two integrals.

Dido525
 3 years ago
Best ResponseYou've already chosen the best response.0So it would be:\[\pi \int\limits_{0}^{4}((\sqrt{x}+5)3)^2((5\sqrt{x})3)^2dx\]

tkhunny
 3 years ago
Best ResponseYou've already chosen the best response.1That should do it. Or, you can do them as separate integrals.

Dido525
 3 years ago
Best ResponseYou've already chosen the best response.0THanks so much! Lets see if I get the right answer...

Dido525
 3 years ago
Best ResponseYou've already chosen the best response.0Yeah it's right :) . Thanks so much!

Dido525
 3 years ago
Best ResponseYou've already chosen the best response.0I know. I did it out on maple.

tkhunny
 3 years ago
Best ResponseYou've already chosen the best response.1Interestingly, 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
 3 years ago
Best ResponseYou've already chosen the best response.1Well, 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!
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