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Is this setup correct?
Q: "Use the method of cylindrical shells to find the volume V of the solid obtained by rotating the region bounded by the given curves about the xaxis."
x = 1 + y\(^2\), x = 0, y = 1, y = 3
\[\huge\int\limits_0^{\sqrt{2}} (2 \pi)(y)((3)  (1+y^2)) dy\]
y=3 is the right function
x=1+y\(^2\) is the left function
right  left
Graph below:
 one year ago
 one year ago
Is this setup correct? Q: "Use the method of cylindrical shells to find the volume V of the solid obtained by rotating the region bounded by the given curves about the xaxis." x = 1 + y\(^2\), x = 0, y = 1, y = 3 \[\huge\int\limits_0^{\sqrt{2}} (2 \pi)(y)((3)  (1+y^2)) dy\] y=3 is the right function x=1+y\(^2\) is the left function right  left Graph below:
 one year ago
 one year ago

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agentx5Best ResponseYou've already chosen the best response.1
dw:1341360986379:dw Look good @KingGeorge ? :)
 one year ago

KingGeorgeBest ResponseYou've already chosen the best response.1
I think you switched some things around. First of all, the lines you drew for y=1,3 are vertical lines when they should be horizontal. In short, check the area you want integrate again.
 one year ago

agentx5Best ResponseYou've already chosen the best response.1
Wait a minute I'm getting zero when I evaluate...
 one year ago

agentx5Best ResponseYou've already chosen the best response.1
Oh... I think believe I see what I did there, let's try that graph again: dw:1341361482202:dw
 one year ago

KingGeorgeBest ResponseYou've already chosen the best response.1
That looks better. So your new integrand and bounds would be...?
 one year ago

agentx5Best ResponseYou've already chosen the best response.1
\[\huge\int\limits_1^3 (2 \pi)(y)((3)  (1+y^2)) dy = \frac{26}{3}\] The shells are growing out from the xaxis but I'm still missing something, a radius change perhaps? But which one? +1 or 1?
 one year ago

agentx5Best ResponseYou've already chosen the best response.1
Seems like 1y is logical for a shift upward, the others evaluate negative
 one year ago

KingGeorgeBest ResponseYou've already chosen the best response.1
The only change I would make is in the integrand again. The right side is now \((1+y)^2\) and the left is 0. Hence, you should have \[\huge\int\limits_1^3 (2 \pi)(y)( (1+y^2)0) dy \]
 one year ago

agentx5Best ResponseYou've already chosen the best response.1
\[\large \int\limits_1^3 (2 \pi)(y)( (1+y^2)0) dy =^? \int\limits_1^3 (2 \pi)(1y)((3)  (1+y^2)) dy = \frac{44}{3}\]
 one year ago

agentx5Best ResponseYou've already chosen the best response.1
No wait the first is 44/3 and the second is 44\(\pi\)/3
 one year ago

KingGeorgeBest ResponseYou've already chosen the best response.1
I think I forgot a close parentheses. \[\huge\int\limits_1^3 (2 \pi)(y)( (1+y)^2) dy\]
 one year ago

agentx5Best ResponseYou've already chosen the best response.1
Just noticed it ran off the page too, sry about breaking margins >_<
 one year ago

agentx5Best ResponseYou've already chosen the best response.1
\[\huge\int\limits_1^3 (2π)(y)((1+y)^2)dy = \frac{248\pi }{3}\] Seems logical, let me give it a try...
 one year ago

agentx5Best ResponseYou've already chosen the best response.1
Alas no, it is incorrect
 one year ago

agentx5Best ResponseYou've already chosen the best response.1
Sheesh these particular problems are being, well, problems tonight :D
 one year ago

agentx5Best ResponseYou've already chosen the best response.1
*looks back over steps*
 one year ago

KingGeorgeBest ResponseYou've already chosen the best response.1
Wait, I misread the question. There shouldn't be a close parentheses there XD Have you already tried it without the close parentheses that I added?
 one year ago

agentx5Best ResponseYou've already chosen the best response.1
(2π)(y)((1+y)^2)dy < literally what I started working with
 one year ago

agentx5Best ResponseYou've already chosen the best response.1
Oh it's on the wrong side!
 one year ago

KingGeorgeBest ResponseYou've already chosen the best response.1
How about \[\large\int\limits_1^3 (2 \pi)(y)( 1+y^2) dy?\]
 one year ago

agentx5Best ResponseYou've already chosen the best response.1
That's a quadratic now by mistake
 one year ago

agentx5Best ResponseYou've already chosen the best response.1
And 48π is perfectly correct, awesome!!!
 one year ago

agentx5Best ResponseYou've already chosen the best response.1
Comes out very nice in the end, no fractions
 one year ago

KingGeorgeBest ResponseYou've already chosen the best response.1
Awesome. Now I just need to read the problems correctly.
 one year ago

agentx5Best ResponseYou've already chosen the best response.1
What am I thinking, making vertical lines when it asked for horizontal lol
 one year ago
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