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anonymous
 3 years 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:
anonymous
 3 years 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:

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anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0dw:1341360986379:dw Look good @KingGeorge ? :)

KingGeorge
 3 years ago
Best ResponseYou've already chosen the best response.1I 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.

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0Wait a minute I'm getting zero when I evaluate...

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0Oh... I think believe I see what I did there, let's try that graph again: dw:1341361482202:dw

KingGeorge
 3 years ago
Best ResponseYou've already chosen the best response.1That looks better. So your new integrand and bounds would be...?

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0\[\huge\int\limits_1^3 (2 \pi)(y)((3)  (1+y^2)) dy = \frac{26}{3}\] The shells are growing out from the xaxis but I'm still missing something, a radius change perhaps? But which one? +1 or 1?

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0Seems like 1y is logical for a shift upward, the others evaluate negative

KingGeorge
 3 years ago
Best ResponseYou've already chosen the best response.1The 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 \[\huge\int\limits_1^3 (2 \pi)(y)( (1+y^2)0) dy \]

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

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0No wait the first is 44/3 and the second is 44\(\pi\)/3

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

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0Just noticed it ran off the page too, sry about breaking margins >_<

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0\[\huge\int\limits_1^3 (2π)(y)((1+y)^2)dy = \frac{248\pi }{3}\] Seems logical, let me give it a try...

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0Alas no, it is incorrect

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0Sheesh these particular problems are being, well, problems tonight :D

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0*looks back over steps*

KingGeorge
 3 years ago
Best ResponseYou've already chosen the best response.1Wait, 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?

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0(2π)(y)((1+y)^2)dy < literally what I started working with

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0Oh it's on the wrong side!

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

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0That's a quadratic now by mistake

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0And 48π is perfectly correct, awesome!!!

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0Comes out very nice in the end, no fractions

KingGeorge
 3 years ago
Best ResponseYou've already chosen the best response.1Awesome. Now I just need to read the problems correctly.

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0What am I thinking, making vertical lines when it asked for horizontal lol
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