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 x-axis." 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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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 x-axis." 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:

Mathematics
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|dw:1341360986379:dw| Look good @KingGeorge ? :-)
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.
Wait a minute I'm getting zero when I evaluate...

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Oh... I think believe I see what I did there, let's try that graph again: |dw:1341361482202:dw|
That looks better. So your new integrand and bounds would be...?
\[\huge\int\limits_1^3 (2 \pi)(y)((3) - (1+y^2)) dy = -\frac{26}{3}\] The shells are growing out from the x-axis but I'm still missing something, a radius change perhaps? But which one? +1 or -1?
Seems like 1-y is logical for a shift upward, the others evaluate negative
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 \[\huge\int\limits_1^3 (2 \pi)(y)( (1+y^2)-0) dy \]
\[\large \int\limits_1^3 (2 \pi)(y)( (1+y^2)-0) dy =^? \int\limits_1^3 (2 \pi)(1-y)((3) - (1+y^2)) dy = \frac{44}{3}\]
No wait the first is 44/3 and the second is 44\(\pi\)/3
I think I forgot a close parentheses. \[\huge\int\limits_1^3 (2 \pi)(y)( (1+y)^2) dy\]
Just noticed it ran off the page too, sry about breaking margins >_<
\[\huge\int\limits_1^3 (2π)(y)((1+y)^2)dy = \frac{248\pi }{3}\] Seems logical, let me give it a try...
Alas no, it is incorrect
Sheesh these particular problems are being, well, problems tonight :D
*looks back over steps*
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?
(2π)(y)((1+y)^2)dy <-- literally what I started working with
Oh it's on the wrong side!
How about \[\large\int\limits_1^3 (2 \pi)(y)( 1+y^2) dy?\]
That's a quadratic now by mistake
And 48π is perfectly correct, awesome!!!
Comes out very nice in the end, no fractions
Awesome. Now I just need to read the problems correctly.
Me too lol
What am I thinking, making vertical lines when it asked for horizontal lol

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