Find the volume of the solid formed by rotating the region bounded by the graph of y equals 1 plus the square root of x, the y-axis, and the line y = 3 about the y-axis.

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Find the volume of the solid formed by rotating the region bounded by the graph of y equals 1 plus the square root of x, the y-axis, and the line y = 3 about the y-axis.

Mathematics
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https://gyazo.com/2ae985a854b5c3224e6e7c6c9393f624
Ooo solids o revolution :) These are so fun
eh, I'd rather play a game of tag

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|dw:1444438841566:dw|Ok so you can see the region that we're dealing with.
|dw:1444439029610:dw|We're integrating `from x=0` and `to x= the intersection of these two curves`.
\[\large\rm y=3,\qquad\qquad y=1+\sqrt x\]\[\large\rm 3=1+\sqrt x\]So what is our upper bound on x?
4?
\[\large\rm V=\int\limits_0^4 dv\]Ok that takes care of that. Now let's cut a little slice into that region, and see if we can come up with an equation for the volumen of the shape that is spun around.
|dw:1444439321599:dw|We're going to cut a little slice, we'll say that it has "thickness" dx
|dw:1444439387037:dw|Looks like we get a cylindrical shell, ya? We have some information we need to figure out.
So we need to find r and h?
Volume of a cylindrical shell is \(\large\rm v=(Circumference)(height)(Thickness)\) Good, yes. We already know the thickness,\[\large\rm dv=(Circumference)(height)(dx)\]
|dw:1444439652366:dw|For this particular problem, r is the easy one, h is going to be a little more difficult to figure out.
It looks like our radius is simply our x-coordinate, ya?
\[\large\rm dv=(2\pi r)(height)(dx)\]\[\large\rm dv=(2\pi x)(height)(dx)\]
|dw:1444439757573:dw|