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# Volume of a solid Consider the region $R$ bounded by the graph of $f(x)=\sqrt{x^{2}+1}$ and the $x$ -axis on the interval $[0,2] .$ Find the volume of the solid formed when $R$ is revolved about the $y$ -axis.

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in this problem, we want to look at the volume of ah region. Sorry. The volume of a solid formed by rotating F of X equal to the square root of X squared plus one about the why access on zero Teoh he wrote it to. So we use the shell method. We'll have two pi acts of axe gxe bond 0 to 2 equal to our volume. Well, William f of X bring the two pi in front. So here, if we use you equal Teoh X squared plus one Since d you will have to x t x We can get rid of the X in front of the square root when we do the U substitution. So soup I times are limits. Become 1 to 5 get 1/2 route you do you We bring the 1/2 in front will have pie. If we evaluate this war, get 2/3 you to the three halves from 1 to 5 This is going to be pie 2/3 pi times so five to the three halfs minus one It's right that out so we'll have to root 2/3 by times 55 Finest one

California State University - Long Beach

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