The region under the curves #y=sqrt((2x)/(x+1)), 0<=x<=1# is rotated about the x axis. How do you sketch the region and find the volumes of the two solids of revolution?

Let's start off with the sketch. The easiest way to get an idea of what this curve looks like without a graphing calculator is to plot points and connect them with a smooth curve. So let's pick some x-values and find their y-values:

#x = 0" "" "" " y = 0#
#x = 1/4" "" "" "y = sqrt((2*1/4)/(1/4+1)) = sqrt((1/2)/(5/4)) = sqrt(2/5) ~~ 0.632#
#x = 1/2" "" "" "y = sqrt((2*1/2)/(1/2+1)) = sqrt(1/(3/2)) = sqrt(2/3) ~~ 0.816#
#x = 1" "" "" "y = sqrt((2*1)/(1+1)) = sqrt(2/2) = 1#

If we plot these four points on a graph, we get:

And we know that this is a radical expression, so we can connect the four points with a "square-root" shaped curve, like this:

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Now that we've sketched this curve, we can move on to finding the area under the volume created by rotating it around the x-axis. To do this, remember that we're essentially integrating the area of the circle that this curve (#sqrt((2x)/(x+1))#) sweeps out between #x=0# and #x=1#, our two bounds.

The area of a circle is #pir^2#

The radius #r#, in this case, is #sqrt((2x)/(x+1))#, since that is the distance of the curve from the center (the x-axis) at any given point.

Therefore, to find the volume of the curve created by rotating this curve around the x-axis from #x=0# to #x=1#, we use the formula:

#V = int_0^1 pir^2 dx#

#V = int_0^1 pi(sqrt((2x)/(x+1)))^2 dx#

#V = int_0^1 (2pix)/(x+1) dx#

#V = 2pi int_0^1 x/(x+1) dx#

#V = 2pi int_0^1 ((x+1)-1)/(x+1) dx#

#V = 2pi int_0^1 (1 - 1/(x+1)) dx#

#V = 2pi [x - ln|x+1|]_color(red)0^color(blue)1#

We're done with all of the calculus now. All that's left is to simplify the expression and make a conclusion.

#V = 2pi [(color(blue)1 - ln(color(blue)1+1)) - (color(red)0 - ln(color(red)0+1))]#

#V = 2pi [(1 - ln2) - (0 - ln1)]#

#V = 2pi [(1 - ln2) - (0 - 0)]#

#V = 2pi(1-ln2)#

Final Answer