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?
1 Answer
Explanation:
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 (
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
#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
