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 xvalues and find their yvalues:
#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 "squareroot" shaped curve, like this:
~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
Now that we've sketched this curve, we can move on to finding the area under the volume created by rotating it around the xaxis. 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 xaxis) at any given point.Therefore, to find the volume of the curve created by rotating this curve around the xaxis 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  lnx+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(1ln2)# Final Answer
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To sketch the region under the curve ( y = \sqrt{\frac{2x}{x+1}} ) for ( 0 \leq x \leq 1 ) and find the volumes of the two solids of revolution when rotated about the xaxis, follow these steps:

Sketch the curve ( y = \sqrt{\frac{2x}{x+1}} ) for ( 0 \leq x \leq 1 ). Identify the region bounded by the curve and the xaxis.

Divide the region into subregions if necessary to better understand the geometry.

Determine the axis of rotation (in this case, it's the xaxis).

For each subregion, use the method of disks or washers to find the volume of revolution.

Integrate to find the total volume by summing the volumes of all subregions.

Compute the volumes of the solids of revolution.

Add or subtract the volumes as necessary to find the total volume.

Provide the final answer.
Ensure to correctly set up the integral based on the method chosen (disks or washers) and the limits of integration.
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When evaluating a onesided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.
When evaluating a onesided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.
When evaluating a onesided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.
When evaluating a onesided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.
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