# 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?

Jan 24, 2018

$V = 2 \pi \left(1 - \ln 2\right)$

#### 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:

$x = 0 \text{ "" "" } y = 0$
$x = \frac{1}{4} \text{ "" "" } y = \sqrt{\frac{2 \cdot \frac{1}{4}}{\frac{1}{4} + 1}} = \sqrt{\frac{\frac{1}{2}}{\frac{5}{4}}} = \sqrt{\frac{2}{5}} \approx 0.632$
$x = \frac{1}{2} \text{ "" "" } y = \sqrt{\frac{2 \cdot \frac{1}{2}}{\frac{1}{2} + 1}} = \sqrt{\frac{1}{\frac{3}{2}}} = \sqrt{\frac{2}{3}} \approx 0.816$
$x = 1 \text{ "" "" } y = \sqrt{\frac{2 \cdot 1}{1 + 1}} = \sqrt{\frac{2}{2}} = 1$

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: ~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~

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 ($\sqrt{\frac{2 x}{x + 1}}$) sweeps out between $x = 0$ and $x = 1$, our two bounds.

The area of a circle is $\pi {r}^{2}$

The radius $r$, in this case, is $\sqrt{\frac{2 x}{x + 1}}$, 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:

$V = {\int}_{0}^{1} \pi {r}^{2} \mathrm{dx}$

$V = {\int}_{0}^{1} \pi {\left(\sqrt{\frac{2 x}{x + 1}}\right)}^{2} \mathrm{dx}$

$V = {\int}_{0}^{1} \frac{2 \pi x}{x + 1} \mathrm{dx}$

$V = 2 \pi {\int}_{0}^{1} \frac{x}{x + 1} \mathrm{dx}$

$V = 2 \pi {\int}_{0}^{1} \frac{\left(x + 1\right) - 1}{x + 1} \mathrm{dx}$

$V = 2 \pi {\int}_{0}^{1} \left(1 - \frac{1}{x + 1}\right) \mathrm{dx}$

$V = 2 \pi {\left[x - \ln | x + 1 |\right]}_{\textcolor{red}{0}}^{\textcolor{b l u e}{1}}$

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

$V = 2 \pi \left[\left(\textcolor{b l u e}{1} - \ln \left(\textcolor{b l u e}{1} + 1\right)\right) - \left(\textcolor{red}{0} - \ln \left(\textcolor{red}{0} + 1\right)\right)\right]$

$V = 2 \pi \left[\left(1 - \ln 2\right) - \left(0 - \ln 1\right)\right]$

$V = 2 \pi \left[\left(1 - \ln 2\right) - \left(0 - 0\right)\right]$

$V = 2 \pi \left(1 - \ln 2\right)$