Application of integration. (Volume) Find the points of intersection of the curves $y^2=4x$ and $y^2=\frac{4}{x}$.  Show that the volume of solid generated when the region bounded by the curves and the straight line $x=3$ is revolved through $\pi$ radians about the x-axis is $2\pi (1+2\ln 3)units^3$.
My attempt, 
$$V=4\pi\int_1^3 (x-\frac{1}{x})\,dx=4\pi\left[\frac{1}{2}x^2-\ln(x)\right]_1^3=4\pi\left(4-\ln(3)\right)$$
What did I do wrong?
 A: You did nothing wrong. The region that is being rotated was not well described. You took one (legitimate) interpretation, and the person who produced the answer took another interpretation, which also looks reasonable. (If I were forced to choose between the two, I would lean towards the one used by the person who produced the answer. But there is genuine ambiguity.)
Draw a picture.  For $x=0$ to $x=1$, look at the region $K_1$ bounded above by $y=\sqrt{4x}$, and below by $y=-\sqrt{4x}$. For $x=1$ to $x=3$, look at the region $K_2$ between $y=\sqrt{4/x}$ and $y=-\sqrt{4/x}$. The problem solver was rotating the region $K$ which is the union of $K_1$ and $K_2$. Equivalently, the solver is rotating the part of $K$ which is above the $x$-axis. The volume is 
$$\int_0^1 \pi(4x)\,dx+\int_1^3 \pi\left(\frac{4}{x}\right)\,dx.$$
Calculate. You will get $2\pi+4\pi\ln 3$.
A: The person producing the answer should have better described boundaries like the
green area
$$ y^2 = 4 x, y^2= \frac{4}{x}, (x-axis, \; y=0), x=3. $$ 
or grey area
$$ y^2 = 4 x, y^2= \frac{4}{x}, x=3. $$
Right? And also, is not the Volume halved with $\pi$ rotation?

