Calculating the volume of rotation between $y=4x-16$, $x^2-y^2=9$ around $x$ axis Having the following equations:
$y=4x-16$, 
$x^2-y^2=9$
I calculated the mutual points - $(53/15,-28/15),(5,4)$
How can I find the volume of rotation of the area in between them around the $x$ axis? Should it be split into two different volumes?
I've tried calculating the volume above axis x and the volume below it separately. 
Above axis x I'm getting $2\pi$ :
$$V_1 = \pi\cdot \int^5_4(4x-16)^2dx - \pi \int^4_3(\sqrt{x^2-9})^2dx = 2\pi$$
and below I got $1.446\pi$:
$$V_2 =\pi\cdot \int ^4_{53/15}(4x-16)^2dx + \pi\int^{53/15}_{3}(\sqrt{x^2-9}^2dx = 1.446\pi$$ 
So In total I get a final result of $V= V_1+V_2 = 3.446\pi$. Is there anything I did wrong in my calculation?
 A: The volume can indeed be found by splitting it into different sub-volumes. The first volume is obtained by rotating the hyperbola around the $x$ axis from $x=3$ (the apex of the hyperbola) to $x=4$ (the point at which the line crosses the $x$ axis.  The second volume can be found from the difference between the two rotated curves -- onwards from $x=4$ to $x=5$ (the second intersection point).
I.e. we can write it as 
$\qquad \pi \int_3^4 x^2-9 \, dx + \pi \int_4^5 
x^2-9 \,dx - \pi \int_4^5 (4x - 16)^2 \, dx
$
which simplifies to
$\qquad \pi \int_3^5 x^2-9 \, dx
- \pi \int_4^5 (4x - 16)^2 \, dx
$
I make the result $28\pi/3$. In the integral between $x=3$ and $x=4$, the rotated hyperbola will `fill the gap' below the line $y=4x-16$. 
I have interpreted the region between the curves as extending all the way from the apex of the hyperbola, so that the first integral starts at 3 (rather than 53/15). If you want to start at the intersection, then the first limit in the first integral can be changed.
 
In the image, the red/green show the positive/negative halves of the hyperbola, the vertical line shows the split between areas that are rotated to form the volume.
A: You can perform the calculation via the method of cylindrical shells or with washers. 
First, a sketch of the region in question:

This shows that you have an issue if this region is to be rotated about the $x$-axis, since the relevant cross-section is the portion enclosed by the two curves above the $x$-axis (else you will double-count the volume from the portion of the region below the $x$-axis when you revolve it).  Hence, by the method of cylindrical shells, the radius of the shell ranges from $y = 0$ to $y = 4$, and the height of the shell is given by $$h(y) = (y+16)/4 - \sqrt{y^2+9}.$$  Thus a representative shell has differential volume $$dV = 2 \pi y h(y) \, dy = 2 \pi y \left(\frac{y}{4} + 4 - \sqrt{y^2+9}\right) \, dy,$$ and the total volume is $$V = 2\pi \int_{y=0}^4 \frac{y^2}{4} + 4y - y\sqrt{y^2+9} \, dy = 2\pi \left[\frac{y^3}{12} + 2y^2 - \frac{(y^2+9)^{3/2}}{3}\right]_{y=0}^4 = \frac{28}{3}\pi.$$
If we perform the calculation using washers, we have from $x = 3$ to $x = 4$ the volume $$V_1 = \int_{x=3}^4 \pi( x^2-9) \, dx = \pi \left[\frac{x^3}{3} - 9x\right]_{x=3}^4 = \frac{10}{3}\pi,$$ and from $x = 4$ to $x = 5$, $$V_2 = \int_{x=4}^5 \pi \left(x^2 - 9 - (4x-16)^2\right) \, dx = \pi \left[-5x^3 + 64x^2 - 265x\right]_{x=4}^5 = 6 \pi.$$
From this, the total volume is again $$\frac{28}{3}\pi.$$
