This question is similar to this one (same web-based homework site?), so I'll summarize the process.
The intersection points are found from $ \ x^{1/4} \ - \ \frac{1}{64} x \ = \ 0 \ \ \Rightarrow \ \ x^{1/4} \ ( 1 \ - \ \frac{1}{64} x^{3/4} ) \ = \ 0 \ , $ making the intersections $ \ (0, 0) \ $ and $ \ (64^{4/3} = 256, 4) \ . $
The first integral you wrote is correct for using the "shell method", in that you are taking "slices" parallel to the "vertical" rotation axis, so the "radius arm" of the cylinders is $ \ r = x \ $ and the "heights" are given by $ \ x^{1/4} - \frac{1}{64} x \ $ . But that makes the "thickness" of the shells $ \ dx \ , $ so you must integrate over the interval bounded by the $ \ x-$ coordinates of the intersections, $ \ [0, 256] \ , \ $ which is what Ron Gordon is commenting on.
You can as easily for this volume use the "disk method", which would take the integration over the interval in $ \ y \ , \ [0, 4] \ $ , since the "thickness" of the disks is $ \ dy \ , $ the "slices" being perpendicular to the rotation axis. So the "outer radius" is $ \ x_{outer} = 64y \ , $ the inner radius is $ \ x_{inner} = y^4 \ , $ and the volume integral is $ \ \pi \ \int_0^4 \ \ (64y)^2 - (y^4)^2 \ \ dy \ . $
By either method, we obtain the (enormous) volume
$$ 2 \pi \ \cdot \ 4^9 \ (\frac{4}{9} - \frac{1}{3}) \ = \ \pi \ \cdot \ 4^9 \ (\frac{1}{3} - \frac{1}{9}) \ = \ \frac{524,288 \cdot \pi}{9} \ . $$
As this just got bumped from the homepage and @AlexR' answer is not correct, here the correct answer
\begin{align*}
A & = \pi \int_0^{0.7} (e^x + 2)^2\ dx = \pi \left( \int_0^{0.7} e^{2x}\ dx + 4 \int_0^{0.7} e^x\ dx + 4\int_0^{0.7}dx\right)\\
& = \pi \left( \int_0^{0.7} e^{2x}\ dx + 4 \int_0^{0.7} e^x\ dx + 4\cdot0.7 \right)\\
& = \pi \left( \frac12 \int_0^{1.4} e^x\ dx + 4\int_0^{0.7} e^x\ dx + 2.8\right)\qquad \qquad(u=2x,du=2dx)\\
& = \pi \left( \frac12 \left(e^{1.4}-e^0\right) + 4\left(e^{0.7}-e^0\right) + 2.8 \right) \\
& \approx 26.3347\ldots
\end{align*}
Best Answer
It is difficult for me to produce a diagram for you, but a diagram is critical. Draw the hyperbola $xy=1$ (the part in the first quadrant will be enough). Draw the curve $x=y^{1/2}$. Note that this is the first quadrant part of the familiar parabola $y=x^2$. The two curves meet at the point $(1,1)$.
Now draw the line $y=2$. The region we are rotating is below the line $y=2$, to the right of $xy=1$, and to the left of the parabola $y=x^2$. It is a very roughly triangular region, except that two sides of the "triangle" are curvy.
If you have trouble with the graph, perhaps Wolfram Alpha, or another graphing program, or a graphing calculator, would help, but the diagram is really not difficult.
Now imagine rotating this region about the $y$-axis. We get a solid with a "hole" in it.
A cross-section of the solid at $y$ is a "washer," a circle with a circular hole in it. The outer radius of the washer is given by the "$x$" of the parabola, that is, by $y^{1/2}$. The inner radius is the $x$ of the hyperbola, it is $\frac{1}{y}$. So the area of cross-section at $y$ is $$\pi\left((y^{1/2})^2-\frac{1}{y^2}\right).$$ For the volume, "add up" (integrate) from $y=1$ to $y=2$. The integration will be very straightforward.