$$\text{Region:}\;\; y=2-(1/2)x,\; y=0, \;x=1,\; x=2;\;\;\text{rotated about the x-axis}$$
Your procedure (approaching the problem) is correct. But I think both you and the "answer" are incorrect: perhaps a simple mistep in expanding the square, or a miscalculation. The answer you refer to may very well be a typo or an error.
$$\begin{align} \pi \int_1^2 \left(2 - \frac x2\right)^2\,dx & = \pi\int_1^2 \left(4 - 2x + \dfrac{x^2}{4} \right)\,dx \\ \\
& = \pi\Big[4x - x^2 + \dfrac{x^3}{12}\Big]_1^2 \\ \\
& = \pi\left[\left(8 - 4 + \frac{8}{12}\right) - \left( 4 - 1 + \frac{1}{12}\right)\right]\\ \\
& = \pi\left(1 + \dfrac 7{12} \right)\\ \\
& = \dfrac {19\pi}{12}\end{align}$$
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
$$ 2 \pi \ \cdot \ 4^9 \ (\frac{4}{9} - \frac{1}{3}) \ = \ \pi \ \cdot \ 4^9 \ (\frac{1}{3} - \frac{1}{9}) \ = \ \frac{524,288 \cdot \pi}{9} \ . $$
Best Answer
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*}