Find the volume of the solid formed by rotating the region in the 1st quadrant enclosed by the curves $y = x^{1/4}$ and $y = x/64$ about the $y$-axis
I did both the equations below but got an incorrect answer for both. I'm not sure what I'm doing wrong. I got the upper limit by solving $x^{1/4}=x/64$
$\displaystyle\int_0^4 2\pi x\left(x^{1/4}-\frac x{64}\right)dx$
$\displaystyle\int_0^4 2\pi \left(x^{1/4}-\frac x{64}\right)$
$y=x^{1/4}$ and $y=\frac x{64}$
Best Answer
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
$$ 2 \pi \ \cdot \ 4^9 \ (\frac{4}{9} - \frac{1}{3}) \ = \ \pi \ \cdot \ 4^9 \ (\frac{1}{3} - \frac{1}{9}) \ = \ \frac{524,288 \cdot \pi}{9} \ . $$