Find the volume using revolutions The parabola $y=\dfrac{x^2}{16}$ and the line $y=2$
The textbook says to use the washer method. I know that the parabola is facing up and very wide. It intersects the line $y=2$ at $x=-4\sqrt{2}$ and $x=4\sqrt{2}$. So my limits of integration is from $-4\sqrt{2}$ to $4\sqrt{2}$. I used the radius $2-\dfrac{x^2}{16}$ , I found this by looking at the curve. My integral looks like this $\int_{-4\sqrt{2}} ^{4\sqrt{2}}(2-\dfrac{x^2}{16})^2dx$. I got the volume = $\dfrac{256\pi\sqrt{2}}{15}$. However, the online homework says that the answer is wrong and to use the washer method. I cannot find the inner and outer radius from looking at this graph because there doesn't seem to be one. Can someone explain what I am missing here?
 A: The "disk/washer method" always requires making slices perpendicular to the axis of revolution.  Since it appears that your axis is the $ \ y-$axis, the slices will be along the $ \ y-$axis, so you must write the parabola as a function of $ \ y \ $ here.  The limits of integration will be $ \ y = 0 \ $ to $ \ y = 2 \ $ , the differential will be $ \ dy \ $ , and the function becomes  $ \ x = \sqrt{16y} \ $ , making the area of each "disk" $ \ \pi \ \cdot \ 16y \ . $
What you have been doing is portions of the "shell" method, where "slices" are made parallel to the axis of revolution.  The volume of a shell is $ \ 2 \pi \ \cdot \ r \ \cdot \ h \ \ dr \ $ .  With the integration being done along the $ \ x-$axis, your shell will have radii $ \ r = x \ $ and the "height" of each shell is $ \ h(x) \ $ .  You would integrate from $ \ x = 0 \ $ to $ \ x = 4√2 \ $ , using the differential $ \ dx \ $ , and the height of each shell is what you've been calling a "radius", $ \ h(x) = 2 - \frac{x^2}{16} \ . $
So the "disk" integral is 
$$16\pi \ \int_0^2 \ y \ \ dy \ \  ,   $$
while the "shell" integral is
$$ 2 \pi \ \int_0^{4√2} \ x \ \cdot ( \ 2 - \frac{x^2}{16} \ ) \ \ dy \ \ .  $$
You should get the same answer by either method...
A: You did not write down the question. Because of the approach you took, I will assume that you are asked to rotate about the $x$-axis. In that case, your outer radius is $2$, and your inner radius is $\frac{x^2}{16}$. In that case, the volume can be written as
$$\int_{-4\sqrt{2}}^{4\sqrt{2}} \pi\left(4-\frac{x^4}{16^2}\right)\,dx.$$
I would probably use symmetry, and integrate from $0$ to $4\sqrt{2}$ instead, and double the result. 
