How to find volume from Washer method? So, I want to use Washer method to find 
$$\pi \int_{y=0}^2 (\sqrt{4y})^2-(\sqrt{4y}-0.1)^2 dy.$$
I've literally spent a day trying to do so, and when I treat each part separately and then subtract the answer keeps resulting in 
a) a negative
or 
b) two drastically far apart numbers
Hence, I think something is wrong in the initial set up.  Any thoughts on what is wrong?
 A: From your graph it seems that you are considering the solid obtained by rotating the curve about the $y$-axis. Is it correct?
Why don't you simplify before integrating?
$$(\sqrt{4y})^2-(\sqrt{4y}-0.1)^2=0.4 \sqrt{y}-0.01.$$
It remains to evaluate
$$\pi\left(0.4 \int_0^2 \sqrt{y}\,dy -0.01 \int_0^2 \,dy\right)=
\pi\left(0.4 \left[\frac{2y^{3/2}}{3}\right]_0^2 -0.01 \cdot 2\right).$$
What is the final result?
A: If you're rotating about the $x$-axis, you need functions in terms of $x$ for the first quadrant:
$$
x=\sqrt{4y}\implies y=\frac{x^2}{4},\\
x=\sqrt{4y}-0.1\implies y=\frac{(x+0.1)^2}{4}.
$$
Your integral should look like this where $2$ is supposedly the upper bound of integration:
$$
V=\pi\int_{0}^{2}\left[\left(\frac{x^2}{4}\right)^2-\left(\frac{[x+0.1]^2}{4}\right)^2\right]\,dx.
$$
If you're rotating about the $y$-axis, then your integral looks fine. There must have been a mistake in your calculations:
$$
V=\pi\int_{0}^{2}\left[\left(\sqrt{4}y\right)^2-\left(\sqrt{4y}-0.1\right)^2\right]\,dy=\\
\pi\int_{0}^{2}\left[4y-\left(4y-0.2\sqrt{4y}+0.01\right)\right]\,dy=\\
\pi\int_{0}^{2}\left(4y-4y+0.2\cdot 2\sqrt{y}-0.01\right)\,dy=\\
\pi\int_{0}^{2}\left(0.4\sqrt{y}-0.01\right)\,dy=\\
0.4\pi\int_{0}^{2}\sqrt{y}\,dy-0.01\pi\int_{0}^{2}\,dy=\\
0.4\pi\left(\frac{2\sqrt{2^3}}{3}-\frac{2\sqrt{0^3}}{3}\right)-0.01\pi(2-0)=\\
0.8\pi\frac{\sqrt{2^2\cdot 2}}{3}-0.02\pi=\\
0.8\pi\frac{2\sqrt{2}}{3}-\pi\frac{0.02\cdot 3}{3}=\\
\pi\left(\frac{0.8\cdot 2\sqrt{2}}{3}-\frac{0.06}{3}\right)=\\
\pi\frac{1.6\sqrt{2}-0.06}{3}\approx 2.31\ \text{cubic units}.
$$
This is easy to integrate. Just remember that $\int\sqrt{x}\,dx=\frac{2\sqrt{x^3}}{3}+C$ and $\int\,dx=x+C$.
