Washer method confusion Compute volume of solid created by revolving the area bounded by $y=\sqrt x$, $y=0$ $x=1$, $x=4$,  around the y axis.
I understand that I can find the volume by integration of the $A(y)$ from $0$ to $2$ since these are the $y$ values of the intercepts of $x=1$ and $x=4$.
My understanding is that I would compute the area $(A(y) $by:
$\pi$ $(R^2)-r^2)$ where $R$ is the outer radius and $r$ is the inner radius. Then I integrate this as $\int_0^2 A(y)dy$.
So, I compute the outer radius $R$ by calculating the $x$ distance from the right most boundary, in this case $x=4$ and the $y$ axis. Thus $R=4$. Is this the way to calculate $R$ even if part of the $R$ $x$ distance isn't inside the boundary?? Ugh,confused on the definition here.
The inner radius is the $x$ distance from the function to the axis of rotation (the $y$ axis).  So, $r=y^2$.
But, it I would think that this x distance is ONLY within the region of $x=1$ and $x=4$. So, since I should always calculate from right to left, wouldn't this radius be $y^2-1$?? But seems like I am using the $R$ by NOT considering the area in the bounded region and I AM using the $r$ by considering the bonded region.
I cannot seem to understand this washer problem even though I can work others and I can see that I don't fully understand the definition of how to computer the inner and outer radii.
Can someone clarify?  Bottom line is that I can't seem to be able to compute the inner and outer radii of this problem correctly.
 A: First, draw a diagram. This step is mandatory and to be skipped only if you have a clear mental picture of how it will look without drawing it.

Since we're rotating about the $y$-axis the washers are horizontal sections through the shaded areas -- and the radius is just the $x$ coordinate of each point in the figure. We can see on the diagram that the range of relevant $y$s is $[0,2]$, and the inner and outer radiuses will be:
$$ {\rm inner}(y) = \begin{cases} 1 & \text{for }0\le y \le 1 \\ y^2 & \text{for } 1 \le y \le 2 \end{cases} \qquad \qquad {\rm outer}(y) = 4 $$
Can you take it from here?

By the way, the shape of the area suggests it would be much easier to compute with the shell method. You wouldn't need any piecewise defined functions and could do everything in one integral.
A: Cross section


Shell method

$$V = \int\limits_{r=1}^4 2 \pi r \sqrt{r} dr = \frac{124 \pi }{5}$$

Washer method

$$V = \int\limits_{z=0}^1 \pi (4^2 - 1^2)\ dz + \int\limits_{z=1}^2 \pi (4^2 - z^2)\ dz = \frac{124 \pi }{5}$$
A: The inner radius would be $1$ for $0\leq y<1$ and $y^2$ for $1\leq y\leq2$.
The outer radius would be 4 for $0\leq y\leq2$
Then, the set up for the volume would be
$$V=\pi\int_{0}^{1}(4^2-1^2)~dy+\pi\int_{1}^{2}(4^2-y^4)~dy=\frac{124\pi}{5}$$

