Disk Method Problem, where Axis of Rotation is Shifted from the Y-Axis

Question

Could somebody please check my work here?

The textbook answer is 224pi/15, but I'm getting something different. Is it a set up error?

Question: Find the volume of the solid generated by revolving the region by y = x^(1/2) and the lines y = 2 and x = 0 about the line x = 4.

I'm supposed to use the disk method to solve this.

In general, volume the definite integral from a to b of A(x) with respect to x, where A(x) = pi*r^2.

Is this the following the right set up?:

pi * definite integral from 0 to 2 of (4-y^2)^2 with respect to y.

I assumed that the radius would have to be 4 minus the function. Or is that wrong?

When I work out that, I get the answer of 256pi/15.

Thanks for your time.

Answer 1

Draw a picture. The region we are rotating is a vaguely triangular region with corners $(0,0)$, $(4,2)$, and $(0,2)$.

We are taking slices perpendicular to the $y$-axis. Here $y$ ranges from $0$ to $2$. At height $y$, the cross-section is a "washer," a disk (circle) with a circular hole in it.

The outer radius of the washer is always $4$. The inner radius is $4-x$, where $y=\sqrt{x}$. So the area of the washer is $\pi(4^2-(4-x)^2)$, which is $\pi(4^2-(4-y^2)^2)$.

That is what we need to integrate. It is useful to simplify before integrating.

Answer 2

Below is a picture of the region of the $x,y$ plane bounded by the lines $y = 0,$ $y = 2,$ $x = 0,$ and $x = 4.$ The function $y = x^{1/2}$ is also plotted.

The region bounded by $y = 2,$ $x = 0,$ and $y = x^{1/2}$ is the upper left portion of this rectangle, above and to the left of $y = x^{1/2}.$ When you rotate this region around the line $x = 4,$ you don't get a stack of disks. You get a stack of annuli (plural of annulus), that is, disks with circular holes cut in them.

What you have computed is the volume swept by rotating the lower right portion of the rectangle around the line $x = 4.$ That is, you've computed the volume of the holes that were cut out of the disks to make the annuli.

The good news is, the rectangle sweeps out a cylindrical region whose volume is easy to calculate, and the volume of the object you computed, subtracted from the volume of the cylinder, is the volume of the desired solid. That means that after setting up your integral, you performed the integration correctly. In fact it seems to me that subtracting your answer from the volume of the cylinder is a legitimate solution method.

Answer 3

The error in your part is that you found the region that is the inverse of the $4$ x $2$ rectangle (take a look at the picture shown by David. K).

Therefore, you should be using the washer method. the outer radius is $4$, and the inner radius is $(4-x)$. So your region integrated is $4^2 - (4-y^2)^2$

So the integral is:

$$\pi\int_0^2 \left(4^2 - (4-y^2)^2\right)dy = \frac{224\pi}{15}$$

Comment if you have questions.