Volume by Shell and Washer Methods (Part Two)

[rodneyspencer]

1. The problem statement, all variables and given/known data

Find the volume generated by rotating the given region about the given line using the Shell method and the Washer method.

x = 4y [y = x/4], y = 0, x = 0, x = 8 about x=8

2. Relevant equations

Washer method (about x):
V = pi \(\displaystyle \int^b_a\) ((R[sub:1v5l8jbv]top[/sub:1v5l8jbv][sup:1v5l8jbv]2[/sup:1v5l8jbv](x) - r[sub:1v5l8jbv]bottom[/sub:1v5l8jbv][sup:1v5l8jbv]2[/sup:1v5l8jbv](x))dx

Shell method (about x):
V = 2pi \(\displaystyle \int^d_c\) (y[f(y)-g(y)])dy

3. The attempt at a solution

Here, my issue is how to set up the washer method to take into account the new rotation. As I understand it, because my shell is already with respect to x, I multiply the integrand by (8-x).

So, in this case because the Washer method is integrated with respect to y I tried converting the new location (x=8) to it's y-axis equivalent (8=4y ==> y = 2) and multiplied (2 - y) into the Washer integrand. Again, I have different answers.

I've also tried it without altering y but I don't see how that could be correct because the new shape is a cone and this generates a sort of bowl.

[In general, am I correct in that for quadrants I&IV I multiply the integrand by (x-value - x) and for quadrants II&III, (x-value + x)?]

Here's my work:

Washer Test 1:
V = \(\displaystyle \int^2_0\) (pi(2-y)((8)[sup:1v5l8jbv]2[/sup:1v5l8jbv]-(4y)[sup:1v5l8jbv]2[/sup:1v5l8jbv]))dy = 320pi/3

Washer Test 2:
V = \(\displaystyle \int^2_0\) (pi((8)[sup:1v5l8jbv]2[/sup:1v5l8jbv]-(4y)[sup:1v5l8jbv]2[/sup:1v5l8jbv]))dy = 256pi/3

Shell:
V = 2pi \(\displaystyle \int^8_0\) (x(8-x)(x/4))dx = 512pi/3

 

[galactus]

Shells:

\(\displaystyle 2\pi\int_{0}^{8}\frac{x}{4}(8-x)dx\)

Washers:

\(\displaystyle \pi\int_{0}^{2}(4y-8)^{2}dy\)

 

[rodneyspencer]

I appreciate your fast reply. I wasn't clear enough in my original post. I'm having trouble using Shells and Washers when the region is revolved about a line. I included a graph to show what I mean.

The integrals I've tried are in my original post. What I'm really curious about is setting up the integral when the integral must be taken with respect to the parallel axis. For instance, on this one, how do I take x=8 as the line of rotation into account when setting up the Washer method since the Washer method is with respect to y.

Also, am I correct that with the shell method for rotating about a vertical line in quadrants I&IV I multiply the integrand by (lineValue - x) and for quadrants II&III, I multiply the integrand by (lineValue + x)?

 

[Deleted member 4993]

I appreciate your fast reply. I wasn't clear enough in my original post. I'm having trouble using Shells and Washers when the region is revolved about a line. I included a graph to show what I mean.

The integrals I've tried are in my original post. What I'm really curious about is setting up the integral when the integral must be taken with respect to the parallel axis. For instance, on this one, how do I take x=8 as the line of rotation into account when setting up the Washer method since the Washer method is with respect to y.

Not really! Washer method can be applied for any axes.

In the case you have indicated - the washer radius would be (8-4y) and the integration would look something like:

\(\displaystyle V \ = \ \int_0^2\pi * (8-4y)^2 dy\)

and that's it.....


Also, am I correct that with the shell method for rotating about a vertical line in quadrants I&IV I multiply the integrand by (lineValue - x) and for quadrants II&III, I multiply the integrand by (lineValue + x)?