# Volume of Region R Using Washers & Shells Method

• clipzfan611
In summary, my teacher assigned this homework without teaching it and I'm kind of lost. I tried to do washers and shells with respect to X, but I'm not quite clear on what that means. I know the answer will be the same, but I'd like to learn both methods. My school has a block schedule so I'm not sure what that has to do with my teacher assigning homework before the material is covered in class.

## Homework Statement

Find the volume of the solid generated when region R is resolved about the x-axis using the method of "washers" AND "shells" with respect to X.

## Homework Equations

Let R be the region enclosed by the graphs of y=(64x)^(1/4) and y=x

## The Attempt at a Solution

My teacher assigned this without teaching it since my school has block schedule so I'm kind of lost.
Formula for washer: pie*(R^2-r^2) So pie*((64x)^1/4)^2-(X^2)) ?
Formula for shells: 2pie(r)(h)delta(r) No clue.
I know the answer will be same but I'd like to learn both methods.

I'm not quite clear on what having a "block schedule" has to do with your teacher assigning homework before the material is covered in class.

Formula for washer: pie*(R^2-r^2) So pie*((64x)^1/4)^2-(X^2)) ?
Formula for shells: 2pie(r)(h)delta(r)
Are what? You don't say what those are supposed to equal! Don't just copy formulas without understanding them.

The idea of using "washers" is this: imagine "slicing" your graph with lines parallel to the y-axis. Rotated around the x-axis The section of the line between the graphs y= (64x)1/4 and y= x forms a "washer" (a disk with the middle, from y= 0 to y= (64x)1/4 removed). y measures the radius. The entire disk, from y= 0 to y= x, would have area $\pi y^2= \pi x^2$ while the interior that is removed, from y= 0 to y=(64x)1/4 has area $\pi((64x)^{1/4})^2= \pi(64x)^{1/2}= 8\pi x^{1/2}$. Subtracting those gives the formula you have, which is only part of the formula for volume< the area of such a "washer" is [\pi x^2- 8\pi x^{1/2}= \pi (x^2- 8x^{1/2})[/itex]. Now imagine that line as having "infinitesmal" thickness $\Delta x$ (since the line is parallel to the y-axis, its "thickness" is perpendicular to that axis and so parallel to the x-axis). The volume of that thin washer is $\pi (x^2- 8x^{1/2})\Delta x$. Now, let the thickness go to 0 as the number of slices goes to infinity and that Riemann sum becomes the integral
$$\pi\int_0^4 (x^2- 8x^{1/2})dx$$
I'll leave it to you to figure out where I got the limits of integration.

Do do shells, turn everything 90 degrees! Imagine now drawing a line parallel to the x-axis and rotating that about the x-axis. That gives a very very thing cylindrical shell. The volume of that will be the circumference of the circle ($2\pi r= 2\pi y$ (the radius of our circle is again measured by y) times the thickness, dy (since now our line is parallel to the x-axis it's thickness is parallel to the y axis), times the length of the cylinder which is the difference between the x-values at the ends of the line: x= y4/64 and y= x. The volume of that infinitesmal cylinder is $2\pi y(y^4/64- y)dy$. Integrate that:
$$2\pi \int_0^8 (\frac{y^5}{64}- y^2)dy$$
Do you see why the limits of integration are the same in both integrals, even though one is with respect to x and the other y?

"I'll leave it to you to figure out where I got the limits of integration."
Because its the highest point where they intersect?

"Do you see why the limits of integration are the same in both integrals, even though one is with respect to x and the other y?"
Because the 2nd one has 2pi so you just multiply it to the upper limit?

For the first one I got a negative volume, which doesn't make sense since its in the 1st quadrant...-64/3pi.

Also how does the dynamic change when you rotate it about the y-axis with respect to x and y?

clipzfan611 said:
"I'll leave it to you to figure out where I got the limits of integration."
Because its the highest point where they intersect?
Well, yes and no- there were two limits of integration.

"Do you see why the limits of integration are the same in both integrals, even though one is with respect to x and the other y?"
Because the 2nd one has 2pi so you just multiply it to the upper limit?
No, that has nothing to do with the limits of integration! What are the points where y= x and y= (64x)1/4 intersect?

For the first one I got a negative volume, which doesn't make sense since its in the 1st quadrant...-64/3pi.

Also how does the dynamic change when you rotate it about the y-axis with respect to x and y?
Well, you have to swap x and y, don't you? (While keeping the same functions.)

As for the "negative volume"- that's my mistake! I was thinking of 4th power rather than 1/4. The y= x graph is below the y= (64x)1/4 graph. Reverse the two:
$$\int_0^4 (8x^{1/2}- x^2)dx$$

Well, you have to swap x and y, don't you? (While keeping the same functions.)
So you just have to make it x= instead of y= right? In terms of the actual equation?

So, for shells:
2pi(intergral)((64x)1/4-x)dx, x=0..4
For washers:
pi(integral)((y4/64)2-y2)dy, y=0..4
...when you rotate it about the y-axis with respect to x and y?

Last edited:
Yes, that is the idea.

$$2\pi \int_0^8 (\frac{y^5}{64}- y^2)dy$$

Also how do you integrate that? I tried u-sub and it didnt work. Anti differntiation keeps giving me 512/3pi which is wrong because I know it has to be equivalent to the answer to my first question--64/3pi

$$\frac{\pi}{32}\int_{y=0}^4 y^5 dy- 2\pi\int_{y= 0}^4 y^2 dy$$