# How to set up equations to find the volume for solid of revolution?

Let $$R$$ be the region bounded by the graphs of $$y =\sqrt x$$, $$y=2$$, and $$x=0$$.

Find the volume of the solid generated by revolving $$R$$ about the $$x$$-axis;
Find the volume of the solid generated by revolving $$R$$ about the $$y$$-axis;
Find the volume of the solid generated by revolving $$R$$ about the line $$x = 4$$;
Find the volume of the solid generated by revolving $$R$$ about the line $$y = 2$$.

I am trying to set up these integrals, I know how to solve them but I always have a hard time setting them up, any help would be greatly appreciated :)

• I would suggest to draw a picture. Then sketch what each of the solids look like. – Matti P. Dec 7 '20 at 11:43
• It is easier if we see how you are setting them up and point out if there are mistakes or you show your effort and tell us where you are stuck. Sketching should help as called out by Matti P. – Math Lover Dec 7 '20 at 12:16

## 1 Answer

First off, a (correct-ish) drawing always helps. Always.

Also, practice. There is no way around it: in order to really learn how to do this well enough to be confident you can do it on an exam, you just have to do it a bunch of times. No amount of answers on math forums or videos on youtube is going to change that.

With that said, here is a quick breakdown of the standard method:

1. Decide whether you want to integrate with respect to $$x$$ or $$y$$ (the final form of the integrand depends on this choice in relation to the axis of rotation)
2. Divide whatever axis you chose to integrate with respect to into little pieces
3. Slice up the plane region we're after into strips according to the division form point 2
4. Rotate these slices the same way as the whole region, find the volume of each small corresponding figure, this is your integrand

The dependence mentioned in point 1 determines the basic shape of the figures in point 4, and from here the two names "washer method" and "(cylindrical) shell method" are born. I'll do the first two problems, both times integrating with respect to $$x$$, to illustrate both methods.

So, without further ado, here is a drawing of the relevant region of the plane: This you should be able to do on a piece of paper. If not exactly, then at least a close enough sketch to recognize the basic shape, to determine distances, and to determine intersection points. It is also more important than the two other figures in this answer.

The actual volumes are a bit harder to draw, but if you can sort-of imagine them from the above figure, you'll be in pretty good shape. Anyways, here is the result of taking the above figure, and rotate it about the $$x$$-axis: The volume we are after is the region that is between the green parabola and the blue plane, and inside the red cylinder.

Since we are integrating with respect to $$x$$, we divide the $$x$$-axis into many small pieces of width $$\Delta x$$ (for simplicity, say they are all equally wide). For each narrow piece, there is a corresponding strip of the planar region. Now rotate such a strip about the $$x$$ axis. It will look a like a coin with a hole in it (except perhaps the very first and the very last slice). If you've ever been to a hardware store and bought washers, that's exactly what they look like. Hence the name for this method: the washer method.

We want to add together the volume of each little washer-shaped slice. If the slices are thin enough, it won't matter much that the inner hole in the wacher has a slanted edge. We can just pretend that it is a flat cylinder with a cylindrical piece missing. The volume of such a slice is the volume of the large cylinder minus the volume of the small cylinder.

More symbolically, let's take the slice at some particular value $$x$$ along the $$x$$-axis. The slice has width $$\Delta x$$, the outer radius of the cylinder is $$2$$, and the inner radius is $$\sqrt x$$. The volume of the slice is therefore $$\pi\cdot 2^2\cdot \Delta x - \pi\cdot \sqrt x^2\cdot \Delta x = \pi\left(2^2 - \sqrt x^2\right)\Delta x$$ This is basically all done now. The range of valid $$x$$-values (i.e. the $$x$$-values for which our volume exists) is $$0$$ to $$4$$. As we take thinner and thinner slices, the sum of the volume of all of them turns into an integral as $$\Delta x$$ turns into $$dx$$, and we get $$\int_0^4\pi\left(2^2-\sqrt x^2\right)dx$$

What about rotation about the $$y$$ axis? In that case, the region ends up looking something like this: Which is to say, the volume we're after is the upside-down mountaintop that lies between the green bowl and the red plane. Again, we integrate with respect to $$x$$, so we divide the $$x$$-axis into small pieces of width $$\Delta x$$. Again slice the planar region into vertical strips, but this time rotate them about the $$y$$-axis. The resulting figure will now be a thin-walled tube / pipe, rather than a flat washer. This is where the name comes from: the cylindrical shell method.

We are after the volume of each shell, and from there want to sum them up to get the volume of the entire figure. So, take an arbitrary $$x$$-value. What is the volume of the corresponding shell? The shell is a cylinder with a smaller cylinder taken out. The volume is the difference between these two.

More symbolically, they both have height $$2-\sqrt x$$, the inner cylinder has radius $$x$$, and the outer cylinder has radius $$x + \Delta x$$. So its volume is $$\pi\left(x + \Delta x\right)^2(2-\sqrt x) - \pi\cdot x^2\cdot(2-\sqrt x) = \pi\left(2x\Delta x + \Delta x^2\right)(2-\sqrt x)$$ And here we get to another part of integral calculus that might seem hand-wavey: If $$\Delta x$$ is small enough, then $$\Delta x^2$$ is vanishingly small compared to $$2x\Delta x$$ for all but the innermost few shells. So as we pass to the limit of $$\Delta x\to 0$$, we can just ignore the squared $$\Delta x^2$$ term. And thus we have our integrand. Again, the bounds for $$x$$ is $$0$$ to $$4$$. So the volume becomes $$\int_0^4\pi\cdot 2x\cdot(2-\sqrt x)dx$$