We can choose to rotate our shapes around different axes and of course the volumes will be very different. If we choose the shape defined by the area in between the functions y=x^2 and y=x and we choose to rotate it around the vertical line x=-1, we get the shape on the left hand side of this figure:

sqrtxm1

The cross-section is shown here:

crosssection

Again, we have to ask what the cylinder is going to look like at height y (ask yourself what a thin slice of the shape would look like at height y: It’s going to be an annulus). This time both the inner and the outer radii of the annulus are going to change as we change y. Now the inner radius is going to be 1+y and the outer radius is going to be 1+\sqrt{y}.
Thus we can see that the annulus at height y is going to have area \pi ((1+\sqrt{y})^2-(1+y)^2) (the 1 is because we are going from the point x=-1 to the point y and \sqrt{y}). The volume is just this multiplied by \Delta y. Thus, again, adding up all the pieces, we get a total volume of:

 

Volume=\int_0^1 \pi ((1+\sqrt{y})^2-(1+y)^2)dy=\frac{\pi}{2}

 

We have come up with some examples, but let’s try and come up with a general expression for this method, when we rotate a function about the x axis and when we rotate it about the y axis.

Let’s first consider the case when we revolve a function f(x) in the range between 0 and x' about the x axis. We can take cylindrical slices through it in the (z, x) plane. The radius of a shell at x is f(x) and so its area is \pi f(x)^2. Thus, its volume is \Delta x \pi f(x)^2 and then, after adding up all disks and taking the limit that their number goes to infinity gives:

 

Volume=\int_0^{x'}\pi f(x)^2dx

 

If we can calculate this integral, then we can work out the volume.
How about if we rotate about the y-axis. In this case we take disks at a given position in the y direction and thickness \Delta y. If you are taking the shape to be the area below the function, swept out around the $x$ axis you will have a set of annuli, each with outer radius x' and inner radius given by f^{-1}(y). The slices will go from 0 to f(x') (as we are taking them in the y direction. Thus, the total volume of the shape is:

 

Volume=\int_0^{f(x')}\pi ({x'}^2-f^{-1}(y)^2)dy

 

We may be able to do this, but we can only do so if we know the inverse function. Note also that the above is only true when we have a monotonically increasing function in the region of interest. If we do not then the inverse function is not single valued and so you have to be more careful with the general definition.

Calculating the inverse function is not always easy and so sometimes it is easier to perform a calculation by taking the cylinders in another way. This is the method of cylindrical shells which we will see in the next section.

Sometimes you will see an example which isn’t given by a function but where the shape is simply described. For instance: Imagine taking a circle of radius r and drawing parallel lines across that circle. Make each of these parallel lines be the base of a triangle. We are free to choose the height of the triangle, so let’s start by saying that the heights take the form of a circle, also of radius r as shown here:

triangcirc

We have to work out, for a given slice, a distance x from the origin, what the area of a triangle is going to be.

triangcirc2The green line marks the distance x from the centre of the circle, so the height of the triangle is \sqrt{r^2-x^2}. The radius of the base is going to 2\sqrt{r^2-x^2}, so the area of the triangle is given by \frac{1}{2}\sqrt{r^2-x^2}2\sqrt{r^2-x^2}=r^2-x^2. The volume of a small slice like this is thus (r^2-x^2)\Delta x and so to get the total volume of the shape we’ve come up with here we have:

 

V=\int_{-r}^r dx (r^2-x^2)=\frac{4r^3}{3}

 

Let’s take another form, this time a shape with base, a disk of radius R and a smaller top disk of radius r, where the sides of the shape are straight surfaces between the bottom and top disks. We can actually plot this shape as here, just by taking the surface in between the blue curves and rotating it about the x-axis.

frost
The radius of each disk taken vertically will be R-\frac{R-r}{h}x so we can calculate the total volume of the form to be:

 

Volume=\int_0^h \pi\left(R-\frac{R-r}{h}\right)^2 dx=\frac{\pi h}{3}\frac{R^3-r^3}{R-r}

How clear is this post?