# Using Volume of Revolution to calculate the volume of a doughnut

I'm working through a problem at the moment and I'll walk through everything I've done so far. I hit a point where I'm stuck and I don't exactly know how to proceed. So here it goes.

A circle of radius $r$ centred on $(s,t)$ is given by the equation $(x-s)^2+(y-t)^2=r^2$. A circle of radius r has an area $\pi r^2$.

so $\int_{-r}^{r} \sqrt{r^2-x^2} dx$ = $\frac{1}{2}\pi r^2$

I know this to be true. But why exactly is it? Beyond the integration I don't know geometrically why this is so. So if anyone could shed light on this it would be appreciated.

Now I thought it would be good to calculate the volume of a sphere first before a doughnut because it would give me something to go off, so this is how I did it and hopefully it's correct.

$V_x$ = $\int_{-r}^{r}(r^2-x^2)dx$ = $\pi\Big|r^2x-\frac{x^3}{3}\Big|_{-r}^r$

$V_x= \pi\Big[\Big(r^3-\frac{r^3}{3}\Big)-\Big(-r^3+\frac{r^3}{3}\Big)\Big] = \frac{4\pi r^3}{3}$

Okay so that's my sphere done. Now to the doughnut, which should be similar but with a different revolution, maybe about the y axis?

The distance from the centre of the doughnut hole to the centre of the circular arm is $R.$ The circular arm has internal radius $r < R.$ So I have this to work with.

We should have an inner and outer radius. So two separate equations.

inner radius = $x=R-\sqrt{r^2-y^2}$

outer radius = $x=R+\sqrt{r^2-y^2}$

So cross-sectional area should be $A(y)=\pi$ (outer radius)$^2$ - $\pi$ (inner radius)$^2$

$\pi \Big[\Big(R+\sqrt{r^2-y^2}\Big)^2-\Big(R-\sqrt{r^2-y^2}\Big)^2\Big]$

So I'm just stumped here. Rotating about the x-axis for a volume is done with the equation

$V=\pi \int_{a}^{b}[f(x)]^2dx$

We don't rotate about the x axis here, plus a doughnut is just confusing to work with I'm not really sure how else to proceed to find the Volume of Revolution for this. Any help solutions appreciated. :)

• Update: I think I need to try and find area of the cross section properly, as in fully evaluate it. And think about moving from there. I'll try to do that now. – 99 Fishing Sep 19 '17 at 4:21
• You can try cylindrical shell method, it works for this example – Triatticus Sep 19 '17 at 6:54
• You should redefine the equation of the circle, if it's intended to be the cross-section of the doughnut. – hypergeometric Sep 20 '17 at 15:51

For the given dimensions of the doughnut i.e. inner radius $r$ and outer radius $R$, and assuming that it is centred on the origin and lies perpendicular to the $x-y$ plane, the equation of the cross section on the positive $x$-axis side is $$(x-R)^2+y^2=r^2$$ which is a circle with radius $r$ and centre $(R,0)$, with area $\pi r^2$.
By Pappus' Second Centroid Theorem, the volume of the doughnut is the cross sectional area multiplied by the distance travelled by the centre of the cross-section, i.e. $$V=\pi r^2\cdot 2\pi R=2\pi^2Rr^2$$