Calculus Integral Rotation Circle

Question

Is this equation the equation for a circle with radius r and shifted down R?

How would i find the bounds of the integral for this problem? or is the integral going to be in terms of r?

Answer 1

It is a circle of radius $r$ shifted up by $R$. To see that it is shifted up, note that compared with $x^2+y^2=r^2$, you must increase $y$ by $R$ to get a solution.

Your solution will be in terms of both $r$ and $R$.

The volume will be a torus. $R$ is the radius of the centerline circle and $r$ is the radius of the circle that has been swept around to create the torus. The bounds on the integral depend on which direction you are integrating. $x$ ranges from $-r$ to $r$, while $y$ ranges from $-r-R$ to $r+R$

Answer 2

The standard circle is shifted up by $R$.

The solid you get is a torus (doughnut). Its volume depends on the parameters $r$ and $R$.

For the volume, there are various ways to proceed. For example, we can do slicing perpendicular to the $y$-axis, or use the method of cylindrical shells.

Let's do slicing. We have $(y-R)^2=r^2-x^2$, and therefore $y=R\pm\sqrt{r^2-x^2}$.

The expression with the $+\sqrt{r^2-x^2}$ is the top half of the circle, and the other expression gives the bottom half of that circle. So we find the volume obtained by rotating the top half, and subtract the volume obtained by rotating the bottom half. Then our volume is $$\int_{-r}^r \pi \left((R+\sqrt{r^2-x^2})^2-(R-\sqrt{r^2-x^2})^2\right)\,dx.$$

A little algebra simplifies the integrand very considerably: there is a lot of cancellation. And the integral we end up with has a familiar geometric interpretation, so we can give the answer without actually integrating.