Can you compute the volume of a 4-Ball by using solids of revolution on a 3-Ball? So in first year calculus most people at some point will be asked to compute the volume of the unit sphere in 3 dimensions, by rotating $y=\sqrt{1-x^2}$ about the x axis in $\mathbb{R}^3$. and integrating over it.
Im wondering if something similar can be used to compute the volume of the 4-ball in $\mathbb{R}^4$ by taking $z=\sqrt{1-x^2-y^2}$ and rotating it about the x,y plane in $\mathbb{R}^4$. Its been a while since ive done this kind of calculus so im very rusty, but it seems like it should be the same thing were you just multiply the function by $2\pi$ and do the double integral normally to get your answer, however I feel like there must be a mistake here. Also is there a way to get the 4 dimensional case by doing a revolution on the 2 dimensional case twice, so that you can solve it while doing a single variable integral?
 A: Well I feel a bit silly for asking this now, but I figured it out, so I'll answer my own question incase someone else has this question in the future.
The answer is yes, all you need to do is remember exactly how we found the volume of the sphere in the 3 dimensional case. What we did was found a formula for the area of a disk of a fixed radius and integrated over that function. In particular, if out sphere is $x^2+y^2+z^2\leq r^2$ then for each value of $x$ we consider the set of all $(x,y,z)$ in the circle with that particular x value. that set will be a disk of radius $\sqrt{r^2-x^2}$ and then we integrate over all such disks. the area of a disk of radius $r$ is $\pi r^2$ so we get the volume by solving $\int_{-r}^r\pi(\sqrt{r^2-x^2})^2dx$ which indeed gives us $\frac{4}{3}\pi r^3$
In the 4 dimensional case, we do the same thing, except instead of having an area of a disk at each x value, we have a volume of a sphere at each x value. so we just substitute the area formula for the volume formula and find get the integral $\int_{-r}^r\frac{4}{3}\pi(\sqrt{r^2-x^2})^3dx$ which is a harder integral to solve, but it does give us the right answer.
