Answer
$V=\frac{4\pi r^3}{3}$
Work Step by Step
We can represent the volume of a sphere with the triple integral:
$$V=\iiint_{R}dV$$
R represents the region inside a sphere centered at the origin with radius r.
We can rewrite this expression as an iterated integral as follows:
$$V=\int_{-r}^{r}\int_{-\sqrt{r^2-x^2}}^{\sqrt{r^2-x^2}}\int_{-\sqrt{r^2-x^2-y^2}}^{\sqrt{r^2-x^2-y^2}}dz\,dy\,dx$$
Solving the innermost integral, we get:
$$V=\int_{-r}^{r}\int_{-\sqrt{r^2-x^2}}^{\sqrt{r^2-x^2}}2\sqrt{r^2-x^2-y^2}\,dy\,dx$$
Using the substitution $y=\sqrt{r^2-x^2}\sin\theta\,d\theta$, we can rewrite the expression as:
$$V=\int_{-r}^{r}\int_{-\sqrt{r^2-x^2}}^{\sqrt{r^2-x^2}}2\sqrt{r^2-x^2-(r^2-x^2)\sin^2\theta}\sqrt{r^2-x^2}\cos\theta\,dx$$
Simplifying, we get:
$$V=\int_{-r}^{r}\int_{-\pi/2}^{\pi/2}2(r^2-x^2)\cos^2\theta\,d\theta\,dx$$
Using symmetry and a bit of rearranging, we can rewrite the expression as:
$$V=4\int_{-r}^{r}(r^2-x^2)\,dx\int_{0}^{\pi/2}\cos^2\theta\,d\theta$$
After using some more symmetry and solving the integral on the right, we get:
$$V=2\pi\int_{0}^{r}(r^2-x^2)\,dx$$
Finally, solving this integral, we get:
$$V=2\pi\bigg[r^2x-\frac{x^3}{3}\bigg]_{0}^{r}=\frac{4\pi r^3}{3}$$
