The problem asks us to determine the volume of a sphere with radius $r$ using integration.

1. Problem Description

The problem asks us to determine the volume of a sphere with radius $r$ using integration.

2. Solution Steps

We can determine the volume of a sphere by integrating the area of circular cross-sections. Imagine slicing the sphere horizontally. Each slice will be a circle.

Consider a sphere with radius $r$ centered at the origin. The equation of the sphere is $x^2 + y^2 + z^2 = r^2$. We can solve for $x^2$ as $x^2 = r^2 - y^2 - z^2$.

If we fix a value for x, say $x=x_0$, we get a circular cross-section of the sphere. Its radius, let's call it $R$, satisfies $x_0^2 + y^2 + z^2 = r^2$. Then $y^2 + z^2 = r^2 - x_0^2$. So, the radius of the circular cross-section at $x=x_0$ is $R = \sqrt{r^2 - x_0^2}$. The area of that circle is $A(x_0) = \pi R^2 = \pi (r^2 - x_0^2)$.

We can integrate the area of these circular cross-sections along the x-axis from $-r$ to $r$ to find the volume of the sphere.

$$V = \int_{-r}^{r} A(x) dx = \int_{-r}^{r} \pi (r^2 - x^2) dx$$

Now, let's evaluate the integral:

$$V = \pi \int_{-r}^{r} (r^2 - x^2) dx = \pi \left[ r^2x - \frac{x^3}{3} \right]_{-r}^{r}$$

$$V = \pi \left[ (r^3 - \frac{r^3}{3}) - (-r^3 - \frac{(-r)^3}{3}) \right]$$

$$V = \pi \left[ r^3 - \frac{r^3}{3} + r^3 - \frac{r^3}{3} \right]$$

$$V = \pi \left[ 2r^3 - \frac{2r^3}{3} \right] = \pi \left[ \frac{6r^3 - 2r^3}{3} \right]$$

$$V = \pi \left[ \frac{4r^3}{3} \right] = \frac{4}{3} \pi r^3$$

3. Final Answer

The volume of the sphere is $\frac{4}{3} \pi r^3$.