The expression $${\text{V}} = \int_0^{\text{H}} {\pi {{\text{R}}^2}{{\left( {1 – \frac{{\text{h}}}{{\text{H}}}} \right)}^2}} {\text{dh}}$$ for the volume of a cone is equal to A. $$\int_0^{\text{R}} {\pi {{\text{R}}^2}{{\left( {1 – \frac{{\text{h}}}{{\text{H}}}} \right)}^2}} {\text{dr}}$$ B. $$\int_0^{\text{R}} {\pi {{\text{R}}^2}{{\left( {1 – \frac{{\text{h}}}{{\text{H}}}} \right)}^2}} {\text{dh}}$$ C. $$\int_0^{\text{H}} {2\pi {\text{rH}}{{\left( {1 – \frac{{\text{r}}}{{\text{R}}}} \right)}^2}} {\text{dh}}$$ D. $$4\int_0^{\text{R}} {\pi {\text{rH}}{{\left( {1 – \frac{{\text{r}}}{{\text{R}}}} \right)}^2}} {\text{dr}}$$

The correct answer is $\boxed{\text{B}}$.

The volume of a cone is given by the formula $$V = \frac{1}{3}\pi r^2 h$$ where $r$ is the radius of the base and $h$ is the height.

In the given expression, $R$ is the radius of the base and $H$ is the height. The expression can be rewritten as $$V = \int_0^H \pi R^2 \left(1 – \frac{h}{H}\right)^2 dh$$

The following is a brief explanation of each option:

• Option A: This option is incorrect because it integrates with respect to $r$, not $h$.
• Option B: This option is correct because it integrates with respect to $h$ and the integrand is the volume of a cone with radius $R$ and height $h$.
• Option C: This option is incorrect because it integrates with respect to $r$ and the integrand is not the volume of a cone.
• Option D: This option is incorrect because it is four times the correct answer.