Math Snap

STEP 1

1. The region is bounded by the curves $y = \sqrt{x}$, $y = 0$, and $x = 9$.
2. The region is revolved around the vertical line $x = 9$.

STEP 2

1. Set up the integral for the volume using the method of cylindrical shells.
2. Determine the limits of integration.
3. Evaluate the integral to find the volume.

STEP 3

Set up the integral for the volume using the method of cylindrical shells. The formula for the volume using cylindrical shells is:
$$V = 2\pi \int_{a}^{b} (r(x) \cdot h(x)) \, dx$$ where $r(x)$ is the radius of the shell and $h(x)$ is the height of the shell.
For this problem:
- The radius $r(x)$ is the distance from the line $x = 9$ to the curve $x$, which is $9 - x$.
- The height $h(x)$ is the function $y = \sqrt{x}$.

STEP 4

Determine the limits of integration. The region is bounded by $x = 0$ and $x = 9$.
Thus, the limits of integration are from $x = 0$ to $x = 9$.

Evaluate the integral to find the volume:
$$V = 2\pi \int_{0}^{9} (9 - x) \cdot \sqrt{x} \, dx$$ Simplify and evaluate the integral:
$$V = 2\pi \int_{0}^{9} (9\sqrt{x} - x\sqrt{x}) \, dx$$ $$= 2\pi \left( \int_{0}^{9} 9\sqrt{x} \, dx - \int_{0}^{9} x^{3/2} \, dx \right)$$ Evaluate each integral separately:
$$\int 9\sqrt{x} \, dx = 9 \cdot \frac{2}{3}x^{3/2} = 6x^{3/2}$$ $$\int x^{3/2} \, dx = \frac{2}{5}x^{5/2}$$ Substitute back and evaluate from 0 to 9:
$$V = 2\pi \left[ 6x^{3/2} \Big|_{0}^{9} - \frac{2}{5}x^{5/2} \Big|_{0}^{9} \right]$$ Calculate:
$$= 2\pi \left[ 6(9^{3/2}) - \frac{2}{5}(9^{5/2}) \right]$$ $$= 2\pi \left[ 6(27) - \frac{2}{5}(243) \right]$$ $$= 2\pi \left[ 162 - \frac{486}{5} \right]$$ $$= 2\pi \left[ 162 - 97.2 \right]$$ $$= 2\pi (64.8)$$ $$= 129.6\pi$$ Thus, the volume of the solid is:
$$\boxed{129.6\pi}$$