The region between the curve $$ y=\sqrt{x}, 0 \leq x \leq 9 $$ and the $$ y $$-axis is revolved about the $$ y $$-axis to generate a solid. Find the exact value of its volume. $$ [?] \pi $$

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We use the shell method. The radius of a typical cylindrical shell is $$ x $$ and its height is given by the $$ y $$-value of the curve, namely $$ \sqrt{x} $$. The volume is then

$$
V = 2\pi \int_{0}^{9} x \sqrt{x}\, dx.
$$

Since

$$
x \sqrt{x} = x \cdot x^{\frac{1}{2}} = x^{\frac{3}{2}},
$$

we have

$$
V = 2\pi \int_{0}^{9} x^{\frac{3}{2}}\, dx.
$$

To evaluate the integral, compute

$$
\int x^{\frac{3}{2}}\, dx = \frac{2}{5} x^{\frac{5}{2}} + C.
$$

Thus,

$$
\int_{0}^{9} x^{\frac{3}{2}}\, dx = \frac{2}{5} \left[ x^{\frac{5}{2}} \right]_{0}^{9} = \frac{2}{5} \left( 9^{\frac{5}{2}} - 0 \right).
$$

Since

$$
9^{\frac{5}{2}} = \left( \sqrt{9} \right)^5 = 3^5 = 243,
$$

it follows that

$$
\int_{0}^{9} x^{\frac{3}{2}}\, dx = \frac{2}{5} \cdot 243 = \frac{486}{5}.
$$

Therefore, the volume is

$$
V = 2\pi \cdot \frac{486}{5} = \frac{972\pi}{5}.
$$

The exact volume of the solid is

$$
\boxed{\frac{972\pi}{5}}.
$$
The volume of the solid is $$ \frac{972\pi}{5} $$.

The region between the curve

and the -axis is revolved about
the -axis to generate a solid.
Find the exact value of its volume.

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The region between the curve and the -axis is revolved about the -axis to generate a solid. Find the exact value of its volume.