To find the volume of the solid generated by revolving the region between the curve [tex]\( y = \sqrt{x} \)[/tex], for [tex]\( 0 \leq x \leq 9 \)[/tex], and the [tex]\( y \)[/tex]-axis around the [tex]\( y \)[/tex]-axis, we can use the disk method.
Step-by-step solution:
1. Express [tex]\( x \)[/tex] in terms of [tex]\( y \)[/tex]:
Given the curve [tex]\( y = \sqrt{x} \)[/tex], we first express [tex]\( x \)[/tex] in terms of [tex]\( y \)[/tex]:
[tex]\[
y = \sqrt{x} \implies x = y^2
\][/tex]
2. Determine the limits for [tex]\( y \)[/tex]:
Since [tex]\( y = \sqrt{x} \)[/tex] and [tex]\( x \)[/tex] ranges from 0 to 9, [tex]\( y \)[/tex] will range from [tex]\( \sqrt{0} \)[/tex] to [tex]\( \sqrt{9} \)[/tex], i.e., from 0 to 3. Hence, the limits for [tex]\( y \)[/tex] are 0 and 3.
3. Volume of the solid using the disk method:
The disk method uses the formula for volume:
[tex]\[
V = \pi \int_{a}^{b} [R(y)]^2 \, dy
\][/tex]
where [tex]\( R(y) \)[/tex] is the radius of the disk at height [tex]\( y \)[/tex].
In this case, [tex]\( R(y) \)[/tex] is given by the function [tex]\( x = y^2 \)[/tex]. Thus, the radius [tex]\( R(y) = y^2 \)[/tex].
4. Set up the integral:
[tex]\[
V = \pi \int_{0}^{3} (y^2)^2 \, dy
\][/tex]
5. Simplify the expression:
[tex]\[
V = \pi \int_{0}^{3} y^4 \, dy
\][/tex]
6. Evaluate the integral:
To evaluate the integral, we find the antiderivative of [tex]\( y^4 \)[/tex]:
[tex]\[
\int y^4 \, dy = \frac{y^5}{5} + C
\][/tex]
Using the limits from 0 to 3, we get:
[tex]\[
\left[\frac{y^5}{5}\right]_{0}^{3} = \left( \frac{3^5}{5} \right) - \left( \frac{0^5}{5} \right)
= \frac{3^5}{5} - 0
= \frac{243}{5}
\][/tex]
7. Multiply by [tex]\(\pi\)[/tex]:
[tex]\[
V = \pi \cdot \frac{243}{5} = \frac{243\pi}{5}
\][/tex]
Thus, the exact value of the volume of the solid is:
[tex]\[
\boxed{\frac{243\pi}{5}}
\][/tex]
