Q
  is the region bounded by the graph of  
u
(
y
)
=
y
2
9
 , the  
y
 -axis,  
y
=
0
 , and  
y
=
6
 . Find the volume of the solid of revolution formed by revolving  
Q
  around the  
x
 -axis.

Submit an exact answer in terms of  
π
 .

To find the volume of the solid of revolution formed by revolving the region \( Q \) around the \( x \)-axis, we use the **disk method**. The general formula for the disk method when revolving around the \( x \)-axis is:

\[
V = \pi \int_{a}^{b} \left[ f(y) \right]^2 \, dy
\]

Given:
- The function is \( u(y) = \frac{y^2}{9} \).
- The region is bounded by \( y = 0 \) and \( y = 6 \).

We need to set up the integral as follows:

\[
V = \pi \int_{0}^{6} \left( \frac{y^2}{9} \right)^2 \, dy
\]

Now, simplify the integrand:

\[
\left( \frac{y^2}{9} \right)^2 = \frac{y^4}{81}
\]

Thus, the volume integral becomes:

\[
V = \pi \int_{0}^{6} \frac{y^4}{81} \, dy
\]

Factor out the constant \( \frac{1}{81} \):

\[
V = \frac{\pi}{81} \int_{0}^{6} y^4 \, dy
\]

Now, integrate \( y^4 \):

\[
\int y^4 \, dy = \frac{y^5}{5}
\]

So, we substitute this into the integral:

\[
V = \frac{\pi}{81} \left[ \frac{y^5}{5} \right]_{0}^{6}
\]

Now, evaluate the definite integral:

\[
V = \frac{\pi}{81} \left( \frac{6^5}{5} - \frac{0^5}{5} \right)
\]
\[
V = \frac{\pi}{81} \times \frac{7776}{5}
\]
\[
V = \frac{\pi \times 7776}{405}
\]
\[
V = \frac{2592 \pi}{135}
\]
\[
V = \frac{192 \pi}{5}
\]

Thus, the volume of the solid of revolution is:

\[
V = \frac{192 \pi}{5}
\]

Let me know if you'd like more details or if you have any questions!

Here are 5 related questions to expand on this topic:
1. How does the disk method differ from the washer method?
2. What happens to the volume if the region is revolved around the \( y \)-axis instead?
3. Can the shell method be used to find this volume? If so, how?
4. What is the geometric interpretation of the volume obtained?
5. How would the result change if the region were revolved around a line other than the \( x \)-axis?

**Tip**: Always visualize the solid generated by the revolution to better understand the problem's geometry.

The region bounded by the graph of u(y) = y^2/9, the y-axis, y = 0, and y = 6. Find the volume of the solid of revolution formed by revolving Q around the x-axis. Submit an exact answer in terms of π.

Find the volume of the solid formed by revolving the region bounded by the graph of u(y) = y^2/9, the y-axis, y = 0, and y = 6 around the x-axis. This problem explores the disk method in calculus for volume calculation, suitable for students in Grades 11-12 or college-level courses.

Math Problem Statement

Q is the region bounded by the graph of
u ( y )

y 2 9 , the
y -axis,
y

0 , and
y

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