Define  
Q
  as the region bounded by the functions  
f
(
x
)
=
x
  and  
g
(
x
)
=
3
 √
x
  in the first quadrant between  
y
=
3
  and  
y
=
5
 . If  
Q
  is rotated around the  
y
 -axis, what is the volume of the resulting solid?

Submit an exact answer in terms of  
π
 .

We are tasked with finding the volume of a solid formed by rotating a region \( Q \) around the \( y \)-axis. The region \( Q \) is bounded by the curves:

- \( f(x) = x \)
- \( g(x) = \sqrt[3]{x} \)

This region is between \( y = 3 \) and \( y = 5 \) in the first quadrant. To solve this problem, we'll use the **shell method** since we are rotating around the \( y \)-axis and our functions are given in terms of \( x \).

### Step 1: Express the functions in terms of \( y \)

To use the shell method, we need to rewrite the functions in terms of \( y \).

- From \( f(x) = x \), we have \( x = y \).
- From \( g(x) = \sqrt[3]{x} \), we cube both sides to get \( x = y^3 \).

### Step 2: Set up the integral

The volume \( V \) of the solid formed by rotating the region around the \( y \)-axis using the shell method is given by:

\[
V = 2\pi \int_{y_1}^{y_2} \text{(radius)} \times \text{(height)} \, dy
\]

- The radius of each shell is the distance from the \( y \)-axis, which is \( y \).
- The height of each shell is the difference between the two curves: \( y - y^3 \).

Thus, the volume is:

\[
V = 2\pi \int_{3}^{5} \left( y - y^3 \right) \, dy
\]

### Step 3: Compute the integral

First, break the integral into two parts:

\[
V = 2\pi \left( \int_{3}^{5} y \, dy - \int_{3}^{5} y^3 \, dy \right)
\]

Now, compute each integral:

1. \( \int y \, dy = \frac{y^2}{2} \)
2. \( \int y^3 \, dy = \frac{y^4}{4} \)

Evaluating these from \( y = 3 \) to \( y = 5 \):

\[
\left[ \frac{y^2}{2} \right]_{3}^{5} = \frac{5^2}{2} - \frac{3^2}{2} = \frac{25}{2} - \frac{9}{2} = \frac{16}{2} = 8
\]

\[
\left[ \frac{y^4}{4} \right]_{3}^{5} = \frac{5^4}{4} - \frac{3^4}{4} = \frac{625}{4} - \frac{81}{4} = \frac{544}{4} = 136
\]

### Step 4: Final expression for the volume

Now substitute these results back into the expression for the volume:

\[
V = 2\pi \left( 8 - 136 \right) = 2\pi \times (-128) = -256\pi
\]

Since volume cannot be negative, we take the absolute value. Therefore, the exact volume is:

\[
V = 256\pi
\]

### Conclusion:
The volume of the resulting solid is \( 256\pi \).

---

Do you want any more details or have any questions? Here are some related questions to think about:

1. How would the volume change if the region was rotated around the \( x \)-axis instead?
2. What happens if the bounds of the region were different?
3. Can the washer method be used for this problem?
4. What if the functions \( f(x) \) and \( g(x) \) intersected within the given bounds?
5. How would this change if we rotated around the line \( x = 2 \) instead of the \( y \)-axis?

**Tip**: Always check if your setup is correct before solving the integral—this avoids mistakes in limits or function expressions.

Define Q as the region bounded by the functions f(x) = x and g(x) = √3(x) in the first quadrant between y = 3 and y = 5. If Q is rotated around the y-axis, what is the volume of the resulting solid? Submit an exact answer in terms of π.

Learn how to calculate the volume of a solid formed by rotating the region bounded by f(x) = x and g(x) = √3(x) around the y-axis. This detailed solution walks through the use of the shell method and integral calculus to find the volume in terms of π. Suitable for college-level calculus students.

Math Problem Statement

Define
Q as the region bounded by the functions
f ( x )

x and
g ( x )

3 √ x in the first quadrant between
y

3 and
y

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