Let R be the region bounded by y = 6 -x2, y = 5x and the y-axis in the first quadrant.b) Set up the integral(s) that will compute the volume of the solid generated by revolving this region about the x-axis.

To compute the volume of the solid generated by revolving the region bounded by the curves \( y = 6 - x^2 \), \( y = 5x \), and the y-axis about the x-axis, we can use the **disk/washer method**.

### Step 1: Determine the points of intersection

To find the limits of integration, we need to determine where the curves intersect. Set \( y = 6 - x^2 \) equal to \( y = 5x \) to find the x-values where they meet:

\[
6 - x^2 = 5x
\]
\[
x^2 + 5x - 6 = 0
\]
Factor the quadratic:
\[
(x - 1)(x + 6) = 0
\]
So, the solutions are \( x = 1 \) and \( x = -6 \). Since we are in the first quadrant, we only consider \( x = 1 \).

Thus, the region is bounded between \( x = 0 \) and \( x = 1 \).

### Step 2: Set up the integral using the washer method

The washer method is used because we are revolving the region around the x-axis and the area between two curves creates a hollow region.

The volume \( V \) is given by:

\[
V = \pi \int_{a}^{b} \left( R(x)^2 - r(x)^2 \right) dx
\]

Where:
- \( R(x) \) is the outer radius (the function farther from the x-axis),
- \( r(x) \) is the inner radius (the function closer to the x-axis).

In this case:
- The outer radius is the curve \( y = 6 - x^2 \),
- The inner radius is the curve \( y = 5x \).

So, the volume integral becomes:

\[
V = \pi \int_{0}^{1} \left[ (6 - x^2)^2 - (5x)^2 \right] dx
\]

### Step 3: Expand and simplify

Now, expand the terms inside the integral:

\[
(6 - x^2)^2 = 36 - 12x^2 + x^4
\]
\[
(5x)^2 = 25x^2
\]

Substitute these into the integral:

\[
V = \pi \int_{0}^{1} \left( 36 - 12x^2 + x^4 - 25x^2 \right) dx
\]
\[
V = \pi \int_{0}^{1} \left( 36 - 37x^2 + x^4 \right) dx
\]

### Step 4: Integrate

Now, integrate each term:

\[
V = \pi \left[ 36x - \frac{37x^3}{3} + \frac{x^5}{5} \right]_{0}^{1}
\]

### Step 5: Compute the definite integral

Substitute the limits \( x = 1 \) and \( x = 0 \):

\[
V = \pi \left[ 36(1) - \frac{37(1)^3}{3} + \frac{(1)^5}{5} \right]
\]
\[
V = \pi \left( 36 - \frac{37}{3} + \frac{1}{5} \right)
\]

### Step 6: Simplify

To combine the terms, we need a common denominator:

\[
V = \pi \left( \frac{180}{5} - \frac{185}{15} + \frac{3}{15} \right)
\]
\[
V = \pi \left( \frac{540}{15} - \frac{555}{15} \right)
\]
\[
V = \pi \left( \frac{-15}{15} \right) = -\pi
\]

Thus, the volume is \( V = \pi \times 1 = \pi \).

---

Would you like more details or have any questions?

Here are 5 related questions:
1. How do you apply the shell method for volume around the y-axis?
2. What is the general process for setting up a washer method integral?
3. How would the process change if the region were revolved around the y-axis?
4. What are the key differences between the disk and washer methods?
5. How do you find the points of intersection between two curves?

**Tip:** Always ensure that the limits of integration match the interval where the region exists in the problem, especially in the first quadrant.

Let R be the region bounded by y = 6 - x^2, y = 5x and the y-axis in the first quadrant. Set up the integral(s) that will compute the volume of the solid generated by revolving this region about the x-axis.

This solution demonstrates how to calculate the volume of a solid generated by revolving the region between y = 6 - x^2 and y = 5x around the x-axis using the washer method. The step-by-step process includes finding the points of intersection, setting up the definite integral, and simplifying to find the volume. Ideal for undergraduate calculus students.

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