Certainly! Let's find the volume formed by rotating the region enclosed by the curves [tex]\(x = 2.5y\)[/tex] and [tex]\(x = y^3\)[/tex] with [tex]\(y \geq 0\)[/tex] about the [tex]\(y\)[/tex]-axis.
### Step 1: Find the Points of Intersection
First, we need to determine where the two curves intersect. To find these points, we set the equations equal to each other:
[tex]\[ 2.5y = y^3 \][/tex]
Rearranging the equation, we get:
[tex]\[ y^3 - 2.5y = 0 \][/tex]
Factor out [tex]\(y\)[/tex]:
[tex]\[ y(y^2 - 2.5) = 0 \][/tex]
This gives us the solutions:
[tex]\[ y = 0 \quad \text{or} \quad y^2 = 2.5 \][/tex]
Solving for [tex]\(y^2 = 2.5\)[/tex], we get:
[tex]\[ y = \sqrt{2.5} \quad \text{or} \quad y = -\sqrt{2.5} \][/tex]
However, since we are only considering [tex]\(y \geq 0\)[/tex]:
[tex]\[ y = 0 \quad \text{and} \quad y = \sqrt{2.5} \][/tex]
Note that [tex]\(\sqrt{2.5}\)[/tex] can be simplified further to:
[tex]\[ y = \sqrt{2.5} = \sqrt{2.5} = \sqrt{ \frac{25}{10}} = \frac{5}{\sqrt{10}} = \frac{5 \cdot \sqrt{10}}{10} = \frac{\sqrt{25}}{\sqrt{4}} = 1.58113883008419 \][/tex]
Thus, the points of intersection are:
[tex]\[ y = 0 \quad \text{and} \quad y = 1.58113883008419 \][/tex]
### Step 2: Set Up the Integral for the Washer Method
Next, we rotate the region around the [tex]\(y\)[/tex]-axis. This requires us to employ the washer method to find the volume. The volume formula for the washer method when rotating about the [tex]\(y\)[/tex]-axis is given by:
[tex]\[ V = \pi \int_{a}^{b} \left( R(y)^2 - r(y)^2 \right) dy \][/tex]
Here, [tex]\(R(y)\)[/tex] is the outer radius, and [tex]\(r(y)\)[/tex] is the inner radius for the given [tex]\(y\)[/tex]. In our case:
- The outer radius is given by the line [tex]\(x = 2.5y\)[/tex].
- The inner radius is given by the curve [tex]\(x = y^3\)[/tex].
So, [tex]\(R(y) = 2.5y\)[/tex] and [tex]\(r(y) = y^3\)[/tex]. Therefore, the volume integral becomes:
[tex]\[ V = \pi \int_{0}^{1.58113883008419} \left[ (2.5y)^2 - (y^3)^2 \right] dy \][/tex]
Simplify the integrand:
[tex]\[ (2.5y)^2 = 6.25y^2 \][/tex]
[tex]\[ (y^3)^2 = y^6 \][/tex]
Thus, the integral is:
[tex]\[ V = \pi \int_{0}^{1.58113883008419} (6.25y^2 - y^6) \, dy \][/tex]
### Step 3: Compute the Integral
Now, integrate the function within the bounds:
[tex]\[ \int_{0}^{1.58113883008419} (6.25y^2 - y^6) \, dy \][/tex]
Find the indefinite integral first:
[tex]\[ \int (6.25y^2 - y^6) \, dy = 6.25 \int y^2 \, dy - \int y^6 \, dy \][/tex]
This gives:
[tex]\[ 6.25 \cdot \frac{y^3}{3} - \frac{y^7}{7} = \frac{6.25}{3}y^3 - \frac{y^7}{7} \][/tex]
Now, we evaluate this expression from 0 to 1.58113883008419:
[tex]\[ \left[ \frac{6.25}{3} y^3 - \frac{y^7}{7} \right]_{0}^{1.58113883008419} \][/tex]
At [tex]\(y = 1.58113883008419\)[/tex]:
[tex]\[ \frac{6.25}{3} (1.58113883008419)^3 - \frac{(1.58113883008419)^7}{7} = 4.70577032763152 \][/tex]
At [tex]\(y = 0\)[/tex], the terms vanish. So:
[tex]\[ \left[ \frac{6.25}{3} y^3 - \frac{y^7}{7} \right]_{0}^{1.58113883008419} = 4.70577032763152 \][/tex]
### Step 4: Calculate the Volume
Finally, multiply by [tex]\(\pi\)[/tex]:
[tex]\[ V = \pi \times 4.70577032763152 = 14.7836134907680 \][/tex]
Therefore, the volume formed by rotating the given region about the [tex]\(y\)[/tex]-axis is:
[tex]\[ \boxed{14.7836134907680} \][/tex]
