Section 6.2: Problem 2 (1 point) Find the volume formed by rotating the region enclosed by: $$ x=9 y \text { and } y^{3}=x \text { with } y \geq 0 $$ about the $$ y $$-axis. Volume $$ =-\frac{\pi}{2} $$

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To find the volume of the region enclosed by $$ x = 9y $$ and $$ y^3 = x $$ for $$ y \geq 0 $$ when rotated about the $$ y $$-axis, we'll use the **washer method**. Here's a step-by-step solution:

1. **Determine the Points of Intersection:**
   $$
   9y = y^3 \implies y^3 - 9y = 0 \implies y(y^2 - 9) = 0 \implies y = 0 \text{ or } y = \pm 3
   $$
   Since $$ y \geq 0 $$, the points of intersection are $$ y = 0 $$ and $$ y = 3 $$.

2. **Set Up the Washer Method:**
   The volume $$ V $$ is given by:
   $$
   V = \pi \int_{0}^{3} \left[ (x_{\text{outer}})^2 - (x_{\text{inner}})^2 \right] dy
   $$
   Here:
   - $$ x_{\text{outer}} = 9y $$
   - $$ x_{\text{inner}} = y^3 $$

3. **Substitute and Integrate:**
   $$
   V = \pi \int_{0}^{3} \left[ (9y)^2 - (y^3)^2 \right] dy = \pi \int_{0}^{3} \left[ 81y^2 - y^6 \right] dy
   $$
   $$
   V = \pi \left[ \frac{81y^3}{3} - \frac{y^7}{7} \right]_0^3 = \pi \left[ 27y^3 - \frac{y^7}{7} \right]_0^3
   $$
   Evaluating from 0 to 3:
   $$
   V = \pi \left[ 27(3)^3 - \frac{(3)^7}{7} \right] = \pi \left[ 729 - \frac{2187}{7} \right]
   $$
   $$
   V = \pi \left[ \frac{5103}{7} - \frac{2187}{7} \right] = \pi \left[ \frac{2916}{7} \right] = \frac{2916\pi}{7}
   $$

4. **Final Answer:**
   The volume is $$ \frac{2916\pi}{7} $$.

**Note:** The initial answer you provided was negative, which isn't possible for a volume. Volumes are always non-negative quantities.

**Final Answer:**
The volume is 2916 × π divided by seven. In symbols,

Volume = 2916 π ⁄ 7
The volume is $$ \frac{2916\pi}{7} $$.

Section 6.2: Problem 2 (1 point) Find the volume formed by rotating the region enclosed by: \[ x=9 y \text { and } y^{3}=x \text { with } y \geq 0 \] about the \( y \)-axis. Volume \( =-\frac{\pi}{2} \)

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