Use the method of cylindrical shells to find the volume generated by rotating the region bounded by the given curves about the specified axis y = x³, y = 8, x = 0; about x = 5

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**Calculus: Volume Using Cylindrical Shells**

To find the volume generated by rotating the region bounded by the given curves about a specified axis, we use the method of cylindrical shells.

**Given Curves:**
\[ y = x^3 \]
\[ y = 8 \]
\[ x = 0 \]

**Rotation Axis:**
\(\text{about } x = 5 \)

To solve this problem, follow these steps:

1. **Visualize the Bounded Region:**
   - \( y = x^3 \) is a cubic curve.
   - \( y = 8 \) is a horizontal line.
   - \( x = 0 \) is the y-axis.

2. **Determine the Intersection Points:**
   - The region is bounded on the left by \( x = 0 \), on the right by the point where \( y = x^3 \) meets \( y = 8 \).

3. **Set up the Integral:**
   - Use cylindrical shells to find the volume when the region is rotated about \( x = 5 \).

The volume \( V \) is given by:

\[ V = 2\pi \int_{a}^{b} ( \text{radius} ) ( \text{height} ) \, dx \]

In this case:
- The radius is the distance from the axis of rotation \( x = 5 \) to a point \( x \) on the curve, which is \( 5 - x \).
- The height is given by \( y = x^3 \).

**So, the integral becomes:**
\[ V = 2\pi \int_{0}^{2} (5 - x)(x^3) \, dx \]

4. **Evaluate the Integral:**

\[ V = 2\pi \int_{0}^{2} (5x^3 - x^4) \, dx \]
\[ V = 2\pi \left[ \frac{5x^4}{4} - \frac{x^5}{5} \right]_{0}^{2} \]

5. **Simplify:**
\[ V = 2\pi \left\{ \left( \frac{5(2)^4}{4} - \frac{(2)^5}{5} \right) - \left( \frac{5(0)^4}{4} -
Transcribed Image Text:**Calculus: Volume Using Cylindrical Shells** To find the volume generated by rotating the region bounded by the given curves about a specified axis, we use the method of cylindrical shells. **Given Curves:** \[ y = x^3 \] \[ y = 8 \] \[ x = 0 \] **Rotation Axis:** \(\text{about } x = 5 \) To solve this problem, follow these steps: 1. **Visualize the Bounded Region:** - \( y = x^3 \) is a cubic curve. - \( y = 8 \) is a horizontal line. - \( x = 0 \) is the y-axis. 2. **Determine the Intersection Points:** - The region is bounded on the left by \( x = 0 \), on the right by the point where \( y = x^3 \) meets \( y = 8 \). 3. **Set up the Integral:** - Use cylindrical shells to find the volume when the region is rotated about \( x = 5 \). The volume \( V \) is given by: \[ V = 2\pi \int_{a}^{b} ( \text{radius} ) ( \text{height} ) \, dx \] In this case: - The radius is the distance from the axis of rotation \( x = 5 \) to a point \( x \) on the curve, which is \( 5 - x \). - The height is given by \( y = x^3 \). **So, the integral becomes:** \[ V = 2\pi \int_{0}^{2} (5 - x)(x^3) \, dx \] 4. **Evaluate the Integral:** \[ V = 2\pi \int_{0}^{2} (5x^3 - x^4) \, dx \] \[ V = 2\pi \left[ \frac{5x^4}{4} - \frac{x^5}{5} \right]_{0}^{2} \] 5. **Simplify:** \[ V = 2\pi \left\{ \left( \frac{5(2)^4}{4} - \frac{(2)^5}{5} \right) - \left( \frac{5(0)^4}{4} -
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