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
Optimization
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Integration
Integration means to sum the things. In mathematics, it is the branch of Calculus which is used to find the area under the curve. The operation subtraction is the inverse of addition, division is the inverse of multiplication. In the same way, integration and differentiation are inverse operators. Differential equations give a relation between a function and its derivative.
Application of Integration
In mathematics, the process of integration is used to compute complex area related problems. With the application of integration, solving area related problems, whether they are a curve, or a curve between lines, can be done easily.
Volume
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Area
Area refers to the amount of space a figure encloses and the number of square units that cover a shape. It is two-dimensional and is measured in square units.
![**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} -](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F1c8b7e9c-3342-48f1-b698-8a1892e2f614%2F1ad95277-643a-42cb-b70e-4819a10f021c%2Fxyj0bh_processed.jpeg&w=3840&q=75)
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