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The problem asks for the volume of a sphere with radius \( r = 4 \) inches using the method of shells. The provided answer seems incorrect.

First, let's find the correct volume of a sphere using the standard formula:

\[
V = \frac{4}{3} \pi r^3
\]

Given \( r = 4 \) inches, the volume is:

\[
V = \frac{4}{3} \pi (4)^3 = \frac{4}{3} \pi \times 64 = \frac{256}{3} \pi \, \text{cubic inches}.
\]

So the correct volume should be \( \frac{256 \pi}{3} \) cubic inches, not \( \frac{512 \pi}{3} \), which is double the correct value.

Would you like a detailed explanation of the method of shells or have any other questions?

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Here are 5 related questions for you:
1. How is the volume of a sphere derived using calculus?
2. Can the method of disks be used instead of shells to find the volume of a sphere?
3. How does the method of shells work for other 3D objects like cones?
4. What is the geometric interpretation of the method of shells for a sphere?
5. How would the volume change if the radius of the sphere doubled?

**Tip:** When solving problems, always check if the given radius or diameter is used correctly in volume formulas to avoid errors.

Use the method of shells to find the volume (in in.³) of a sphere of radius r = 4 in.

Calculate the volume of a sphere using the method of shells with a radius of 4 inches. This calculus problem highlights the use of integral techniques to solve for the volume of a three-dimensional object. Learn how to apply the formula V = (4/3)πr³ for accurate results.

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