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A circular cylindrical hole is bored through a solid sphere, the axis of the hole being a diameter of the sphere. The volume of the remaining solid is V = 2 L 2p 0 L 23 0 L 24-z 2 1 r dr dz du. a. Find the radius of the hole and the radius of the sphere. b. Evaluate the integral?
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A circular cylindrical hole is bored through a solid sphere, the axis ...
**a. Finding the Radius of the Hole and the Radius of the Sphere**

To find the radius of the hole and the radius of the sphere, we need to analyze the given volume expression and make use of the properties of a sphere and a cylindrical hole.

Let's break down the given volume expression:

V = 2 L^2 * pi * 0 * L^2 * 3 * 0 * L^2 * 4 - z^2 * 1 * r * dr * dz * du

From the expression, we can identify several variables:

- L: Represents the length of the cylindrical hole.
- r: Represents the radius of the hole.
- z: Represents the distance from the center of the sphere to the top of the cylindrical hole.
- u: Represents the angle of rotation around the axis of the hole.

From this, we can conclude that the remaining solid after the cylindrical hole is bored is symmetric about the axis of the hole. Thus, we can express the volume in terms of the cylindrical coordinates (r, z, u).

Now, let's analyze the given expression further:

V = 2 L^2 * pi * 0 * L^2 * 3 * 0 * L^2 * 4 - z^2 * 1 * r * dr * dz * du

We can simplify this expression to:

V = 2 * pi * L^2 * (L^2 * 3 * L^2 * 4 - z^2) * ∫(r * dr) * ∫(dz) * ∫(du)

To find the radius of the hole, we can equate the expression inside the parentheses to zero:

L^2 * 3 * L^2 * 4 - z^2 = 0

Simplifying this equation, we get:

L^6 = z^2

Taking the square root of both sides, we have:

L^3 = z

Therefore, the radius of the hole is L^3.

Now, to find the radius of the sphere, we need to consider the original volume of the sphere without the hole. Let's denote the radius of the sphere as R. The volume of the sphere is given by:

V_sphere = (4/3) * pi * R^3

We can equate this volume to the remaining volume after the hole is bored:

V_sphere = V

(4/3) * pi * R^3 = 2 * pi * L^2 * (L^2 * 3 * L^2 * 4 - z^2) * ∫(r * dr) * ∫(dz) * ∫(du)

Canceling out the common terms, we get:

(4/3) * R^3 = 2 * L^2 * (L^2 * 3 * L^2 * 4 - z^2) * ∫(r * dr) * ∫(dz) * ∫(du)

Simplifying further, we have:

(4/3) * R^3 = 2 * L^2 * (L^2 * 3 * L^2 * 4 - L^6) * ∫(r * dr) * ∫(dz) * ∫(du)

(4/3) * R^3 = 2 * L
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A circular cylindrical hole is bored through a solid sphere, the axis of the hole being a diameter of the sphere. The volume of the remaining solid is V = 2 L 2p 0 L 23 0 L 24-z 2 1 r dr dz du. a. Find the radius of the hole and the radius of the sphere. b. Evaluate the integral?
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A circular cylindrical hole is bored through a solid sphere, the axis of the hole being a diameter of the sphere. The volume of the remaining solid is V = 2 L 2p 0 L 23 0 L 24-z 2 1 r dr dz du. a. Find the radius of the hole and the radius of the sphere. b. Evaluate the integral? for Class 12 2024 is part of Class 12 preparation. The Question and answers have been prepared according to the Class 12 exam syllabus. Information about A circular cylindrical hole is bored through a solid sphere, the axis of the hole being a diameter of the sphere. The volume of the remaining solid is V = 2 L 2p 0 L 23 0 L 24-z 2 1 r dr dz du. a. Find the radius of the hole and the radius of the sphere. b. Evaluate the integral? covers all topics & solutions for Class 12 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for A circular cylindrical hole is bored through a solid sphere, the axis of the hole being a diameter of the sphere. The volume of the remaining solid is V = 2 L 2p 0 L 23 0 L 24-z 2 1 r dr dz du. a. Find the radius of the hole and the radius of the sphere. b. Evaluate the integral?.
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