How can I find the volume of the solid that lies within the cylinder defined by x² + y² ≤ 1 and the sphere defined by x² + y² + z² ≤ 4?

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Finding the Volume of the Solid

To find the volume of the solid that lies within both the cylinder defined by the equation x² + y² ≤ 1 and the sphere defined by x² + y² + z² ≤ 4, we need to first understand the boundaries set by these geometric shapes.

Understanding the Cylinder and Sphere

The cylinder:

  • has a radius of 1 (since it’s defined by x² + y² ≤ 1) and extends infinitely in the z-direction.

The sphere:

  • has a radius of 2 (as determined from x² + y² + z² ≤ 4, which implies z² ≤ 4 – x² – y²).

Setting Up the Volume Integral

To compute the volume of the solid, we will use cylindrical coordinates, where we set:

  • α (radius) goes from 0 to 1
  • θ (angle) goes from 0 to 2π
  • z goes from the bottom of the sphere to the top, constrained by the equation of the sphere.

The volume integral in cylindrical coordinates can be expressed as:

Volume Integral

We need to compute:

V = ∫∫∫ r \, dz \, r \, dr \, dθ

Where:

  • r is the radial coordinate (which ranges from 0 to 1).
  • θ is the angular coordinate (which ranges from 0 to 2π).
  • z limits are given by the sphere’s equation rearranged as: z = ± sqrt{4 – r²} (resulting from

Calculating the Volume

The limits for z, from -sqrt(4 – r²) to +sqrt(4 – r²), and integrating:

V = ∫_{0}^{2π} ∫_{0}^{1} ∫_{-sqrt{4 - r²}}^{sqrt{4 - r²}} r \, dz \, r \, dr \, dθ

Carrying out the integration, we find:

V = ∫_{0}^{2π} ∫_{0}^{1} (2sqrt{4 - r²}) \, r \, dr \, dθ

Evaluating the Integrals

First, we can handle the integration with respect to z top:

= ∫_{0}^{2π} ∫_{0}^{1} 2r*sqrt{4 - r²} \ dr \, dθ

Next, we will compute:

2 ∫_{0}^{2π} dθ = 4π

Then moving to the r integral:

∫_{0}^{1} 2r*sqrt{4 - r²} \ dr = 2(&frac{1}{3}(4 - r²)^{3/2}) \frac{1}{2}|_{0}^{1}

Final Calculation

Putting it all together, we have:

V = 4π * (from r integral results)

Thus, the volume of the solid contained within the specified cylinder and sphere comes out to be:

Final Result

V = &frac{8}{3}π

This result provides us with the volume of the solid enclosed by the cylinder and the sphere. Utilizing integration techniques and understanding the geometric boundaries is essential to arrive at this answer.

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