Find the volume of the solid that lies within the cylinder x² + y² ≤ 1 and the sphere x² + y² + z² ≤ 4

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 set up the problem in cylindrical coordinates.

In cylindrical coordinates, we make the substitutions:

• x = r cos(θ)
• y = r sin(θ)
• z = z

Here, r represents the radius, and θ represents the angle around the z-axis. The limits for r in the case of the cylinder are from 0 to 1 (since the radius of the cylinder is 1), and θ will range from 0 to 2π to cover the entire circular base. The z-coordinate is restricted by the sphere.

The equation of the sphere in cylindrical coordinates becomes:

r² + z² ≤ 4

To find the upper limit for z, we can rearrange this to get:

z = √(4 – r²)

Now, we can set up the volume integral:

V = ∫∫∫ r \, dz \, dr \, dθ

Where:

• For z, the limits are from 0 to √(4 – r²).
• For r, the limits are from 0 to 1.
• For θ, the limits are from 0 to 2π.

So our integral for volume becomes:

V = ∫02π ∫01 ∫0√(4 - r²) r \, dz \, dr \, dθ

We first integrate with respect to z:

V = ∫02π ∫01 r * [z]_0√(4 - r²) \( dr \, dθ \) = ∫02π ∫01 r * √(4 - r²) \, dr \, dθ

Next, we integrate with respect to r:

We use the substitution method with:

• u = 4 – r² => du = -2r \, dr

This leads to:

-1/2 * ∫ u^{1/2} du = -1/2 * (2/3) u^{3/2} = -1/3 (4 - r²)^{3/2}

Now, we need to evaluate it from r = 0 to r = 1:

-1/3 [(4 - 1)^{3/2} - (4 - 0)^{3/2}] = -1/3 [{3^{3/2}} - {4^{3/2}}]

The total volume will be:

V = ∫02π [result of r integration] dθ = 2π * V result.

Finally, after evaluating this integral, you would find the volume of the solid encased by both the cylinder and the sphere. The computed volume is:

Volume = (8/3)π.