Consider a sphere of radius 2 centred around the origin. Calculate the volume of the part of the sphere that is sandwiched between the xy xy -plane and the plane at z=1 z=1 parallel to the xy xy -plane. The answer is

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BlackTom AI · Accepted Answer

We are given a sphere of radius 2 centered at the origin and asked for the volume of the portion of the sphere that lies between the xy-plane (z = 0) and the plane z = 1.
First, note that the cross-section of the sphere at a fixed z is a circle of radius r(z) where r(z)^2 = 4 - z^2 (since x^2 + y^2 + z^2 = 4).
Therefore, the area of the cross-section at height z is A(z) = π r(z)^2 = π(4 - z^2).
To find the volume of the region between z = 0 and z = 1, integrate A(z) with respect to z from 0 to 1:
V = ∫_{0}^{1} π(4 - z^2) dz = π ∫_{0}^{1} (4 - z^2) dz.
Compute the integral: ∫(4 - z^2) dz = 4z - z^3/3. Evaluate from 0 to 1:
V = π[ (4(1) - 1^3/3) - (0) ] = π(4 - 1/3) = π(11/3) = 11π/3.
Now examine the answer options:
- Option A: [math: 73π] 7/3 π equals 7π/3. This does not match 11π/3; numerically it is smaller, so this option represents a volume that is too small and does not equal the computed value.
- Option B: π π equals π^2, which is about 9.87, whereas 11π/3 is about 11.52. This option is dimensionally inconsistent with a volume in terms of π and does not match the computed value.
- Option C: 11/3 π equals 11π/3, which exactly matches the value obtained from the integral. This is the correct numerical result for the requested volume.
Thus, comparing each option to the computed volume, the first and second options do not equal 11π/3, while the third option matches the calculation precisely.

PHAS0009_24-25 2.3 Quiz 2 [0.25 hrs]

Consider a sphere of radius 2 centred around the origin. Calculate the volume of the part of the sphere that is sandwiched between the xy xy -plane and the plane at z=1 z=1 parallel to the xy xy -plane. The answer is

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