S be the unit sphere x ^ 2 y ^ 2 z ^ 2 = 1 Then the value of surface integral int s [(2x ^ 2 3x) - y ^ 2 5z ^ 2] * ds is (a) 2π (c) 8π (b) 4pi (d) 12π?

Question

S be the unit sphere x ^ 2   y ^ 2   z ^ 2 = 1 Then the value of surface integral int s [(2x ^ 2   3x) - y ^ 2   5z ^ 2] * ds is (a) 2π (c) 8π (b) 4pi (d) 12π?

Answer

Solution:

The given surface integral can be written as:

∫_S [(2x^2 + 3x) - y^2 + 5z^2] * ds

where S is the unit sphere x^2 + y^2 + z^2 = 1.

To solve this, we will use the parameterization of the unit sphere. Let:
x = sinθcosφ
y = sinθsinφ
z = cosθ

where θ varies from 0 to π and φ varies from 0 to 2π.

Step 1:
Compute the cross product of the partial derivatives of the parameterization:

∂r/∂θ = (cosθcosφ, cosθsinφ, -sinθ)
∂r/∂φ = (-sinθsinφ, sinθcosφ, 0)

∂r/∂θ × ∂r/∂φ = (cosθcosφ, cosθsinφ, -sinθ) × (-sinθsinφ, sinθcosφ, 0)
= (-cosθsin^2φ, -cosθcos^2φ, -sinθcosφ)

Step 2:
Compute the magnitude of the cross product:

|∂r/∂θ × ∂r/∂φ| = √[(-cosθsin^2φ)^2 + (-cosθcos^2φ)^2 + (-sinθcosφ)^2]
= √[cos^2θsin^4φ + cos^2θcos^4φ + sin^2θcos^2φ]
= √[cos^2θ(sin^4φ + cos^4φ) + sin^2θcos^2φ]
= √[cos^2θ + sin^2θcos^2φ]
= √[1 + (1 - sin^2θ)cos^2φ]
= √[1 + cos^2φ - sin^2θcos^2φ]
= √[1 + cos^2φ - sin^2θ(1 - sin^2φ)]
= √[1 + cos^2φ - sin^2θ + sin^2θsin^2φ]
= √[2 - sin^2θ + sin^2θsin^2φ]
= √[2 - sin^2θ(1 - sin^2φ)]
= √[2 - sin^2θcos^2(1 - sin^2φ)]
= √[2 - sin^2θcos^2θ]
= √[2(1 - sin^2θcos^2θ)]
= √[2(1 - sin^2θ(1 - sin^2θ))]
= √[2(1 - sin^2θ + sin^4θ)]
= √[2(1 + sin^2θ)]
= √[2(1 + (1 - cos^2θ))]
= √[4 - 2cos^2θ]
= √[4sin^2θ]
= 2sinθ

Step 3:
Substitute the parameterization and the magnitude of the cross product into the

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