Find the surface area of the part of the sphere x² + y² + z² = 81 that lies above the cone z = √(x² + y²)

Solution:

We need to find the surface area of the portion of the sphere x² + y² + z² = 81 above z = √ (x² + y²)

z=√(x² + y²) can be written as x² + y² = z²

Now, x² + y² + z² = 81 can be written as

z² = 81 - x² - y²

z = √(81 - x² - y²)

Taking first derivative,

f(x) = -x/√(81 - x² - y²)

f(y) = -y/√(81 - x² - y²)

The formula to find surface area is given by

dS = ∬√(1 + [f(x)]² + [f(y)]²)

Converting to polar, 

Parameterize the part of the sphere by 

s(u,v) = (9 cos u sin v, 9 sin u sin v 9 cos v)

With 0 ≤ u ≤ 2𝜋 and 0 ≤ v ≤ 𝜋/4.

Now, dS = ∥su×sv∥

dS = 81 sin v du dv

So, the area S is given by

\(\\\int \int ds=81\int_{v=0}^{v=\pi /4}\int_{u=0}^{u=2\pi }sin v du dv \\ \\\int \int ds=81\int_{v=0}^{v=\pi /4}2\pi sin v dv \\ \\\int \int ds=81(2\pi)\int_{v=0}^{v=\pi /4} sin v dv\)

∬dS = 81(2𝜋)(1 - 1/√2)

On simplification,

∬dS = 162𝜋(1 - 1/√2)

= 162𝜋 - 162𝜋/√2

= 162𝜋 - 81√2𝜋

Taking out common term,

∬dS = 𝜋(162 - 81√2)

Therefore, surface area of the part of the sphere is 𝜋(162 - 81√2).

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Find the surface area of the part of the sphere x² + y² + z² = 81 that lies above the cone z = √ (x² + y²)

Summary:

The surface area of the portion of the sphere x² + y² + z² = 81 above the cone z = √ (x² + y²) is 𝜋(162 - 81√2).