Question

Compute the surface area of the portion of the sphere with center the origin and radius 4 that lies inside the cylinder x^2+y^2=12

Answer 1

16π

Explanation:

Given that:

The sphere of the radius =
`x^2 + y^2 +z^2 = 4^2`

`z^2 = 16 -x^2 -y^2`

`z = √(16-x^2-y^2)`

The partial derivatives of
`Z_x = (-2x)/(2 √(16-x^2 -y^2))`

`Z_x = (-x)/(√(16-x^2 -y^2))`

Similarly;

`Z_y = (-y)/(√(16-x^2 -y^2))`

∴

`dS = √(1 + Z_x^2 +Z_y^2) \ \ . dA`

`=\sqrt{1 + (x^2)/(16-x^2-y^2) + (y^2)/(16-x^2-y^2) }\ \ .dA`

`=\sqrt{ (16)/(16-x^2-y^2) }\ \ .dA`

`=(4)/(√( 16-x^2-y^2) )\ \ .dA`

Now; the region R = x² + y² = 12

Let;

x = rcosθ = x; x varies from 0 to 2π

y = rsinθ = y; y varies from 0 to
`√(12)`

dA = rdrdθ

∴

The surface area
`S = \int \limits _R \int \ dS`

`= \int \limits _R\int \ (4)/(√( 16-x^2 -y^2) ) \ dA`

`= \int \limits^(2 \pi)_(0) } \int^(√(12))_(0) (4)/(√(16-r^2)) \ \ rdrd \theta`

`= 2 \pi \int^(√(12))_(0) \ (4r)/(√(16-r^2))\ dr`

`= 2 \pi * 4 \Bigg [ (√(16-r^2))/((1)/(2)(-2)) \Bigg]^(√(12))_(0)`

`= 8\pi ( - √(4) + √(16))`

= 8π ( -2 + 4)

= 8π(2)

= 16π