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
1 answer:
Answer:
16π
Step-by-step explanation:
Given that:
The sphere of the radius = 


The partial derivatives of 

Similarly;

∴




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 
dA = rdrdθ
∴
The surface area 



![= 2 \pi \times 4 \Bigg [ \dfrac{\sqrt{16-r^2}}{\dfrac{1}{2}(-2)} \Bigg]^{\sqrt{12}}_{0}](https://tex.z-dn.net/?f=%3D%202%20%5Cpi%20%5Ctimes%204%20%5CBigg%20%5B%20%5Cdfrac%7B%5Csqrt%7B16-r%5E2%7D%7D%7B%5Cdfrac%7B1%7D%7B2%7D%28-2%29%7D%20%5CBigg%5D%5E%7B%5Csqrt%7B12%7D%7D_%7B0%7D)

= 8π ( -2 + 4)
= 8π(2)
= 16π
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