Answer:
16π
Step-by-step explanation:
Given that:
The sphere of the radius = [tex]x^2 + y^2 +z^2 = 4^2[/tex]
[tex]z^2 = 16 -x^2 -y^2[/tex]
[tex]z = \sqrt{16-x^2-y^2}[/tex]
The partial derivatives of [tex]Z_x = \dfrac{-2x}{2 \sqrt{16-x^2 -y^2}}[/tex]
[tex]Z_x = \dfrac{-x}{\sqrt{16-x^2 -y^2}}[/tex]
Similarly;
[tex]Z_y = \dfrac{-y}{\sqrt{16-x^2 -y^2}}[/tex]
∴
[tex]dS = \sqrt{1 + Z_x^2 +Z_y^2} \ \ . dA[/tex]
[tex]=\sqrt{1 + \dfrac{x^2}{16-x^2-y^2} + \dfrac{y^2}{16-x^2-y^2} }\ \ .dA[/tex]
[tex]=\sqrt{ \dfrac{16}{16-x^2-y^2} }\ \ .dA[/tex]
[tex]=\dfrac{4}{\sqrt{ 16-x^2-y^2} }\ \ .dA[/tex]
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 [tex]\sqrt{12}[/tex]
dA = rdrdθ
∴
The surface area [tex]S = \int \limits _R \int \ dS[/tex]
[tex]= \int \limits _R\int \ \dfrac{4}{\sqrt{ 16-x^2 -y^2} } \ dA[/tex]
[tex]= \int \limits^{2 \pi}_{0} } \int^{\sqrt{12}}_{0} \dfrac{4}{\sqrt{16-r^2}} \ \ rdrd \theta[/tex]
[tex]= 2 \pi \int^{\sqrt{12}}_{0} \ \dfrac{4r}{\sqrt{16-r^2}}\ dr[/tex]
[tex]= 2 \pi \times 4 \Bigg [ \dfrac{\sqrt{16-r^2}}{\dfrac{1}{2}(-2)} \Bigg]^{\sqrt{12}}_{0}[/tex]
[tex]= 8\pi ( - \sqrt{4} + \sqrt{16})[/tex]
= 8π ( -2 + 4)
= 8π(2)
= 16π
