We are to find the surface area of part of a hemisphere. On the hemisphere
z = f ( x , y ) = a 2 − x 2 − y 2 f x ( x , y ) = − x a 2 − x 2 − y 2 f y ( x , y ) = − y a 2 − x 2 − y 2 \begin{aligned}
z=f(x,y)=\sqrt{a^2-x^2-y^2}\qquad
f_x(x,y)=-\frac{x}{\sqrt{a^2-x^2-y^2}}\qquad
f_y(x,y)=-\frac{y}{\sqrt{a^2-x^2-y^2}}
\end{aligned} z = f ( x , y ) = a 2 − x 2 − y 2 f x ( x , y ) = − a 2 − x 2 − y 2 x f y ( x , y ) = − a 2 − x 2 − y 2 y
so that
d S = 1 + f x ( x , y ) 2 + f y ( x , y ) 2 d x d y = 1 + x 2 a 2 − x 2 − y 2 + y 2 a 2 − x 2 − y 2 d x d y = a 2 a 2 − x 2 − y 2 d x d y \begin{aligned}
\mathrm{d}S&=\sqrt{1+f_x(x,y)^2+f_y(x,y)^2}\,\mathrm{d}x\,\mathrm{d}y
=\sqrt{1+\frac{x^2}{a^2-x^2-y^2}+\frac{y^2}{a^2-x^2-y^2}}\,\mathrm{d}x\,\mathrm{d}y \\
&=\sqrt{\frac{a^2}{a^2-x^2-y^2}}\,\mathrm{d}x\,\mathrm{d}y
\end{aligned} d S = 1 + f x ( x , y ) 2 + f y ( x , y ) 2 d x d y = 1 + a 2 − x 2 − y 2 x 2 + a 2 − x 2 − y 2 y 2 d x d y = a 2 − x 2 − y 2 a 2 d x d y
In polar coordinates, this is d S = a a 2 − r 2 r d r d θ \mathrm{d}S=\frac{a}{\sqrt{a^2-r^2}}\,r\,\mathrm{d}r\,\mathrm{d}\theta d S = a 2 − r 2 a r d r d θ x 2 − a x + y 2 = 0 x^2-ax+y^2=0 x 2 − a x + y 2 = 0 r 2 − a r cos θ = 0 r^2-ar\cos\theta=0 r 2 − a r cos θ = 0 r = a cos θ r=a\cos\theta r = a cos θ
So the
S u r f a c e A r e a = 2 ∫ 0 π / 2 d θ ∫ 0 a cos θ d r r a a 2 − r 2 = 2 a ∫ 0 π / 2 d θ [ − a 2 − r 2 ] 0 a cos θ = 2 a ∫ 0 π / 2 d θ [ a − a sin θ ] = 2 a 2 [ θ + cos θ ] 0 π / 2 = a 2 [ π − 2 ] \begin{aligned}
{\rm Surface\ Area}
&= 2\int_0^{\pi/2}\mathrm{d}\theta\int_0^{a\cos\theta}\mathrm{d}r\ r
\frac{a}{\sqrt{a^2-r^2}}
= 2a\int_0^{\pi/2}\mathrm{d}\theta\ \Big[-\sqrt{a^2-r^2}\,\Big]_0^{a\cos\theta}
\\
&= 2a\int_0^{\pi/2}\mathrm{d}\theta\ \big[a-a\sin\theta\big] \\
&= 2a^2\Big[\theta+\cos\theta\Big]_0^{\pi/2}
=a^2[\pi-2]
\end{aligned} Surface Area = 2 ∫ 0 π /2 d θ ∫ 0 a c o s θ d r r a 2 − r 2 a = 2 a ∫ 0 π /2 d θ [ − a 2 − r 2 ] 0 a c o s θ = 2 a ∫ 0 π /2 d θ [ a − a sin θ ] = 2 a 2 [ θ + cos θ ] 0 π /2 = a 2 [ π − 2 ]
