Finding Surface Area
Find the surface area of the part of the sphere \(x^{2}+y^{2}+z^{2}=a^{2}\) that lies above the \(xy\)-plane and is contained within the cylinder \( x^{2}+y^{2}=ax,\) \(a\gt 0\).
The partial derivatives of \(z\) \(=f(x,y)\) are \[ f_{x}(x,y)=-\dfrac{x}{\sqrt{a^{2}-x^{2}-y^{2}}}\qquad f_{y}(x,y)=- \dfrac{y}{\sqrt{a^{2}-x^{2}-y^{2}}} \]
The projection of the cylinder onto the \(xy\)-plane is the region \(R\) enclosed by the circle \(x^{2}+y^{2}=ax\), \(a\gt 0.\) Then the surface area \(S\) that we seek is \[ S=\displaystyle\iint\limits_{\kern-3ptR}\sqrt{\dfrac{x^{2}}{a^{2}-x^{2}-y^{2}}+\dfrac{y^{2}}{ a^{2}-x^{2}-y^{2}}+1}\,\,{\it dA}=\displaystyle\iint\limits_{\kern-3ptR}\sqrt{\dfrac{a^{2}}{ a^{2}-x^{2}-y^{2}}}\,{\it dA} \]
Since the integrand involves the expression \(x^{2}+y^{2}\) and the boundary of \(R\) is the circle \(x^{2}+y^{2}=ax\), we use polar coordinates \((r,\theta)\). We begin by finding the limits of integration. Since \begin{eqnarray*} x^{2}+y^{2} &=&ax \\ r^{2} &=&ar\cos \theta \\ r &=&a\cos \theta \end{eqnarray*}
we find that \(r\) varies from \(0\) to \(a\cos \theta\).
Since half the area is to the right of the \(xz\)-plane and half is to the left of the plane, we can let \(\theta\) vary from \(0\) to \(\dfrac{\pi }{2}\) and double the area. Then the surface area \(S\) is \[ \begin{array}{rcl} S& =&\int\limits\int_{\kern-11ptR}\sqrt{\dfrac{a^{2}}{a^{2}-x^{2}-y^{2}}}\,d\!A \underset{\underset{{\color{#0066A7}{\hbox {a>0}}}}{\color{#0066A7}{\uparrow }}}{=} \int\limits\int_{\kern-11ptR}\dfrac{a}{\sqrt{a^{2}-r^{2}}} \,r\,dr\,d\theta \underset{\underset{{\color{#0066A7}{\hbox{Double the area}}}}{\color{#0066A7}{\uparrow}}}{=} 2\int_{0}^{\pi /2}\!\!\int_{0}^{a\cos \theta }\dfrac{a}{\sqrt{a^{2}-r^{2}}} \,r\,dr\,d\theta \notag \\ \quad \\ &=&-2a\int_{0}^{\pi /2}\Big[ \sqrt{a^{2}-r^{2}}\Big] _{0}^{a\cos \theta }d\theta =-2a\int_{0}^{\pi /2}(a\sin \theta -a)\,d\theta\\ &=&-2a^{2}\big[-\cos \theta -\theta \big] _{0}^{\pi /2}=a^{2}(\pi -2) \end{array} \]