Now because our surface is a sphere we’ll need to parameterize it and use the following formula for the surface integral.

$$\iint\limits_{S}{{f\left( {x,y,z} \right)\,dS}} = \iint\limits_{D}{{f\left( {\vec r\left( {u,v} \right)} \right)\left\| {{{\vec r}_u} \times {{\vec r}_v}} \right\|\,dA}}$$

where $u$ and $v$ will be chosen as needed when doing the parameterization.

We saw how to parameterize a sphere in the previous section so we won’t go into detail for the parameterization. The parameterization is,

$$\vec r\left( {\theta ,\varphi } \right) = \left\langle {3\sin \varphi \cos \theta ,3\sin \varphi \sin \theta ,3\cos \varphi } \right\rangle \hspace{0.25in}\,\,\,\,\,\frac{1}{2}\pi \le \theta \le \pi ,\,\,\,\,\,0 \le \varphi \le \frac{1}{2}\pi$$

We needed the restriction on $\varphi$ to make sure that we only get a portion of the upper half of the sphere (i.e.$z \ge 0$). Likewise the restriction on $\theta$ was needed to get only the portion that was in the 2^nd quadrant of the $xy$-plane (i.e. $x \le 0$ and $y \ge 0$).

Next, we’ll need to compute the cross product.

$${\vec r_\theta } = \left\langle { - 3\sin \varphi \sin \theta ,3\sin \varphi \cos \theta ,0} \right\rangle \hspace{0.25in}\hspace{0.25in}{\vec r_\varphi } = \left\langle {3\cos \varphi \cos \theta ,3\cos \varphi \sin \theta , - 3\sin \varphi } \right\rangle$$ $$\begin{align*}{{\vec r}_\theta } \times {{\vec r}_\varphi } & = \left| {\begin{array}{*{20}{c}}{\vec i}&{\vec j}&{\vec k}\\{ - 3\sin \varphi \sin \theta }&{3\sin \varphi \cos \theta }&0\\{3\cos \varphi \cos \theta }&{3\cos \varphi \sin \theta }&{ - 3\sin \varphi }\end{array}} \right|\\ & = - 9{\sin ^2}\varphi \cos \theta \,\vec i - 9\sin \varphi \cos \varphi {\sin ^2}\theta \vec k - 9\sin \varphi \cos \varphi {\cos ^2}\theta \vec k - 9{\sin ^2}\varphi \sin \theta \vec j\\ & = - 9{\sin ^2}\varphi \cos \theta \,\vec i - 9{\sin ^2}\varphi \sin \theta \vec j - 9\sin \varphi \cos \varphi \left( {{{\sin }^2}\theta + {{\cos }^2}\theta } \right)\vec k\\ & = - 9{\sin ^2}\varphi \cos \theta \,\vec i - 9{\sin ^2}\varphi \sin \theta \vec j - 9\sin \varphi \cos \varphi \vec k\end{align*}$$

The magnitude of the cross product is,

$$\begin{align*}\left\| {{{\vec r}_\theta } \times {{\vec r}_\varphi }} \right\| & = \sqrt {{{\left( { - 9{{\sin }^2}\varphi \cos \theta } \right)}^2} + {{\left( { - 9{{\sin }^2}\varphi \sin \theta } \right)}^2} + {{\left( { - 9\sin \varphi \cos \varphi } \right)}^2}} \\ & = \sqrt {81{{\sin }^4}\varphi \left( {{{\cos }^2}\theta + {{\sin }^2}\theta } \right) + 81{{\sin }^2}\varphi {{\cos }^2}\varphi } \\ & = \sqrt {81{{\sin }^2}\varphi \left( {{{\sin }^2}\varphi + {{\cos }^2}\varphi } \right)} \\ & = 9\left| {\sin \varphi } \right|\\ & = 9\sin \varphi \end{align*}$$

The integral is then,

$$\iint\limits_{S}{{xz\,dS}} = \iint\limits_{D}{{\left( {3\sin \varphi \cos \theta } \right)\left( {3\cos \varphi } \right)\left( {9\sin \varphi } \right)\,dA}} = \iint\limits_{D}{{81\cos \varphi {{\sin }^2}\varphi \cos \theta \,dA}}$$

Don’t forget to plug the $x$ and $z$ component of the surface parameterization into the integrand and $D$ is just the limits on $\theta$ and $\varphi$ we noted above in the parameterization.

Show Step 3

Now all that we need to do is evaluate the double integral and that shouldn’t be too difficult at this point.

The integral is then,

$$\begin{align*}\iint\limits_{S}{{xz\,dS}} & = \iint\limits_{D}{{81\cos \varphi {{\sin }^2}\varphi \cos \theta \,dA}}\\ & = \int_{{\frac{1}{2}\pi }}^{\pi }{{\int_{0}^{{\frac{1}{2}\pi }}{{81\cos \varphi {{\sin }^2}\varphi \cos \theta \,d\varphi }}\,d\theta }}\\ & = \int_{{\frac{1}{2}\pi }}^{\pi }{{\left. {\left( {27{{\sin }^3}\varphi \cos \theta \,d\varphi } \right)} \right|_0^{\frac{1}{2}\pi }\,d\theta }}\\ & = \int_{{\frac{1}{2}\pi }}^{\pi }{{27\cos \theta \,d\theta }}\\ & = \left. {\left( {27\sin \theta } \right)} \right|_{\frac{1}{2}\pi }^\pi = \require{bbox} \bbox[2pt,border:1px solid black]{{ - 27}}\end{align*}$$