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### Section 6-2 : Parametric Surfaces

8. Determine the surface area of the portion of $${x^2} + {y^2} + {z^2} = 25$$ with $$z \le 0$$.

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We first need to parameterize the sphere and we’ve already done a sphere in this problem set so we won’t go into great detail with the parameterization here.

The parameterization for the full sphere is,

$\vec r\left( {\theta ,\varphi } \right) = \left\langle {5\sin \varphi \cos \theta ,5\sin \varphi \sin \theta ,5\cos \varphi } \right\rangle$

We don’t want the full sphere of course. We only want the lower half of the sphere, i.e. the portion with $$z \le 0$$. This means that we’ll need to restrict $$\varphi$$ to $$\frac{1}{2}\pi \le \varphi \le \pi$$. Recall that $$\varphi$$ is the angle points make with the positive $$z$$-axis and because we only want points below the $$xy$$-plane we’ll need the range of $$\frac{1}{2}\pi \le \varphi \le \pi$$.

We want the full lower half and so we’ll use $$0 \le \theta \le 2\pi$$ for our $$\theta$$ range.

Show Step 2

Next, we need to compute $${\vec r_\theta } \times {\vec r_\varphi }$$. Here is that work.

${\vec r_\theta } = \left\langle { - 5\sin \varphi \sin \theta ,5\sin \varphi \cos \theta ,0} \right\rangle \hspace{0.5in}{\vec r_\varphi } = \left\langle {5\cos \varphi \cos \theta ,5\cos \varphi \sin \theta , - 5\sin \varphi } \right\rangle$ \begin{align*}{{\vec r}_\theta } \times {{\vec r}_\varphi } & = \left| {\begin{array}{*{20}{c}}{\vec i}&{\vec j}&{\vec k}\\{ - 5\sin \varphi \sin \theta }&{5\sin \varphi \cos \theta }&0\\{5\cos \varphi \cos \theta }&{5\cos \varphi \sin \theta }&{ - 5\sin \varphi }\end{array}} \right|\\ & = - 25{\sin ^2}\varphi \cos \theta \,\vec i - 25\sin \varphi \cos \varphi {\sin ^2}\theta \vec k - 25\sin \varphi \cos \varphi {\cos ^2}\theta \vec k - 25{\sin ^2}\varphi \sin \theta \vec j\\ & = - 25{\sin ^2}\varphi \cos \theta \,\vec i - 25{\sin ^2}\varphi \sin \theta \vec j - 25\sin \varphi \cos \varphi \left( {{{\sin }^2}\theta + {{\cos }^2}\theta } \right)\vec k\\ & = - 25{\sin ^2}\varphi \cos \theta \,\vec i - 25{\sin ^2}\varphi \sin \theta \vec j - 25\sin \varphi \cos \varphi \vec k\end{align*}

Now, we what we really need is,

\begin{align*}\left\| {{{\vec r}_\theta } \times {{\vec r}_\varphi }} \right\| & = \sqrt {{{\left( { - 25{{\sin }^2}\varphi \cos \theta } \right)}^2} + {{\left( { - 25{{\sin }^2}\varphi \sin \theta } \right)}^2} + {{\left( { - 25\sin \varphi \cos \varphi } \right)}^2}} \\ & = \sqrt {625{{\sin }^4}\varphi \left( {{{\cos }^2}\theta + {{\sin }^2}\theta } \right) + 625{{\sin }^2}\varphi {{\cos }^2}\varphi } \\ & = \sqrt {625{{\sin }^2}\varphi \left( {{{\sin }^2}\varphi + {{\cos }^2}\varphi } \right)} \\ & = 25\left| {\sin \varphi } \right|\\ & = 25\sin \varphi \end{align*}

Note that we can drop the absolute value bars on the sine because we know that sine will be positive in $$\frac{1}{2}\pi \le \varphi \le \pi$$.

Show Step 3

The integral for the surface area is then,

$A = \iint\limits_{D}{{25\sin \varphi \,dA}} = \int_{0}^{{2\pi }}{{\int_{{\frac{1}{2}\pi }}^{\pi }{{25\sin \varphi \,d\varphi }}\,d\theta }}$

As noted in the integral above $$D$$ is just the ranges of $$\theta$$ and $$\varphi$$ we found in Step 1.

Show Step 4

Now we just need to evaluate the integral to get the surface area.

$A = \int_{0}^{{2\pi }}{{\int_{{\frac{1}{2}\pi }}^{\pi }{{25\sin \varphi \,d\varphi }}\,d\theta }} = \int_{0}^{{2\pi }}{{\left. { - 25\cos \varphi } \right|_{\frac{1}{2}\pi }^\pi \,d\theta }} = \int_{0}^{{2\pi }}{{25\,d\theta }} = \require{bbox} \bbox[2pt,border:1px solid black]{{50\pi }}$