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Section 17.2 : 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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Start Solution

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.

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