Paul's Online Notes
Paul's Online Notes
Home / Calculus III / Surface Integrals / Parametric Surfaces
Show Mobile Notice Show All Notes Hide All Notes
Mobile Notice
You appear to be on a device with a "narrow" screen width (i.e. you are probably on a mobile phone). Due to the nature of the mathematics on this site it is best views in landscape mode. If your device is not in landscape mode many of the equations will run off the side of your device (should be able to scroll to see them) and some of the menu items will be cut off due to the narrow screen width.

Section 6-2 : Parametric Surfaces

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

Show All Steps Hide All Steps

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.

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