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Section 6-2 : Parametric Surfaces

10. Determine the surface area of the portion of the surface given by the following parametric equation that lies inside the cylinder \({u^2} + {v^2} = 4\).

\[\vec r\left( {u,v} \right) = \left\langle {2u,vu,1 - 2v} \right\rangle \]

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Start Solution

We’ve already been given the parameterization of the surface in the problem statement so we don’t need to worry about that for this problem. All we really need to do yet is to acknowledge that we’ll need to restrict \(u\) and \(v\) to the disk \({u^2} + {v^2} \le 4\).

Show Step 2

Next, we need to compute \({\vec r_u} \times {\vec r_v}\). Here is that work.

\[{\vec r_u} = \left\langle {2,v,0} \right\rangle \hspace{0.5in}{\vec r_v} = \left\langle {0,u, - 2} \right\rangle \] \[{\vec r_u} \times {\vec r_v} = \left| {\begin{array}{*{20}{c}}{\vec i}&{\vec j}&{\vec k}\\2&v&0\\0&u&{ - 2}\end{array}} \right| = - 2v\vec i + 4\vec j + 2u\vec k\]

Now, we what we really need is,

\[\left\| {{{\vec r}_u} \times {{\vec r}_v}} \right\| = \sqrt {{{\left( { - 2v} \right)}^2} + {{\left( 4 \right)}^2} + {{\left( {2u} \right)}^2}} = \sqrt {4{u^2} + 4{v^2} + 16} = 2\sqrt {{u^2} + {v^2} + 4} \] Show Step 3

The integral for the surface area is then,

\[A = \iint\limits_{D}{{2\sqrt {{u^2} + {v^2} + 4} \,dA}}\]

Where \(D\) is the disk \({u^2} + {v^2} \le 4\).

Show Step 4

Because \(D\) is a disk the best bet for this integral is to use the following “version” of polar coordinates.

\[u = r\cos \theta \hspace{0.25in} v = r\sin \theta \hspace{0.25in} {u^2} + {v^2} = {r^2} \hspace{0.25in} dA = r\,dr\,d\theta \]

The polar coordinate limits for this \(D\) is,

\[\begin{array}{c}0 \le \theta \le 2\pi \\ 0 \le r \le 2\end{array}\]

So, the integral to converting to polar coordinates gives,

\[A = \iint\limits_{D}{{2\sqrt {{u^2} + {v^2} + 4} \,dA}} = \int_{0}^{{2\pi }}{{\int_{0}^{2}{{2r\sqrt {{r^2} + 4} \,dr}}\,d\theta }}\] Show Step 5

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

\[\begin{align*}A & = \int_{0}^{{2\pi }}{{\int_{0}^{2}{{2r\sqrt {{r^2} + 4} \,dr}}\,d\theta }} = \int_{0}^{{2\pi }}{{\left. {\frac{2}{3}{{\left( {{r^2} + 4} \right)}^{\frac{3}{2}}}} \right|_0^2\,d\theta }}\\ & = \int_{0}^{{2\pi }}{{\frac{2}{3}\left( {{8^{\frac{3}{2}}} - {4^{\frac{3}{2}}}} \right)\,d\theta }} = \left. {\frac{2}{3}\left( {{8^{\frac{3}{2}}} - 8} \right)\theta } \right|_0^{2\pi } = \require{bbox} \bbox[2pt,border:1px solid black]{{\frac{{32}}{3}\pi\left( {\sqrt 8 - 1} \right) = 61.2712}}\end{align*}\]