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Section 17.2 : Parametric Surfaces

10. Determine the surface area of the portion of the surface given by the following parametric equation that lies inside the cylinder u2+v2=4.

r(u,v)=2u,vu,12v

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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 u2+v24.

Show Step 2

Next, we need to compute ru×rv. Here is that work.

ru=2,v,0rv=0,u,2 ru×rv=|ijk2v00u2|=2vi+4j+2uk

Now, we what we really need is,

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*}