Section 17.2 : Parametric Surfaces
7. Determine the surface area of the portion of \(2x + 3y + 6z = 9\) that is inside the cylinder \({x^2} + {y^2} = 7\).
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Start SolutionWe first need to parameterize the surface. Because we are wanting the portion that is inside the cylinder centered on the \(z\)-axis it makes sense to first solve the equation of the plane for \(z\) to get,
\[z = \frac{3}{2} - \frac{1}{3}x - \frac{1}{2}y\]The parameterization for the full plane is then,
\[\vec r\left( {x,y} \right) = \left\langle {x,y,\frac{3}{2} - \frac{1}{3}x - \frac{1}{2}y} \right\rangle \]We only want the portion that is inside the cylinder given in the problem statement so we’ll also need to restrict \(x\) and \(y\) to those in the disk \({x^2} + {y^2} \le 7\). This will now give only the portion of the plane that is inside the cylinder.
Show Step 2Next, we need to compute \({\vec r_x} \times {\vec r_y}\). Here is that work.
\[{\vec r_x} = \left\langle {1,0, - \frac{1}{3}} \right\rangle \hspace{0.5in}{\vec r_y} = \left\langle {0,1, - \frac{1}{2}} \right\rangle \] \[{\vec r_x} \times {\vec r_y} = \left| {\begin{array}{*{20}{c}}{\vec i}&{\vec j}&{\vec k}\\1&0&{ - \frac{1}{3}}\\0&1&{ - \frac{1}{2}}\end{array}} \right| = \frac{1}{3}\vec i + \frac{1}{2}\vec j + \vec k\]Now, we what we really need is,
\[\left\| {{{\vec r}_x} \times {{\vec r}_y}} \right\| = \sqrt {{{\left( {\frac{1}{3}} \right)}^2} + {{\left( {\frac{1}{2}} \right)}^2} + 1} = \sqrt {\frac{{49}}{{36}}} = \frac{7}{6}\] Show Step 3The integral for the surface area is then,
\[A = \iint\limits_{D}{{\frac{7}{6}\,dA}}\]In this case \(D\) is just the restriction on \(x\) and \(y\) that we noted in Step 1. So, \(D\) is just the disk \({x^2} + {y^2} \le 7\).
Show Step 4Computing the integral in this case is very simple. All we need to do is take advantage of the fact that,
\[\iint\limits_{D}{{dA}} = {\mbox{Area of }}D\]So, the surface area is simply,
\[A = \iint\limits_{D}{{\frac{7}{6}\,dA}} = \frac{7}{6}\iint\limits_{D}{{dA}} = \frac{7}{6}\left[ {{\mbox{Area of }}D} \right] = \frac{7}{6}\left[ {\pi {{\left( {\sqrt 7 } \right)}^2}} \right] = \require{bbox} \bbox[2pt,border:1px solid black]{{\frac{{49}}{6}\pi }}\]