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Section 4-9 : Surface Area

1. Determine the surface area of the portion of \(2x + 3y + 6z = 9\) that is in the 1st octant.

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Start Solution

Okay, let’s start off with a quick sketch of the surface so we can get a feel for what we’re dealing with.

We’ve given the sketches with a set of “traditional” axes as well as a set of “box” axes to help visualize the surface.

Show Step 2

Let’s first set up the integral for the surface area of this surface. This surface doesn’t force a region \(D\) in any of the coordinates planes so we can work with any of them that we want to.

So, let’s work with \(D\) in the \(xy\)-plane. This, in turn, means we’ll first need to solve the equation of the plane for \(z\) to get,

\[z = \frac{3}{2} - \frac{1}{3}x - \frac{1}{2}y\]

Now, the integral for the surface area is,

\[S = \iint\limits_{D}{{\sqrt {{{\left[ { - \frac{1}{3}} \right]}^2} + {{\left[ { - \frac{1}{2}} \right]}^2} + 1} \,dA}} = \iint\limits_{D}{{\sqrt {\frac{{49}}{{36}}} \,dA}} = \iint\limits_{D}{{\frac{7}{6}\,dA}}\] Show Step 3

Now, we’ll need to figure out the region \(D\) we’ll need to use in our integral. In this case \(D\) is simply the triangle in the \(xy\)-plane that is directly below the plane. Here is a quick sketch of \(D\).

The equation of the hypotenuse can be simply found by plugging \(z = 0\) into the equation for the plane since it is simply nothing more than where the plane intersects with the \(xy\)‑plane.

Here are a set of limits for this \(D\).

\[\begin{array}{c} \displaystyle 0 \le x \le \frac{9}{2}\\ \displaystyle 0 \le y \le - \frac{2}{3}x + 3\end{array}\] Show Step 4

Now, normally we’d proceed to evaluate the integral at this point and we could do that if we wanted to. However, we don’t need to do that in this case.

Let’s do a quick rewrite the surface area integral as follows.

\[S = \frac{7}{6}\iint\limits_{D}{{dA}}\]

At this point we can see that the integrand of the integral is only 1.

Show Step 5

Because the integrand is only 1 we can use the fact that the value of this integral is nothing more than the area of \(D\) and since \(D\) is just a right triangle we can quickly compute the area of \(D\).

The surface area is then,

\[S = \frac{7}{6}\iint\limits_{D}{{dA}} = \frac{7}{6}\left( {{\mbox{Area of }}D} \right) = \frac{7}{6}\left( {\frac{1}{2}} \right)\left( {\frac{9}{2}} \right)\left( 3 \right) = \require{bbox} \bbox[2pt,border:1px solid black]{{\frac{{63}}{8}}}\]

Don’t forget this nice little fact about the value of \(\iint\limits_{D}{{dA}}\) . It doesn’t come up often but when it does and \(D\) is an easy to compute area it can greatly reduce the amount of work for the integral evaluation.