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Section 15.9 : Surface Area

3. Determine the surface area of the portion of \(\displaystyle z = 3 + 2y + \frac{1}{4}{x^4}\) that is above the region in the \(xy\)-plane bounded by \(y = {x^5}\), \(x = 1\) and the \(x\)-axis.

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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.

The surface we are after is the orange portion that is above the \(xy\)-plane and the greenish region in the \(xy\)-plane is the region over which we are graphing the surface, i.e. it is the region \(D\) we’ll use in the integral.

Show Step 2

The integral for the surface area is,

\[S = \iint\limits_{D}{{\sqrt {{{\left[ {{x^3}} \right]}^2} + {{\left[ 2 \right]}^2} + 1} \,dA}} = \iint\limits_{D}{{\sqrt {{x^6} + 5} \,dA}}\] Show Step 3

Now, as mentioned in Step 1 the region \(D\) is shown in the sketches of the surface. Here is a 2D sketch of \(D\) for the sake of completeness.

The limits for this region are,

\[\begin{array}{c}0 \le x \le 1\\ 0 \le y \le {x^5}\end{array}\]

Note as well that the integrand pretty much requires us to do the integration in this order.

Show Step 4

With the limits from Step 3 the integral becomes,

\[S = \iint\limits_{D}{{\sqrt {{x^6} + 5} \,dA}} = \int_{0}^{1}{{\int_{0}^{{{x^5}}}{{\sqrt {{x^6} + 5} \,dy}}\,dx}}\] Show Step 5

Okay, all we need to do then is evaluate the integral.

\[\begin{align*}S & = \int_{0}^{1}{{\int_{0}^{{{x^5}}}{{\sqrt {{x^6} + 5} \,dy}}\,dx}}\\ & = \int_{0}^{1}{{\left. {\left( {y\sqrt {{x^6} + 5} } \right)} \right|_0^{{x^5}}\,dx}}\\ & = \int_{0}^{1}{{{x^5}\sqrt {{x^6} + 5} \,dx}}\\ & = \left. {\frac{1}{9}{{\left( {{x^6} + 5} \right)}^{\frac{3}{2}}}} \right|_0^1 = \require{bbox} \bbox[2pt,border:1px solid black]{{\frac{1}{9}\left( {{6^{\frac{3}{2}}} - {5^{\frac{3}{2}}}} \right) = 0.3907}}\end{align*}\]