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### Section 17.2 : Parametric Surfaces

9. Determine the surface area of the portion of $$z = 3 + 2y + \frac{1}{4}{x^4}$$ 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

Parameterizing this surface is pretty simple. We have the equation of the surface in the form $$z = f\left( {x,y} \right)$$ and so the parameterization of the surface is,

$\vec r\left( {x,y} \right) = \left\langle {x,y,3 + 2y + \frac{1}{4}{x^4}} \right\rangle$

Now, this is the parameterization of the full surface and we only want the portion that lies over the following region. So, to get only the portion of the surface we’ll need to restrict $$x$$ and $$y$$ to the following ranges,

\begin{align*}0 \le x \le 1\\ & 0 \le y \le {x^5}\end{align*}

On a side note we can see that we are in the 1st quadrant here and so we know that $$x \ge 0$$ and $$y \ge 0$$. Therefore, we can see that the surface in the 1st quadrant is always above the $$xy$$-plane and so will in fact always be above the region above as suggested in the problem statement.

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Next, we need to compute $${\vec r_x} \times {\vec r_y}$$. Here is that work.

${\vec r_x} = \left\langle {1,0,{x^3}} \right\rangle \hspace{0.5in}{\vec r_y} = \left\langle {0,1,2} \right\rangle$ ${\vec r_x} \times {\vec r_y} = \left| {\begin{array}{*{20}{c}}{\vec i}&{\vec j}&{\vec k}\\1&0&{{x^3}}\\0&1&2\end{array}} \right| = - {x^3}\vec i - 2\vec j + \vec k$

Now, we what we really need is,

$\left\| {{{\vec r}_x} \times {{\vec r}_y}} \right\| = \sqrt {{{\left( { - {x^3}} \right)}^2} + {{\left( { - 2} \right)}^2} + {{\left( 1 \right)}^2}} = \sqrt {{x^6} + 5}$ Show Step 3

The integral for the surface area is then,

$A = \iint\limits_{D}{{\sqrt {{x^6} + 5} \,dA}} = \int_{0}^{1}{{\int_{0}^{{{x^5}}}{{\sqrt {{x^6} + 5} \,dy}}\,dx}}$

As noted in the integral above $$D$$ is just the ranges of $$x$$ and $$y$$ we found in Step 1.

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

\begin{align*}A & = \int_{0}^{1}{{\int_{0}^{{{x^5}}}{{\sqrt {{x^6} + 5} \,dy}}\,dx}} = \int_{0}^{1}{{\left. {y\sqrt {{x^6} + 5} } \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*}