Calculate the area of the surface 9z= x^{3/2} + y^{3/2} with 0 less than or equal to x less than...

Question:

Calculate the area of the surface {eq}9z= x^{3/2} + y^{3/2} {/eq} with {eq}0 \leq x \leq 4, 0 \leq y \leq x {/eq}.

Surface Area:

Given an explicitly given surface, we can write its area in terms of a double integral, that is, from the projection of the surface on some coordinate plane, usually on the xy-plane:

{eq}\begin{gathered} z = z\left( {x,y} \right)\;,S \to D \hfill \\ A\left( S \right) = \iint\limits_S {dS} = \iint\limits_D {\sqrt {z_x^2 + z_y^2 + 1} \;dxdy} \hfill \\ dS = \sqrt {z_x^2 + z_y^2 + 1} \;dxdy \hfill \\ \end{gathered} {/eq}

Answer and Explanation:

First, calculating the differential, we can express the area of surface as a double integral:

{eq}\begin{gathered} A\left( S \right) = \iint\limits_S {dS} = \iint\limits_D {\sqrt {z_x^2 + z_y^2 + 1} \;dxdy} \hfill \\ dS = \sqrt {z_x^2 + z_y^2 + 1} \;dxdy \hfill \\ \hfill \\ 9z = {x^{3/2}} + {y^{3/2}} \hfill \\ z = \frac{1}{9}\left( {{x^{3/2}} + {y^{3/2}}} \right) \hfill \\ \to \left\{ {\begin{array}{*{20}{c}} {{z_x} = \frac{1}{9}\left( {\frac{3}{2}{x^{1/2}}} \right) = \frac{1}{6}{x^{1/2}}} \\ {{z_y} = \frac{1}{9}\left( {\frac{3}{2}{y^{1/2}}} \right) = \frac{1}{6}{y^{1/2}}} \end{array}} \right. \hfill \\ dS = \sqrt {z_x^2 + z_y^2 + 1} \;dxdy \hfill \\ dS = \sqrt {{{\left( {\frac{1}{6}{x^{1/2}}} \right)}^2} + {{\left( {\frac{1}{6}{y^{1/2}}} \right)}^2} + 1} \;dxdy = \sqrt {\frac{1}{{36}}x + \frac{1}{{36}}y + 1} \;dxdy \hfill \\ dS = \frac{1}{6}\sqrt {x + y + 36} \;dxdy \hfill \\ \hfill \\ A\left( S \right) = \iint\limits_S {dS} = \iint\limits_D {\frac{1}{6}\sqrt {x + y + 36} \;\;dxdy} \hfill \\ D \to 0 \leqslant x \leqslant 4,0 \leqslant y \leqslant x \hfill \\ \end{gathered} {/eq}

Solving the double integral, we have:

{eq}\begin{gathered} A\left( S \right) = \iint\limits_S {dS} = \iint\limits_D {\frac{1}{6}\sqrt {x + y + 36} \;\;dxdy} \hfill \\ D \to 0 \leqslant x \leqslant 4,0 \leqslant y \leqslant x \hfill \\ A\left( S \right) = \frac{1}{6}\int\limits_0^4 {\int\limits_0^x {\sqrt {x + y + 36} } } dydx \hfill \\ = \frac{1}{6}\int\limits_0^4 {\int\limits_0^x {{{\left( {x + y + 36} \right)}^{1/2}}} } dydx \hfill \\ = \frac{1}{6}\int\limits_0^4 {\left. {\frac{{{{\left( {x + y + 36} \right)}^{3/2}}}}{{3/2}}} \right|} _0^xdx \hfill \\ = \frac{1}{9}\int\limits_0^4 {\left[ {{{\left( {2x + 36} \right)}^{3/2}} - {{\left( {x + 36} \right)}^{3/2}}} \right]} dx \hfill \\ = \frac{1}{9}\left. {\left[ {\frac{1}{2}\frac{{{{\left( {2x + 36} \right)}^{5/2}}}}{{5/2}} - \frac{{{{\left( {x + 36} \right)}^{5/2}}}}{{5/2}}} \right]} \right|_0^4 \hfill \\ = \frac{1}{9}\left[ {\left( {\frac{1}{5}{{44}^{\frac{5}{2}}} - \frac{2}{5}{{40}^{5/2}}} \right) - \left( {\frac{1}{5}{{36}^{5/2}} - \frac{2}{5}{{36}^{5/2}}} \right)} \right] \hfill \\ \end{gathered} {/eq}


Learn more about this topic:

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Double Integrals: Applications & Examples

from AP Calculus AB & BC: Help and Review

Chapter 12 / Lesson 14
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