Question

Find the area of the part of the surface \({\rm{z = }}{{\rm{x}}^{\rm{2}}}{\rm{ + 2y}}\) that lies above the triangle with vertices \({\rm{(0,0),(1,0)}}\), and \({\rm{(1,2)}}{\rm{.}}\)

Short answer

The surface area \({\rm{ = }}\frac{{{\rm{27 - 5}}\sqrt {\rm{5}} }}{{\rm{6}}}.\)

Step-by-step solution

Explanation of solution.

Recall that

If the surface \({\rm{S}}\) has the form \({\rm{z = g }}\left( {{\rm{x, y}}} \right)\)

Then

Surface area \(x=\iint_{S}{d}S=\iint_{D}{\sqrt{1+{{\left( {{g}_{x}} \right)}^{2}}+{{\left( {{g}_{y}} \right)}^{2}}}}dA\).

Where \({\rm{D}}\) is the \({\rm{x y}}\) plane projection of the surface.

Right angled triangle.

\({\rm{S}}\) is to be above the triangle with vertices \(\left( {{\rm{0,0}}} \right){\rm{, }}\left( {{\rm{1,0}}} \right){\rm{,}}\)and \(\left( {{\rm{1,2}}} \right){\rm{.\;}}\)

\({\rm{D}}\) is the area within the right-angled triangle shown below:

Equation of the line joining \((0,0)\) and \({\rm{(1,2)}}\) is

\(\begin{aligned}\frac{{{\rm{y - 0}}}}{{{\rm{x - 0}}}}\rm &= \frac{{{\rm{2 - 0}}}}{{{\rm{1 - 0}}}}\\\rm y&= 2 x\end{aligned}\)

Define the \({\rm{D}}\) as

\({\rm{\{ (x,y)}} \in {\rm{D}}\mid {\rm{0}} \le y \le {\rm{2x,}}\;\;\;{\rm{0}} \le x \le 1\} \)

\(\begin{aligned}\frac{{{\rm{y - 0}}}}{{{\rm{x - 0}}}}\rm &= \frac{{{\rm{2 - 0}}}}{{{\rm{1 - 0}}}}\\\rm y&= 2 x\end{aligned}\)\(\begin{aligned}\frac{{{\rm{y - 0}}}}{{{\rm{x - 0}}}}\rm &= \frac{{{\rm{2 - 0}}}}{{{\rm{1 - 0}}}}\\\rm y&= 2 x\end{aligned}\)

Calculation.

Therefore

Surface area

\(\begin{array}{l}{\rm{ = }}\int_{\rm{0}}^{\rm{1}} {\int_{\rm{0}}^{{\rm{2x}}} {\sqrt {{\rm{1 + (2x}}{{\rm{)}}^{\rm{2}}}{\rm{ + (2}}{{\rm{)}}^{\rm{2}}}} } } {\rm{dydx}}\\{\rm{ = }}\int_{\rm{0}}^{\rm{1}} {\int_{\rm{0}}^{{\rm{2x}}} {\sqrt {{\rm{5 + 4}}{{\rm{x}}^{\rm{2}}}} } } {\rm{dydx}}\\{\rm{ = }}\int_{\rm{0}}^{\rm{1}} {\left( {{\rm{y}}\sqrt {{\rm{5 + 4}}{{\rm{x}}^{\rm{2}}}} } \right)_{\rm{0}}^{{\rm{2x}}}} {\rm{dx}}\\{\rm{ = }}\int_{\rm{0}}^{\rm{1}} {\rm{2}} {\rm{x}}\sqrt {{\rm{5 + 4}}{{\rm{x}}^{\rm{2}}}} {\rm{dx}}\end{array}\)

Substitute \({\rm{4}}{{\rm{x}}^{\rm{2}}}{\rm{ + 5 = u}}\) and \({\rm{8 x d x = d u}}\),

\({\rm{ = }}\frac{{\rm{1}}}{{\rm{4}}}\int_{\rm{0}}^{\rm{1}} {\rm{8}} {\rm{x}}\sqrt {{\rm{5 + 4}}{{\rm{x}}^{\rm{2}}}} {\rm{dx}}\)

Integration's limits will change from \(\int_{\rm{ - }} {{{\rm{0}}^{\rm{1}}}} {\rm{ to }}\int_{\rm{ - }} {\rm{5}} {\rm{ + 4 \times }}{{\rm{0}}^{{{\rm{2}}^{{\rm{5 + 4 \times }}{{\rm{1}}^{\rm{2}}}}}}}{\rm{ = }}\mathop \smallint \nolimits_5^9 \)

\(\begin{array}{l}{\rm{ = }}\frac{{\rm{1}}}{{\rm{4}}}\int_{\rm{5}}^{\rm{9}} {{{\rm{u}}^{{\rm{1/2}}}}} {\rm{du}}\\{\rm{ = }}\frac{{\rm{1}}}{{\rm{4}}}\left( {\frac{{\rm{2}}}{{\rm{3}}}{{\rm{u}}^{{\rm{3/2}}}}} \right)_{\rm{5}}^{\rm{9}}\\{\rm{ = }}\frac{{{\rm{9}}\sqrt {\rm{9}} {\rm{ - 5}}\sqrt {\rm{5}} }}{{\rm{6}}}\\{\rm{ = }}\frac{{{\rm{27 - 5}}\sqrt {\rm{5}} }}{{\rm{6}}}.\end{array}\)

Therefore,the surface area \({\rm{ = }}\frac{{{\rm{27 - 5}}\sqrt {\rm{5}} }}{{\rm{6}}}.\)