The part of the surface 2y+4z−x^2=5 that lies above the triangle with vertices (0, 0), (2, 0), and (2, 4) Find the area of the surface.

ddaeeric 2021-09-14 Answered

The part of the surface \(\displaystyle{2}{y}+{4}{z}−{x}^{{2}}={5}\) that lies above the triangle with vertices \(\displaystyle{\left({0},{0}\right)},{\left({2},{0}\right)},{\quad\text{and}\quad}{\left({2},{4}\right)}\) Find the area of the surface.

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Expert Answer

Usamah Prosser
Answered 2021-09-15 Author has 17299 answers

\(\displaystyle{2}{y}+{4}{z}-{x}^{{2}}={5}\)
\(\displaystyle{f{{\left({x},{y}\right)}}}={\frac{{{5}}}{{{4}}}}+{\frac{{{x}^{{2}}}}{{{4}}}}-{\frac{{{2}{y}}}{{{4}}}}\) \(\displaystyle{f}_{{x}}={\frac{{{x}}}{{{2}}}}.{f}_{{x}}=-{\frac{{{1}}}{{{2}}}}{0}\leq{x}\leq{2}\)

\(\displaystyle{S}=\int\int\sqrt{{{1}+{f}{x}^{{2}}+{f}{y}}}{d}{A}{0}\leq{y}\leq{2}{x}\)

\(dA=dydx\)

\(\displaystyle{S}={\int_{{{0}}}^{{{2}}}}{\int_{{{0}}}^{{{2}{x}}}}\sqrt{{{1}+{\frac{{{x}^{{2}}}}{{{4}}}}+{\frac{{{1}}}{{{4}}}}}}{\left.{d}{y}\right.}{\left.{d}{x}\right.}\)
\(\displaystyle{S}={\frac{{{1}}}{{{2}}}}{\int_{{{0}}}^{{{2}}}}{\int_{{{0}}}^{{{2}{x}}}}\sqrt{{{5}+{x}^{{2}}}}{\left.{d}{y}\right.}{\left.{d}{x}\right.}\) \(\displaystyle{S}={\frac{{{1}}}{{{2}}}}{\int_{{0}}^{{2}}}{\left\lbrace{2}{x}\sqrt{{{5}+{x}^{{2}}}}\right\rbrace}{\left.{d}{x}\right.}\)
Let \(\displaystyle{5}+{x}^{{2}}={u}\)
\(\displaystyle{2}{x}{\left.{d}{x}\right.}={d}{u}\) \(\displaystyle{S}={\frac{{{1}}}{{{2}}}}{\int_{{0}}^{{2}}}\sqrt{{{u}}}{d}{u}\)
\(\displaystyle{S}={\frac{{{1}}}{{{3}}}}{\left({5}+{X}^{{2}}\right)}^{{{\frac{{{3}}}{{{2}}}}}}{{\mid}_{{0}}^{{2}}}\)
\(\displaystyle{S}={\frac{{{1}}}{{{3}}}}{\left({27}-{5}\sqrt{{{5}}}\right)}\)

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